1. Compact spaces

Fix a topological space $ X $. We say $ X $ is compact if every open cover $ {U_{\alpha}} $ of $ X $ admits a finite subcover $ U_1, \dots, U_n $.

Heine-Borel theorem.   Let $ X $ be a subset of $ \mathbb{R}^n $. Then $ X $ is compact with respect to the Euclidean metric if and only if $ X $ is closed and bounded.

Proposition 1.1.   Let $ X $ be a topological space. Then $ X $ is compact if and only if there exists a subbase $ \mathcal{B} $ of $ X $ such that every open cover by elements of $ \mathcal{B} $ admits a finite subcover.

Theorem 1.2.   Let $ X $ be a metric space. Then $ X $ is compact if and only if every sequence in $ X $ has a convergent subsequence.

Problems

2019 F P5.   Let $ f \colon \mathbb{R}^n \to \mathbb{R}^m $ be continuous ($ n, m < \infty $), and let $ K \subseteq \mathbb{R}^n $ be compact. Show $ f(K) $ is compact.

Proof. We show this holds for any $ f \colon X \to Y $ continuous. Let $ {U_{\alpha}} $ cover $ f(K) $. Then $ {f^{-1}(U_{\alpha})} $ forms an open cover of $ K $ by (the topological definition of) continuity, and hence admits a finite subcover $ f^{-1}(U_1), \dots, f^{-1}(U_n) $. We conclude that $ U_1, \dots, U_n $ cover $ f(K) $. $ \blacksquare $

Alternative proof. Let $ { y_n } $ be a sequence in $ f(K) $. Then it lifts to a (not necessarily unique) sequence $ {x_n} $ in $ K $, i.e. $ f(x_n) = y_n $. By theorem 1.2, this sequence has a convergent subsequence $ {x_{n_i}} $, whose image $ { y_{n_i} } $ is a convergent subsequence of $ { y_n } $ by continuity, i.e. $ \lim_i f(x_{n_i}) = f(\lim_i x_{n_i}) $. $ \blacksquare $

2019 W P2.   Let $ X $ be a compact metric space with open cover $ {U_{\alpha}} $. Show that for some $ \varepsilon > 0 $ every ball of radius $ \varepsilon $ is contained in some $ U_{\alpha} $.

Proof. Suppose not. Then there exists a sequence $ {x_n} $ such that the ball $ B(x_n, 1/n) $ is not contained in any $ U_{\alpha} $. By theorem 1.2, there exists a subsequence $ {x_{n_i}} $ converging to some $ x \in X $. Choose $ U_{\alpha} $ containing $ x $. Since $ U_{\alpha} $ is open, there exists an $ r > 0 $ such that $ B(x, r) \subseteq U_{\alpha} $. For $ i $ sufficiently large, we have $ d(x_{n_i}, x) < r/2 $ and $ 1/n_i < r/2 $, and hence

\[B(x_{n_i}, 1/n_i) \subseteq B(x, r) \subseteq U_{\alpha},\]

a contradiction. $ \blacksquare $

2017 F P1.   Let $ X $ be the set of sequences $ { a_n } $ with $ a_n \in {0, 1} $. Equip $ X $ with the metric

\[d(\{a_n\}, \{b_n\}) = \begin{cases} 0 & \text{if } \{a_n\} = \{b_n\} \\ 2^{-m} & \text{if } m = \min\{n \mid a_n \neq b_n \}. \end{cases}\]

(a) Prove $ (X, d) $ is compact, and (b) that there exists no isolated points.

Proof. (a) Let $ { a_{n, m}} $ be a sequence in $ X $ indexed by $ m $. At the first coordinate, one of the values $ 0 $ or $ 1 $ occurs for infinitely many $ m $; restrict to an infinite subsequence on which the first coordinate is constant. From this subsequence choose an infinite subsequence on which the second coordinate is constant, and continue inductively. Let the $ j $th term of our final subsequence be the $ j $th term of the subsequence selected at stage $ j $. Then for each $ N $, all sufficiently large terms of this diagonal subsequence agree in their first $ N $ coordinates. Hence their distance from the coordinatewise limiting sequence is at most $ 2^{-N} $, so the diagonal subsequence converges. By theorem 1.2, $ X $ is compact.

(b) Suppose $ {a_n} $ is an isolated point. Then there exists an $ N \gg 0 $ with $ B({a_n}, 2^{-N}) $ only containing $ {a_n} $. But

\[b_n = \begin{cases} a_n & \text{if } n \leq N \\ 1-a_n & \text{if } n > N \end{cases}\]

is an element of $ X $ distinct from $ {a_n} $ contained in $ B({a_n}, 2^{-N}) $. $ \blacksquare $

2014 F P2.   Let $ K $ be a compact subset of $ \mathbb{R}^n $, and let $ f \colon K \to \mathbb{R} $ be continuous. Prove that for every $ \varepsilon > 0 $ there exists an $ M \geq 0 $ such that for all $ x, y \in K $,

\[\|f(x)-f(y)\| \leq M \|x-y\| + \varepsilon.\]

Show this is not necessarily true for $ \varepsilon = 0 $.

Proof. By the Heine-Cantor theorem, there exists a $ \delta > 0 $ such that $ |x-y| < \delta $ implies $ |f(x) - f(y)| < \varepsilon $; consequently, it remains to consider $ |x-y| \geq \delta $. The Heine-Borel theorem and 2019 F P5 show that $ K $ and $ f(K) $ are bounded, i.e. $ |f(x) - f(y)| \leq C $ for a fixed $ C $. Consequently,

\[\|f(x) - f(y) \| \leq C/\delta \|x-y\| + \varepsilon,\]

and we set $ M = C/\delta $.

For a counterexample, consider $ f \colon [0, 1] \to [0, 1] $ with $ f(x) = \sqrt{x} $. We see $ \lim_{x \to 0^+} f'(x) = \infty $ implies no such $ M $ exists. $ \blacksquare $.

2. Sequences and series

Monotone convergence theorem.   Let $ {a_n} $ be a monotonically increasing sequence in $ \mathbb{R} $. Then $ {a_n} $ has a limit if and only if it is bounded.

Theorem 2.1.   In $ \mathbb{R}^n $ every Cauchy sequence converges; that is, $ \mathbb{R}^n $ is a complete metric space.

Ratio Test.   Let $ \sum a_n $ be a series in $ \mathbb{R} $ such that $ a_n \neq 0 $ for all sufficiently large $ n $. Write $ L = \lim_n |a_{n+1}/a_n| $, assuming the limit exists. Then $ L < 1 $ implies absolute convergence; $ L > 1 $ implies divergence; and $ L = 1 $ is inconclusive.

Root Test.   Let $ \sum a_n $ be a series in $ \mathbb{R} $. Write $ r = \limsup_n |a_n|^{1/n} $. Then $ r < 1 $ implies absolute convergence; $ r > 1 $ implies divergence; and $ r = 1 $ is inconclusive.

Integral Test.   Let $ f \colon [1, \infty) \to \mathbb{R}^+ $ be nonnegative and monotonically decreasing with $ f(n) = a_n $. Then $ \sum a_n $ converges if and only if $ \int_1^{\infty} f(x) dx < \infty $.

P-Series Test.   A series of the form $ \sum_n n^{-p} $ converges if and only if $ p > 1 $.

Limit Comparison Test.   Let $ {a_n} $ and $ {b_n} $ be eventually positive sequences in $ \mathbb{R} $. Suppose $ L = \lim a_n / b_n $ exists and $ 0 < L < \infty $. Then $ \sum a_n $ converges if and only if $ \sum b_n $ converges.

Alternating Series Test.   Suppose $ {a_n} $ is a monotonically decreasing sequence in $ \mathbb{R}^+ $ with $ \lim_n a_n = 0 $. Then $ \sum (-1)^n a_n $ converges.

Theorem 2.2.   Let $ {a_n} $ be a sequence in $ \mathbb{R}^+ $. Then $ \prod_n (1+a_n) $ converges if and only if $ \sum_n a_n $ converges. If $ 0 < a_n < 1 $, then $ \prod_n (1-a_n) \neq 0 $ if and only if $ \sum_n a_n $ converges.

Banach fixed point theorem.   Let $ (X, d) $ be a complete metric space, and suppose $ f \colon X \to X $ satisfies $ d(f(x), f(y)) \leq q d(x, y) $ for some $ 0 \leq q < 1 $. Then $ f $ has a unique fixed point $ f(x_\ast) = x_\ast $, which is given by taking an arbitrary $ x_0 \in X $, setting $ x_{n+1} = f(x_n) $, and evaluating $ \lim_n x_n = x_\ast $.

Problems

2023 W P2.   Consider the series $ \sum_{a, b \geq 0} p^{-a} q^{-b} $ for fixed distinct primes $ p, q $. Prove it converges and find its sum.

Proof. For each $ r > 0 $ even, we see

\[\left( \sum_{a=0}^{r/2} p^{-a} \right) \left( \sum_{b=0}^{r/2} q^{-b} \right) \leq \sum_{a+b \leq r} p^{-a} q^{-b} \leq \left( \sum_{a=0}^{r} p^{-a} \right) \left( \sum_{b=0}^{r} q^{-b} \right).\]

Hence, as $ r $ goes to infinity, our sum goes to

\[\left( \sum_{a=0}^{\infty} p^{-a} \right) \left( \sum_{b=0}^{\infty} q^{-b} \right) = \left( \frac{1}{1-p^{-1}} \right) \left( \frac{1}{1-q^{-1}} \right)\]

giving us our limit. $ \blacksquare $

2023 W P5.   Show

\[\lim_n \left( \sum_{k=1}^{n} \frac{\sqrt{k}}{n} - \frac{2}{3} \sqrt{n} \right) = 0.\]

Proof. Since $ x \mapsto \sqrt{x} $ is increasing, we have

\[\int_0^n \frac{\sqrt{x}}{n} dx \leq \sum_{k=1}^{n} \frac{\sqrt{k}}{n} \leq \int_1^{n+1} \frac{\sqrt{x}}{n} dx.\]

Integrating yields

\[\frac{2}{3}\sqrt{n} \leq \sum_{k=1}^{n} \frac{\sqrt{k}}{n} \leq \frac{2}{3n}\left((n+1)^{3/2}-1\right),\]

or equivalently

\[0 \leq \sum_{k=1}^{n} \frac{\sqrt{k}}{n} - \frac{2}{3} \sqrt{n} \leq \frac{2}{3n}\left((n+1)^{3/2}-n^{3/2}-1\right).\]

By the mean value theorem, $ (n+1)^{3/2}-n^{3/2} \leq \frac{3}{2}\sqrt{n+1} $, so the right hand side approaches $ 0 $. This gives us the desired equality. $ \blacksquare $

2022 F P1.   Define $ {x_n} $ by $ 0 < x_1 < 1 $ and $ x_{n+1} = 1 - \sqrt{1-x_n} $. Prove that (a) $ {x_n} $ monotonically decreases to $ 0 $, and that (b) $ \lim_n x_{n+1}/x_n = 1/2 $.

Proof. (a) Set $ y_n = 1-x_n $ so that $ y_{n+1} = \sqrt{y_n} $. Then $ y_1 \neq 0 $ implies $ \lim_n y_n = 1 $ (this is well known). Thus, $ \lim_n x_n = \lim_n 1-y_n = 0 $. To show $ x_n $ is monotonically decreasing, we observe $ y_n $ is monotonically increasing, i.e. $ \sqrt{x} > x $ for $ x \in (0, 1) $.

(b) By L'Hôpital's Rule,

\[\lim_{n \to \infty} \frac{x_{n+1}}{x_n} = \lim_{x \to 0} \frac{1-\sqrt{1-x}}{x} = \lim_{x \to 0} \frac{1}{2 \sqrt{1-x}} = \frac{1}{2},\]

and so the limit is $ 1/2 $. $ \blacksquare $

2022 W P2.   Let $ {a_n} $ be a sequence in $ \mathbb{R} $. If $ \sum_n | a_n - a_{n+1} | < \infty $, then the sequence is convergent.

Proof. Fix $ \varepsilon > 0 $. Since the series in the hypothesis converges, there exists an $ N \gg 0 $ such that

\[\| a_n - a_m \| \leq \sum_{k=n}^{m-1} \| a_k - a_{k+1} \| \leq \sum_{k=N}^{\infty} \| a_k - a_{k+1} \| < \varepsilon\]

We conclude that our sequence is Cauchy, and hence has a limit. $ \blacksquare $

2022 W P7.   Suppose $ {a_n} $ is an unbounded increasing sequence in $ \mathbb{R}^+ $. Show $ \sum_n (a_{n+1}-a_n)/a_n $ diverges.

Proof. Fix $ N \gg 0 $. Since $ {a_n} $ is unbounded, there exists an $ M > N $ such that $ a_{M+1} > 2 a_N $. Hence,

\[\sum_{n=N}^{M} \frac{a_{n+1}-a_n}{a_n} \geq \sum_{n=N}^{M} \frac{a_{n+1}-a_n}{a_{M+1}} = \frac{a_{M+1} - a_N}{a_{M+1}} > 1/2.\]

We conclude that the partial sums do not converge. $ \blacksquare $

Alternative proof. The inequality $ x \geq \log(1+x) $ implies

\[\sum_{n=1}^{N} \frac{a_{n+1}-a_n}{a_n} = \sum_{n=1}^{N} \left(\frac{a_{n+1}}{a_n} - 1\right) \geq \sum_{n=1}^{N} \log(a_{n+1}/a_n) = \log(a_{N+1}) - \log(a_1),\]

which is unbounded as $ N $ varies since $ {a_n} $ is unbounded. $ \blacksquare $

2022 W P6.   Let $ t_0 \in \mathbb{R} $ and set $ t_{n+1} = \sin(\cos(t_n)) $. Prove this sequence converges with limit independent of $ t_0 $.

Proof. Set $ f(x) = \sin(\cos(x)) $ and $ I = [-\sin(1), \sin(1)] $. For every $ t_0 \in \mathbb{R} $ we have $ t_1 = f(t_0) \in I $, and $ f(I) \subseteq I $. Furthermore, for $ x \in I $,

\[\|f'(x)\| = \|\sin(x)\cos(\cos(x))\| \leq \sin(\sin(1)) < 1.\]

Thus, $ f \colon I \to I $ is a contraction. The Banach fixed point theorem shows that $ {t_n} $ converges to the unique fixed point of $ f $, independently of $ t_0 $. $ \blacksquare $

2021 F P6.   Let $ a_0 \in (0, 1) $ and $ a_{n+1} = a_n^3 - a_n^2 + 1 $. Prove that (a) $ {a_n} $ converges and find its limit; and (b) that $ b_n = \prod_{i=1}^{n} a_i $ converges and find its limit.

Proof. (a) We observe $ a_{n+1} > a_n $ is equivalent to $ a_n^3 - a_n^2 - a_n + 1 > 0 $. Writing this as a polynomial $ f(x) = x^3 - x^2 - x + 1 $, we observe $ f(x) = (x-1)^2 (x+1) $ and $ f(0) = 1 $. Hence $ f $ is strictly positive on $ (-1, 1) $, and our sequence is monotonically increasing. We further observe $ a_n^3 - a_n^2 +1 < 1 $ implies $ a_n < 1 $, which is true by assumption, so our sequence is bounded above by $ 1 $; thus, our sequence has a limit. To find the limit $ a_* $, we see

\[\lim_n a_{n+1} = \lim_n a_n^3 - a_n^2 + 1\]

implies

\[a_* = a_*^3 - a_*^2 + 1,\]

and the only root of this polynomial on $ (0, 1] $ is $ 1 $. Thus, $ a _* = 1 $.

(b) Since $ {b_n} $ is a bounded monotonically decreasing sequence, its limit exists. Set $ \delta_n = 1-a_n $. The recurrence gives

\[\delta_{n+1} = 1-a_{n+1} = a_n^2(1-a_n) = a_n^2\delta_n.\]

Consequently,

\[b_n^2 = \prod_{i=1}^{n} a_i^2 = \frac{\delta_{n+1}}{\delta_1}.\]

Part (a) shows $ \delta_n \to 0 $, and hence $ b_n \to 0 $. $ \blacksquare $

2021 F P8.   Is the series

\[\sum_{n = 100}^{\infty} \ln(n)^{-\ln(\ln(n))}\]

convergent?

Proof. The function

\[x\longmapsto \ln(x)^{-\ln(\ln x)} =\exp\left(-(\ln(\ln x))^2\right)\]

is positive and decreasing for $ x\geq100 $. By the integral test, our series converges if and only if the integral

\[I = \int_{100}^{\infty} \ln(x)^{-ln(ln(x))} dx\]

does. Substituting $ u = \ln(x) $ and $ dx = e^u du $, we have

\[I = \int_{ln(100)}^{\infty} u^{-\ln(u)} e^u du.\]

The following inequalities are equivalent:

\[\begin{aligned} e^x & > x^{\ln(x)} \\ \log_x(e^x) & > \ln(x) \\ \frac{\ln(e^x)}{\ln(x)} & > \ln(x) \\ x & > \ln(x)^2. \end{aligned}\]

We conclude that $ u^{-\ln(u)} e^u > 1 $, our interal diverges, and so does our series. $ \blacksquare $

2020 F P1.   Let $ x_0 > 0 $ and $ x_{n+1} = \frac{1}{2} (x_n + \frac{4}{x_n}) $. Show that (a) $ x_{n+1} \geq 2 $ for $ n \geq 0 $, (b) $ x_{n+1} \leq x_n $ if $ n \geq 1 $, (c) $ \lim x_n = x_\ast $ exists, and (d) find $ x_\ast $.

Proof. (a) By the AM-GM inequality (cf. optimization),

\[x_{n+1} = \frac{1}{2} (x_n + \frac{4}{x_n}) \geq \sqrt{x_n \frac{4}{x_n}} = 2\]

for $ n \geq 0 $.

(b) We see $ x_{n+1} \leq x_n $ if and only if

\[\frac{1}{2}(x_n + \frac{4}{x_n}) \leq x_n,\]

which reduces to $ 2 \leq x_n $. This is true by (a).

(c) Our sequence is bounded below and monotonically decreasing, so it has a limit.

(d) Write $ x_* = \lim_n x_n $. We observe

\[\lim_n x_{n+1} = \lim_n \frac{1}{2} (x_n + \frac{4}{x_n})\]

implies

\[x_* = \frac{1}{2} (x_* + \frac{4}{x_*}).\]

Solving this equation shows $ x_* = 2 $ is the only positive solution, and thus our limit.

Alternatively, define $ f \colon [1.5, \infty) \to [1.5, \infty) $ by $ f(x) = \frac{1}{2}(x + \frac{4}{x}) $. Then

\[\sup_{x \geq 1.5}\|f'(x)\| = \sup_{x \geq 1.5}\frac{1}{2}\left\|1-\frac{4}{x^2}\right\| \leq \frac{1}{2},\]

so $ f $ is a contraction. By (a), $ x_1 \geq 2 $, so applying the Banach fixed point theorem to the tail of our sequence shows it converges to the unique fixed point. Solving $ x_\ast = \frac{1}{2} \left( x_{*} + \frac{4}{x_\ast} \right) $ yields $ x_\ast = 2 $. $ \blacksquare $

2020 W P1.   Let $ {a_n} $ be a sequence of nonnegative real numbers with $ \lim a_n = 0 $. Show there exist infinitely many $ N \in \mathbb{N} $ such that $ n \geq N $ implies $ a_n \leq a_N $.

Proof. If the sequence is eventually zero, then every sufficiently large index has the desired property. Otherwise, choose an index $k$ with $a_k>0$. The set $ {n \geq k \mid a_n \geq a_k} $ is finite since $ a_n \to 0 $. Let

\[N_k = \max\{n \geq k \mid a_n \geq a_k\}.\]

Then $ n \geq N_k $ implies $ a_n \leq a_{N_k} $. Because the sequence is not eventually zero, choose a new index $k'>N_k$ with $a_{k'}>0$ and repeat. This produces a strictly increasing sequence of indices with the desired property. $ \blacksquare $

2020 W P2.   Write $ {a_n} $ for a sequence of nonnegative real numbers such that $ \lim_n a_n = 0 $ and $ |a_n - a_{n+1}| \leq n^{-2} $. Prove $ \sum_n (-1)^{n-1} a_n $ converges.

Proof. We show that the partial sums are Cauchy. Fix $ \varepsilon > 0 $, and choose $ N \gg 0 $ such that $ a_n < \varepsilon/4 $ for $ n \geq N $ and

\[\sum_{k=N}^{\infty} \frac{1}{k^2} < \frac{\varepsilon}{2}.\]

(This is possible by the p-series, or integral, test.) For $ m \geq n \geq N $, pair adjacent terms in

\[\sum_{k=n}^{m}(-1)^{k-1}a_k.\]

There may be an unpaired term at each endpoint; their total absolute value is at most $ \varepsilon/2 $. Every pair has the form $ \pm(a_k-a_{k+1}) $, so

\[\left|\sum_{k=n}^{m}(-1)^{k-1}a_k\right| \leq \frac{\varepsilon}{2}+\sum_{k=N}^{\infty}|a_k-a_{k+1}| \leq \frac{\varepsilon}{2}+\sum_{k=N}^{\infty}\frac{1}{k^2} <\varepsilon.\]

Thus, our series converges. $ \blacksquare $

2020 W P4.   Show

\[a_n = \sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}}\]

converges in $ \mathbb{R} $.

Proof. First we observe that our sequence is monotonically increasing, and hence we need to show it is bounded. Rewrite each term as

\[a_n = \prod_{k=2}^{n} k^{2^{-(k-1)}}.\]

Since the logarithm is continuous, we see our sequence is bounded if and only if the sequence

\[\ln(a_n) = \sum_{k=2}^{n} \frac{1}{2^{k-1}} \ln(k)\]

is. Since $ \ln(k) \leq \sqrt{k} $ and $ 2^k \geq k^2 $ for $ k\geq4 $, we have

\[\ln(a_n) \leq \frac{\ln2}{2}+\frac{\ln3}{4} +2\sum_{k=4}^{\infty}\frac{1}{k^{3/2}},\]

which is bounded by the p-series test. Thus, $ {a_n} $ is bounded. $ \blacksquare $

3. Uniform continuity

Let $ f, g \colon X \to Y $ be maps between metric spaces. We say $ f $ is uniformly continuous if for each $ \varepsilon > 0 $, there exists a $ \delta > 0 $ such that $ d_X(x, y) < \delta $ implies $ d_Y(f(x), f(y)) < \varepsilon $ for all $ x, y \in X $.

Heine-Cantor theorem.   Let $ f \colon X \to Y $ be continuous. If $ X $ is compact, then $ f $ is uniformly continuous.

Problems

2020 W P3.   Let $ X $ be the space of sequences $ x = {x_n} $ with $ x_n \in [0, 1] $. Set $ d(x, y) = \sup_n | x_n - y_n | $. Suppose $ f \colon X \to \mathbb{R} $ is uniformly continuous. Prove that $ f $ is bounded.

Proof. Choose a $ \delta > 0 $ such that $ d(x, y) < \delta $ implies $ | f(x)-f(y) | < 1 $. Choose an integer $ M > 1/\delta $. Now fix $ x \in X $, and define $ x_n = \frac{n}{M} x $ for $ 0 \leq n \leq M $. Then $ d(x_{n+1}, x_n) \leq 1/M < \delta $, and hence

\[\|f(x)\| \leq \|f(0)\|+\sum_{n=0}^{M-1} \|f(x_{n+1}) - f(x_n)\| < \|f(0)\|+M.\]

Thus, $ f $ is a bounded function. $ \blacksquare $

2020 W P7.   Let $ f \colon \mathbb{R} \to \mathbb{R} $ be continuous and $ f' \colon \mathbb{R} \to \mathbb{R} $ uniformly continous. If $ \displaystyle \lim_{x \to \infty} f(x) = 0 $, does $ \displaystyle \lim_{x \to \infty} f'(x) $ exist?

Proof. We will show the derivative goes to $ 0 $. Suppose not. Then there exists a sequence $ {x_n} $ such that $ | f'(x_n) | > C $ for some $ C > 0 $ and $ \lim_n x_n = \infty $. Passing to a subsequence, either $ f'(x_n) > C $ for every $ n $ or $ f'(x_n) < -C $ for every $ n $; we treat the first case, and the second is analogous. Since $ f' $ is uniformly continuous, there exists a $ \delta > 0 $ so that $ | x-y | < \delta $ implies $ | f'(x) - f'(y) | < C/2 $. Therefore,

\[\int_{x_n - \delta/2}^{x_n + \delta/2} f'(x) dx \geq \int_{x_n-\delta/2}^{x_n+\delta/2} C/2 dx = \delta C/2,\]

which implies $ f(x_n + \delta/2) \geq f(x_n - \delta/2) + \delta C/2 $. This contradicts $ f $ having limit $ 0 $. $ \blacksquare $

2019 W P5.   Give an example of a continuous function $ f \colon (0, 1] \to \mathbb{R} $ that attains neither a maximum nor a minimum. (b) Show that if $ f $ is uniformly continuous, then it must attain a maximum or a minimum.

Proof. (a) Consider $ f(x) = (1-x) \sin(\frac{1}{x}) $. This is a continuous function which obtains no maximum or minimum, because $ \sin(\frac{1}{x}) $ is oscillating, and $ (1-x) $ causes $ f $ to decrease in absolute value away from $ 0 $.

(b) A uniformly continuous function on $ (0, 1] $ extends continuously to $ [0, 1] $ by setting

\[f(0) = \lim_{x \to 0^+}f(x).\]

The extension attains a maximum and a minimum on $ [0, 1] $. If neither extremum were attained on $ (0, 1] $, then both would have to occur only at $ 0 $. This would imply

\[\max_{[0,1]} f = f(0) = \min_{[0,1]} f,\]

so $ f $ would be constant and would attain both extrema on $ (0, 1] $, a contradiction. $ \blacksquare $

2018 F P5.   Let $ B $ be the closed unit ball in $ \mathbb{R}^2 $. Set $ \rho(x, y) = |x-y| $ if $ x $ and $ y $ are collinear and $ \rho(x, y) = |x| + |y| $ elsewise. This is a metric on $ B $. Suppose $ f \colon (B, \rho) \to \mathbb{R} $ is uniformly continuous. Show $ f $ is bounded.

Proof. Choose $ \delta > 0 $ such that $ \rho(x, y) < \delta $ implies $ |f(x)-f(y)| < 1 $, and choose an integer $ M > 1/\delta $. For any $ x \in B $, the points

\[x_j = \frac{j}{M}x, \qquad 0 \leq j \leq M,\]

are collinear and satisfy $ \rho(x_{j+1},x_j)=|x|/M<\delta $. Therefore,

\[\|f(x)\| \leq \|f(0)\|+\sum_{j=0}^{M-1}\|f(x_{j+1})-f(x_j)\| <\|f(0)\|+M.\]

Thus, $ f $ is bounded. $ \blacksquare $

2017 W P6.   Let $ f \colon \mathbb{R}^n \to \mathbb{R} $ be continuous with $ n < \infty $. Suppose $ \displaystyle \lim_{|x| \to \infty} f(x) = 0 $. Prove that $ f $ is uniformly continuous.

Proof. Fix $ \varepsilon > 0 $. Write $ D_r $ for the closed disk of radius $ r $ at $ 0 $. By assumption, there exists an $ r > 0 $ such that $ x, y \in \mathbb{R}^{n} \setminus D_r $ implies

\[\| f(x) - f(y) \| \leq \|f(x)\| + \|f(y)\| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.\]

Thus, $ f $ is uniformly continuous on $ \mathbb{R}^n \setminus D_r $. Furthermore, $ D_{r+1} $ is compact, so $ f $ is uniformly continuous on $ D_{r+1} $ by the Heine-Borel theorem. Since our desired property is local and every point has a ball of radius $ 1/2 $ contained in $ D_{r+1} $ or $ \mathbb{R}^{n} \setminus D_r $, we get our result. $ \blacksquare $

4. Uniform convergence

Let $ X, Y $ be metric spaces, and let $ f, g \colon X \to Y $ be functions. Define the uniform metric by

\[\rho(f, g) = \sup_{x \in X} d_Y(f(x), g(x)).\]

On the set of all maps this can take the value $+\infty$, so it is an extended metric; it is an ordinary metric on any class of maps lying at finite distance from a fixed map. We say a sequence of functions $ f_n \colon X \to Y $ converges uniformly to $ f_* $ if $\rho(f_n,f_)\to0$. We say $ {f_n} $ is *uniformly equicontinuous if for every $ \varepsilon > 0 $, there exists a $ \delta > 0 $ such that $ d_X(x, y) < \delta $ implies $ d_Y(f_n(x), f_n(y)) < \varepsilon $ for all $ n $.

Now assume $ K $ is a compact subset of $ \mathbb{R}^m $, and our sequence is $ f_n \colon K \to \mathbb{R}^q $ for some fixed $ q $. We say $ {f_n} $ is uniformly bounded if there exists an $ M $ such that

\[\sup_{x \in K, n \in \mathbb{N}} \|f_n(x)\| \leq M.\]

Ascoli-Arzelà theorem.   Let $ K \subset \mathbb{R}^m $ be compact, and let $ f_n \colon K \to \mathbb{R}^q $ be a sequence of functions for some fixed $ q $. If $ {f_n} $ is uniformly bounded and uniformly equicontinuous, then there exists a subsequence which converges uniformly.

Weierstrass M-Test.   Let $ f_n \colon X \to \mathbb{R}^q $ be a sequence of functions defined on a set $ X $, where $ q $ is fixed. Set $ M_n = \sup_x | f_n(x) | $. If $ \sum M_n $ converges, then $ \sum f_n(x) $ converges absolutely and uniformly.

Uniform limit theorem.   Write $ X $ for a topological space and $ Y $ a metric space. If $ f_n \colon X \to Y $ are continuous and converge uniformly to $ f \colon X \to Y $, then $ f $ is continuous. (In other words, the space of continuous functions are closed with respect to $ \rho $.)

Theorem 4.1.   Let $ f_n \colon [a, b] \to \mathbb{R} $ be integrable with uniform limit $ f $. Then $ \int_a^b f dx = \lim_n \int_a^b f_n dx $.

Theorem 4.2.   Suppose $ f_n \colon [a, b] \to \mathbb{R} $ are differentiable such that their derivatives $ f_n' $ converge uniformly on $ [a, b] $. If $ f_n(x_0) $ converges for some point $ x_0 \in [a, b] $, then $ f_n $ converge uniformly to some $ f $, and $ f'(x) = \lim_n f_n'(x) $.

Problems

2023 W P1.   Let $ f_n \colon [0, 1] \to \mathbb{R}^+ $ be continuous functions such that $ f_n(x) \leq f_{n+1}(x) $ for every $ x $, and suppose they have uniform limit $ f $. Prove that

\[\lim_n \int_0^1 \left( \sum_{k=1}^{n} f_k(x)^n \right)^{1/n} dx = \int_0^1 f(x) dx.\]

Proof. Set

\[a_n(x)=\left(\sum_{k=1}^n f_k(x)^n\right)^{1/n}.\]

Since the functions are nonnegative and increasing in $ n $, we have

\[f_n(x) = (f_n(x)^n)^{1/n} \leq a_n(x) \leq \left( \sum_{k=1}^n f_n(x)^n \right)^{1/n} = n^{1/n} f_n(x)\]

for every $ x $. Since $ f_n\to f $ uniformly, the sequence $ {|f_n|_\infty} $ is bounded, and hence

\[\|a_n-f\|_\infty \leq (n^{1/n}-1)\|f_n\|_\infty+\|f_n-f\|_\infty \longrightarrow 0.\]

Our result now follows from theorem 4.1. $ \blacksquare $

2023 W P3.   Let $ f_n \colon [0, 1] \to \mathbb{R} $ be a sequence of continuous functions. Suppose that for some $ f \colon [0, 1] \to \mathbb{R} $, we have

\[\lim_n f_n(x_n) = f(\lim_n x_n)\]

for every convergent sequence $ {x_n} $ in $ [0, 1] $. Show $ {f_n} $ converges uniformly to $ f $ or provide a counterexample.

Proof. We first note that $f$ is continuous. If $x_k\to x$, then for each fixed $k$ the hypothesis applied to the constant sequence at $x_k$ shows that $f_n(x_k)\to f(x_k)$ as $n\to\infty$. Choose increasing indices $m_k$ such that

\[|f_{m_k}(x_k)-f(x_k)|<1/k.\]

Define $y_{m_k}=x_k$ and $y_n=x$ at every other index. Then $y_n\to x$, so the hypothesis gives $f_{m_k}(x_k)\to f(x)$. Hence $f(x_k)\to f(x)$, proving continuity.

Now suppose the convergence is not uniform. There are $\varepsilon>0$, increasing indices $n_k$, and points $x_k\in[0,1]$ such that

\[|f_{n_k}(x_k)-f(x_k)|\geq\varepsilon.\]

After passing to a subsequence, compactness gives $x_k\to x_$. Define $y_{n_k}=x_k$ and $y_n=x_$ otherwise. The hypothesis gives $f_{n_k}(x_k)\to f(x_)$, while continuity gives $f(x_k)\to f(x_)$, contradicting the displayed inequality.

This is false if the domain is not compact: consider $ f_n \colon \mathbb{R} \to \mathbb{R} $ given by $ f_n(x) = x/n $. $ \blacksquare $

2023 W P4.   Does there exist a sequence of continuously differentiable functions on $\mathbb R$ that converges uniformly to a function which is not differentiable at $0$?

Proof. Yes, we see that $ f_n = \sqrt{x^2+1/n} $ is continuously differentiable, but $ f_n $ converge to $ |x| $ uniformly on $ \mathbb{R} $. $ \blacksquare $

2022 F P2.   Suppose $ f_n \colon [a, b] \to \mathbb{R} $ converge pointwise to a continuous function $ f $ and that each $ f_n $ is monotonically increasing. Then $ f_n $ converge uniformly to $ f $.

Proof. Fix $ \varepsilon > 0 $. Observe that this implies $ f $ is monotonically increasing, so we can partition $ [a, b] $ into intervals $ [t_i, t_{i+1}] $ such that

\[\|f(t_{i+1}) - f(t_i) \| < \varepsilon/5.\]

Choose an $ N \gg 0 $ such that for $ n \geq N $ we have

\[\| f_n(t_i) - f(t_i) \| < \varepsilon/5\]

for each $ i $. We observe

\[\begin{aligned} \| f_n(t_{i+1}) - f_n(t_i)\| & \leq \| f_n(t_{i+1}) - f(t_{i+1}) \| + \| f(t_{i+1}) - f(t_i) \| \\ & \quad + \| f(t_i) - f_n(t_i) \| \\ & < \frac{3\varepsilon}{5}. \end{aligned}\]

Therefore, for any $ x \in [t_i, t_{i+1}] $, we have

\[\begin{aligned} \| f_n(x) - f(x) \| & \leq \| f_n(x) - f(t_{i+1}) \| + \| f(t_{i+1}) - f(x) \| \\ & < \| f_n(x) - f(t_{i+1}) \| + \frac{\varepsilon}{5} \\ & \leq \| f_n(x) - f_n(t_{i+1}) \| + \| f_n(t_{i+1}) - f(t_{i+1}) \| + \frac{\varepsilon}{5} \\ & < \| f_n(x) - f_n(t_{i+1}) \| + \frac{2\varepsilon}{5} \\ & \leq \| f_n(t_{i+1}) - f_n(t_i) \| + \frac{2\varepsilon}{5} \\ & < \varepsilon. \end{aligned}\]

Hence, $ f_n $ converges uniformly to $ f $. $ \blacksquare $

2022 F P9.   For $ x>1 $, define

\[F(x) = \sum_{n=1}^{\infty} \frac{1}{n^x}.\]

(a) Prove that $ F $ converges uniformly on $ [1+\delta, \infty) $ for any $ \delta > 0 $. Explain why $ F $ is continuous on $ (1, \infty) $. Does the series define a continuous real-valued function on $ [1, \infty) $? (b) Prove $ F $ is continuously differentiable on $ (1, \infty) $ with

\[F'(x) = - \sum_{n=1}^{\infty} \frac{\ln(n)}{n^x}.\]

Proof. (a) On $ [1+ \delta, \infty) $ we have $ n^{-x} \leq n^{-(1+\delta)} $. Since

\[\sum_{n=1}^{\infty} \frac{1}{n^{1+\delta}}\]

converges by the p-test, the Weierstrass M-test implies $ F $ converges uniformly. $ F $ is continuous by the uniform limit theorem. At $ x=1 $ the defining series is the harmonic series, so it does not define a real-valued function on $ [1,\infty) $.

(b) We see that

\[\frac{d}{dx} \frac{1}{n^x} = - \ln(n) \frac{1}{n^x}.\]

For $ x\geq1+\delta $, we have

\[\left\|\frac{\ln(n)}{n^x}\right\| \leq\frac{\ln(n)}{n^{1+\delta}}.\]

The series of bounds converges. Indeed, after discarding finitely many terms its summand is decreasing, and the integral test gives

\[\int_1^\infty\frac{\ln(x)}{x^{1+\delta}}dx =\int_0^\infty u e^{-\delta u}du =\frac{1}{\delta^2}.\]

Thus, the derivative series converges uniformly on $ [1+\delta,\infty) $. Applying theorem 4.2 on each compact subinterval shows that it is the derivative of $ F $. Since each summand is continuous and the convergence is locally uniform on $ (1,\infty) $, the derivative is continuous. $ \blacksquare $

2022 W P1.   Define $ f_n \colon [0, 1] \to \mathbb{R} $ by

\[f_n(x) = \frac{1+x^n}{1+2^{-n}}.\]

Show $ {f_n} $ is not equicontinuous.

Proof. Let $ x_n = (1/2)^{1/n} $ and $ y_n = (3/4)^{1/n} $. Then $ \lim_n | x_n - y_n | = 0 $, but

\[\lim_n \| f_n(x_n) - f_n(y_n) \| = \lim_n \frac{\|1/2 - 3/4\|}{1 + 2^{-n}} = \frac{1}{4},\]

so our sequence is not equicontinuous. $ \blacksquare $

2021 F P7.   Let $ f_n \colon [0, 1]^2 \to \mathbb{R} $ be a uniformly bounded sequence of continuous functions. Set

\[F_n(x, y) = \int_y^1 \int_x^1 s^{-1/2} t^{-1/3} f_n(s, t) ds dt.\]

(a) Show for each $ n $ that $ F_n(x, y) $ is well defined. (b) Show that $ {F_n} $ has a subsequence $ {F_{n_j}} $ which converges uniformly to a continuous $ F $.

Proof. (a) Since the $ {f_n} $ are uniformly bounded there exists an $ M \geq 0 $ such that

\[\sup_{n \in \mathbb{N}, (x, y) \in [0,1]^2} \| f_n(x, y) \| \leq M.\]

Hence,

\[\begin{aligned} \| F_n(x, y) \| & \leq M \int_y^1 t^{-1/3} dt \int_x^1 s^{-1/2} ds \\ & = M [\frac{3}{2} t^{2/3} \mid_y^1 ] [2 s^{1/2} \mid_x^1 ] \\ & = 3 M (1-y^{2/3})(1-x^{1/2}). \end{aligned}\]

This limit is bounded as $ (x, y) $ approaches $ 0 $, and hence $ F_n $ is well defined.

(b) In (a) we showed that $ {F_n} $ is uniformly bounded by $ 3M $. Letting $ x_2 \geq x_1 $ and $ y_2 \geq y_1 $ without loss of generality, we write the difference as two strips:

\[\begin{aligned} \|F_n(x_1, y_1) - F_n(x_2, y_2)\| &\leq \left\|\int_{y_1}^{y_2}\int_{x_1}^{1}s^{-1/2}t^{-1/3}f_n(s,t)\,ds\,dt\right\| \\ &\quad+\left\|\int_{y_2}^{1}\int_{x_1}^{x_2}s^{-1/2}t^{-1/3}f_n(s,t)\,ds\,dt\right\| \\ &\leq 3M\left(y_2^{2/3}-y_1^{2/3}\right) +3M\left(x_2^{1/2}-x_1^{1/2}\right). \end{aligned}\]

Since both $ x^{1/2} $ and $ y^{2/3} $ are continuous functions, we see that $ {F_n} $ is equicontinuous. Hence, by the Ascoli-Arzelà theorem we can find our desired subsequence. $ \blacksquare $

2020 F P9.   Let $ f_n \colon [0, 1] \to [0, 1] $ converge uniformly to $ f \colon [0, 1] \to [0, 1] $ (a not necessarily continuous function). Suppose the $ f_n $ map compact sets to compact sets. Does $ f $ map compact sets to compact sets?

Proof. Not necessarily. Set

\[f_0(x) = \begin{cases} 0 & \text{if } x = 2^{-n} \text{ for some } n \geq 1 \\ 1 & \text{elsewise}, \end{cases}\]

and inductively define

\[f_{n+1}(x) = \begin{cases} x & \text{if } x = 2^{-(n+1)} \\ f_n(x) & \text{elsewise}. \end{cases}\]

Then the sequence $ {f_n} $ converges uniformly to the function

\[f(x) = \begin{cases} x & \text{if } x = 2^{-n} \text{ for some } n \geq 1 \\ 1 & \text{elsewise}. \end{cases}\]

Each $ f_n $ maps compact sets to compact sets because its image contains only finitely many points. On the other hand, the set

\[K=\{0\}\cup\{2^{-n}\mid n\geq1\}\]

is compact, while

\[f(K)=\{1\}\cup\{2^{-n}\mid n\geq1\}\]

is not closed and hence is not compact. $ \blacksquare $

2020 W P8.   Let $ f \colon \mathbb{R} \to \mathbb{R} $ be continuous with $ f(x+1) = f(x) $. Define $ f_n \colon \mathbb{R} \to \mathbb{R} $ by $ f_1 = f $ and for $ n > 1 $

\[f_n(x) = \frac{1}{2} (f_{n-1}(x-2^{-n}) + f_{n-1}(x+2^{-n})).\]

Show that $ f_n $ converges uniformly on $ \mathbb{R} $.

Proof. Since $ f $ is periodic, it is bounded and uniformly continuous on $ \mathbb{R} $. Let $ \varepsilon > 0 $, and choose an $ n \gg 0 $ such that $ |x-y| < 2^{-n} $ implies $ |f(x)-f(y)| < \varepsilon $. Expanding the recurrence gives

\[f_n(x) =\frac{1}{2^{n-1}} \sum_{\varepsilon_2,\ldots,\varepsilon_n\in\{-1,1\}} f\left(x+\sum_{j=2}^{n}\varepsilon_j2^{-j}\right).\]

For $ k\geq1 $, each term in the corresponding expansion of $ f_{n+k}(x) $ is obtained from one of these translates by an additional shift $ h $ satisfying

\[\|h\| \leq\sum_{j=n+1}^{n+k}2^{-j} <2^{-n}.\]

Pairing each term in the average for $ f_{n+k} $ with its ancestor in the average for $ f_n $ shows

\[\|f_{n+k}(x)-f_n(x)\|<\varepsilon\]

uniformly in $ x $ and $ k $. Thus, $ {f_n} $ is Cauchy with respect to the uniform metric and converges uniformly. $ \blacksquare $

2019 F P9.   Let $ f_n \colon [a, b] \to \mathbb{R} $ be continuous with $ f_n(x) \leq f_{n+1}(x) $. Suppose $ f_n $ converge pointwise to a continuous $ f $. Show they converge uniformly to $ f $.

Proof. We will show $ g_n = f-f_n $ converges uniformly to $ 0 $. Fix $ \varepsilon > 0 $. For each $ x \in [a, b] $, choose an $ N_x \gg 0 $ such that $ g_n(x) < \varepsilon $ for all $ n \geq N_x $. Since our $ g_n $ are continuous, for each $ x $ there exists an $ r_x $ such that $ g_{N_x}(B(x, r_x)) \subseteq [0, \varepsilon) $. Since $ [a, b] $ is compact, we can choose a finite subcover $ B(x_1, r_1), \dots, B(x_{\ell}, r_{\ell}). $ Letting $ N = \max {N_{x_1}, \dots, N_{x_\ell}} $, we get $ g_n(x) < \varepsilon $ for all $ x \in [a, b] $ and $ n \geq N $. $ \blacksquare $

2018 F P2.   Show that

\[F(x) = \sum_{n=1}^{\infty} \frac{\sin(x^n)}{n!}\]

converges uniformly and compute its derivative.

Proof. Since $ | \sin(x^n)/n! | \leq 1/n! $, we have $ F(x) \leq e-1 $, so $ F $ converges uniformly by the Weierstrass M-Test. Likewise

\[\left\| \frac{d}{dx} \frac{\sin(x^n)}{n!} \right\| = \left\| \frac{n x^{n-1} \cos(x^n)}{n!} \right\| \leq \frac{\|x\|^{n-1}}{(n-1)!}.\]

Hence, locally the sum

\[F'(x) = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} \cos(x^n)\]

converges uniformly by the Weierstrass M-Test, and by theorem 4.2 it is the derivative of $ F $. $ \blacksquare $

2018 F P7.   Define $ f_n \colon [0, 2\pi] \to \mathbb{R} $ by $ f_n(x) = e^{\sin(nx)} $ and $ F_n(x) = \int_0^x f_n(y) dy $. Show there exists a subsequence of $ {F_n} $ that converges uniformly to a continuous function.

Proof. We see that

\[\|F_n(x)\| \leq \int_0^{2\pi} \| f_n(y) \| dy \leq \int_0^{2 \pi} e dy = 2 \pi e\]

implies our sequence is uniformly bounded. Assuming $ y \geq x $,

\[\| F_n(y) - F_n(x) \| = \left\| \int_x^y f_n(t) dt \right\| \leq e (y-x)\]

implies our sequence is uniformly equicontinuous since $ x $ is continuous. Thus, we can apply the Ascoli-Arzelà theorem to a get a uniformly convergent subsequence, and the uniform limit theorem to see that the limit of this subsequence is continuous. $ \blacksquare $

2017 W P7.   Let $ f_n \colon [0, 1] \to \mathbb{R} $ converge pointwise to $ f $. Assume $ f_n $ and $ f $ are continuous. (a) Does

\[\lim_n \int_0^1 f_n(x) dx = \int_0^1 f(x) dx?\]

(b) What if $ |f_n(x)| \leq M $ for all $ n $ and $ x $?

Proof. (a) Let

\[f_n(x) = \begin{cases} n \exp \left( - \dfrac{1}{1-(nx-1)^2} \right) & \text{if } 0<x<2/n, \\ 0 & \text{elsewise}. \end{cases}\]

Then our $ f_n $ converge pointwise to $ 0 $ but have constant nonzero integral.

(b) Since $ |f_n(x)| \leq M $ on the finite interval $ [0,1] $ and $ f_n(x)\to f(x) $ pointwise, the dominated convergence theorem gives

\[\lim_n\int_0^1f_n(x)dx=\int_0^1f(x)dx.\]

$ \blacksquare $

5. Derivatives

2022 W P3.   Let $ f \colon \mathbb{R} \to [0, \infty) $ be a differentiable function such that $ f $ is decreasing and $ f' $ is increasing. Show $ \displaystyle \lim_{x \to \infty} f'(x) = 0 $.

Proof. Since $ f $ is decreasing, $ f'(x)\leq0 $. Since $ f' $ is increasing and bounded above by $ 0 $, it has a limit $ L\leq0 $. Suppose $ L<0 $. Then for all sufficiently large $ x $ we have $ f'(x)\leq L/2<0 $, and the mean value theorem gives

\[f(x)\leq f(x_0)+\frac{L}{2}(x-x_0)\]

for $ x\geq x_0 $. This forces $ f(x)\to-\infty $, contradicting $ f\geq0 $. Thus, $ L=0 $. $ \blacksquare $

2022 F P6.   Define $ f \colon \mathbb{R}^2 \to \mathbb{R} $ by

\[f(x, y) = \frac{x^3}{x^2 + y^2}\]

for $ (x, y) \neq 0 $ and $ f(0, 0) = 0 $. (a) Show $ f $ is continuous at $ (0, 0) $. (b) Show $ f $ has every directional derivative at $ (0, 0) $. (c) Decide if $ f $ is differentiable at $ (0, 0) $.

Proof. (a) We observe

\[f(r \cos \theta, r \sin \theta) = r \cos^3 \theta,\]

which approaches $ 0 $ as $ r $ goes to $ 0 $.

(b) The directional derivative in the zero direction is $ 0 $. Now set $ v = (a, b)\neq0 $. Then

\[\begin{aligned} \nabla_v f(0) & = \lim_{h \to 0} \frac{f(0+hv)-f(0)}{h} \\ & = \lim_{h \to 0} \frac{(ah)^3}{(ah)^2 + (bh)^2} \frac{1}{h} \\ & = \lim_{h \to 0} \frac{h^3 a^3}{h^3(a^2+b^2)} \\ & = \frac{a^3}{a^2+b^2}. \end{aligned}\]

(c) Suppose that the derivative $ D f(0) $ exists. Then, by linearity,

\[\nabla_{(1, 1)} f(0) = \nabla_{(1, 0)} f(0) + \nabla_{(0, 1)} f(0)\]

implies $ 1/2 = 1 $, a contradiction. Thus, the derivative does not exist. $ \blacksquare $

2021 F P5.   Let $ f \colon \mathbb{R}^2 \to \mathbb{R} $. Suppose $ \frac{\partial f}{\partial x_1} $ exists at $ (0, 0) $, and that $ \frac{\partial f}{\partial x_2} $ exists in a neighborhood of $ 0 $ and is continuous at $ 0 $. Show $ f $ is differentiable at $ 0 $.

Proof. Write

\[f(x,y) - f(0, 0) = f(x, y) - f(x, 0) + f(x, 0) - f(0, 0)\]

Denote our partials $ f_x $ and $ f_y $. Then, by the MVT, there exists a function $ a(x, y) \colon \mathbb{R}^2 \to \mathbb{R} $ such that

\[\begin{aligned} f(x, y) - f(x, 0) & = f_y(x, a(x, y)) y \\ & = f_y(0, 0) y + \left[ f_y(x, a(x, y)) - f_y(0,0) \right] y \\ & = f_y(0, 0) y + o(y). \end{aligned}\]

Likewise

\[\begin{aligned} f(x, 0) - f(0, 0) &= f_x(0, 0) x + \left[ \frac{f(x, 0) - f(0, 0)}{x} - f_x(0, 0) \right] x \\ & = f_x(0, 0) x + o(x). \end{aligned}\]

Thus,

\[f(x, y) - f(0, 0) = f_x(0, 0) x + f_y(0, 0) y + o(x) + o(y).\]

The coefficients implicit in the two little-$ o $ terms approach $ 0 $ as $ (x,y)\to(0,0) $, so their sum divided by $ \sqrt{x^2+y^2} $ approaches $ 0 $. Thus, $ f $ is differentiable at $ (0, 0) $. $ \blacksquare $

2020 W P6.   Let $ f \colon \mathbb{R} \to (0, \infty) $ be a differentiable function such that $ f'(x) > f(x) $ for every $ x\in \mathbb{R} $. Show there exists a $ k > 0 $ such that $ \displaystyle \lim_{x \to \infty} f(x) e^{-kx} = \infty $, and find the least upper bound on such $ k $.

Proof. We see $ f'(x) > f(x) $ implies

\[\frac{d}{dx}\ln f(x)=\frac{f'(x)}{f(x)}>1.\]

Thus, for $ x>0 $,

\[\ln f(x)-\ln f(0)>x,\]

so $ f(x)>f(0)e^x $. Consequently,

\[\lim_{x\to\infty}f(x)e^{-kx}=\infty\]

for every $ 0<k<1 $. The bound $ 1 $ is sharp: the function

\[f(x)=e^x(2+\arctan x)\]

is positive and satisfies $ f'(x)-f(x)=e^x/(1+x^2)>0 $, but $ f(x)e^{-x}\to2+\pi/2 $. Therefore, the least upper bound on the values of $ k $ guaranteed by the hypothesis is $ 1 $. $ \blacksquare $

2020 F P7.   Let $ f \colon \mathbb{R}^2 \to \mathbb{R}^2 $ be a differentiable map with $ f = (f_1, f_2) $. Suppose that for all $ (x_1, x_2) \in \mathbb{R}^2 $

\[\left| \frac{\partial f_1}{\partial x_1}(x_1, x_2) - 2 \right| + \left| \frac{\partial f_1}{\partial x_2}(x_1, x_2) \right| + \left| \frac{\partial f_2}{\partial x_1}(x_1, x_2) \right| + \left| \frac{\partial f_2}{\partial x_2}(x_1, x_2) - 2 \right| < 1/2.\]

Prove $ f $ is globally injective.

Proof. We start by showing $ Df $ is positive definite at every point. Indeed, set

\[Q(a, b) = \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}.\]

Then

\[Q(a, b) = a^2 \frac{\partial f_1}{\partial x_1} + b^2 \frac{\partial f_2}{\partial x_2} + ab \left( \frac{\partial f_1}{\partial x_2} + \frac{\partial f_2}{\partial x_1} \right).\]

The inequality implies $ \frac{\partial f_1}{\partial x_1}, \frac{\partial f_2}{\partial x_2} \geq 1.5 $, so

\[Q(a,b) \geq 1.5(a^2 + b^2)-\frac{1}{2}\|ab\|.\]

Since $ 2|ab|\leq a^2+b^2 $, we obtain

\[Q(a,b)\geq\frac54(a^2+b^2)>0\]

whenever $ (a,b)\neq(0,0) $. Thus, the quadratic form associated to $ Df $ is positive definite.

Now assume $ f(a) = f(b) $ for $ a \neq b $. We set $ \gamma(t) = a + t(b-a) $ and $ t \in [0, 1] $ and

\[F(t) = \left< f(\gamma(t)) - f(a), b-a \right>.\]

Then $ F(0) = F(1) = 0 $, so the mean value theorem tells us $ F'(t) = 0 $ for some $ t \in (0, 1) $. Hence, by the chain rule,

\[F'(t) = (b-a)^T Df(\gamma(t)) (b-a) = 0\]

contradicts $ Df $ being positive definite. $ \blacksquare $

2019 F P1.   Define $ f \colon \mathbb{R}^2 \to \mathbb{R} $ by

\[f(x, y) = \frac{xy^2}{x^2 + y^4}\]

for $ (x, y) \neq 0 $ and $ f(0, 0) = 0 $. (a) Show that $ f $ has every directional derivative at $ 0 $. (b) Show that $ f $ is not continuous at $ (0, 0) $.

Proof. (a) The directional derivative in the zero direction is $ 0 $. Now set $ v = (a, b) \neq 0 $. Then

\[\begin{aligned} \nabla_v f(0) & = \lim_{h \to 0} \frac{f(0+hv)-f(0)}{h} \\ & = \lim_{h \to 0} \frac{ab^2 h^3}{a^2 h^2 + b^4 h^4} \frac{1}{h} \\ & = \lim_{h \to 0} \frac{ab^2}{a^2 + b^4 h^2}. \end{aligned}\]

If $ a = 0 $ this limit evaluates to $ 0 $; elsewise it equals $ b^2/a $.

(b) For the path $ (t, t) $ we see

\[\lim_{t \to 0} f(t, t) = \lim_{t \to 0} \frac{t^3}{t^2+t^4} = \lim_{t \to 0} \frac{t}{1+t^2} = 0\]

While for the path $ (t, \sqrt{t}) $ as $ t\to0^+ $ we have

\[\lim_{t \to 0} f(t, \sqrt{t}) = \lim_{t \to 0} \frac{t^2}{t^2+t^2} = \frac{1}{2}.\]

Thus, $ f $ is not continuous at the origin. $ \blacksquare $

2019 W P4.   (a) Give an example of an everywhere differentiable function with discontinuous derivative. (b) Let $ f, g \colon \mathbb{R} \to \mathbb{R} $. Suppose for every $ \varepsilon > 0 $ there exists a $ \delta > 0 $ such that $ 0 < | h | < \delta $ implies

\[\left\| \frac{f(x+h)-f(x)}{h} - g(x) \right\| < \varepsilon\]

for all $ x \in \mathbb{R} $. Show that $ f $ is continuously differentiable.

Proof. (a) Consider the function $ f(x) = x^2 \sin(\frac{1}{x}) $ continuously extended by $ f(0) = 0 $. Then $ f $ is everywhere differentiable with $ f'(0) = 0 $, but

\[f'(x)=2x\sin(1/x)-\cos(1/x)\]

does not have a limit as $ x\to0 $.

(b) By definition $ f'(x) = g(x) $. Let $ \varepsilon > 0 $ and choose $ \delta > 0 $ as stated in the problem with $ \varepsilon/2 $ in place of $ \varepsilon $. Then for any $ x, y \in \mathbb{R} $ with $ 0<| x-y | < \delta/2 $, we see $ y = x+h $ with $ | h | < \delta/2 $, and hence

\[\begin{aligned} \| g(x) - g(y) \| & \leq \left\| g(x) - \frac{f(x+h) - f(x)}{h} \right\| + \left\| \frac{f(x+h) - f(x)}{h} - g(y) \right\| \\ & = \left\| \frac{f(x+h) - f(x)}{h} - g(x) \right\| + \left\| \frac{f(y-h)-f(y)}{-h} - g(y) \right\| \\ & < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{aligned}\]

Thus, $ g $ is continuus and so is $ f' $. $ \blacksquare $

2016 W P5.   Assume that a function $ f \colon \mathbb{R}^2 \to \mathbb{R} $ satisfies

\[f(x_1+t, x_2+s) \geq f(x_1, x_2) - s^2 - t^2\]

for each $ (x_1, x_2) \in \mathbb{R}^2 $ and each $ (s, t) \in \mathbb{R}^2 $. Prove $ f $ must be constant.

Proof. Let $ x, y \in \mathbb{R}^2 $ be arbitrary. Then we have

\[f(x) \geq f(y) - \| x-y \|^2\]

and

\[f(y) \geq f(x) - \| x-y \|^2.\]

Hence,

\[\| x-y \|^2 \geq f(x) - f(y) \geq -\| x-y \|^2,\]

so

\[\| x-y \| \geq \frac{\|f(x) - f(y)\|}{\|x-y\|}.\]

Using this to take the derivative of $ f $ at $ x $ we get

\[\lim_{\| h \| \to 0} \frac{\|f(x+h) - f(x)\|}{\| h \|} \leq \lim_{\| h \| \to 0} \| h \| = 0.\]

We conclude that $ f $ has zero derivative at every point, and therefore is constant. $ \blacksquare $

6. Optimization

AM-GM Inequality.   Let $ x_1, \dots, x_n \in \mathbb{R}^+ $ be positive reals. Then

\[\frac{1}{n} \sum_{i=1}^{n} x_i \geq \left( \prod_{i=1}^{n} x_i \right)^{1/n}.\]

This is an equality if and only if $ x_1 = \cdots = x_n $.

Jenson's Inequality.   Let $ f \colon \mathbb{R} \to \mathbb{R} $ be convex. For $ x_1, \dots, x_n \in \mathbb{R} $ and $ a_1, \dots, a_n \in \mathbb{R}^+ $ we have

\[f\left( \frac{\sum a_i x_i}{\sum a_i} \right) \leq \sum a_i \frac{f(x_i)}{\sum a_i}.\]

Problems

2016 F P7. Let $ \Omega = {(x, y) \in \mathbb{R}^2 \mid y > 0 } $, and define $ f \colon \Omega \to \mathbb{R} $ by

\[f(x, y) = \frac{2 + \sqrt{(1+x)^2 + y^2} + \sqrt{(1-x)^2 +y^2}}{\sqrt{y}}.\]

Show $ f $ achieves its minimum at a unique point $ (x_0, y_0) \in \Omega $ and find $ (x_0, y_0) $.

Proof. Fix $ y>0 $. Then

\[\sqrt{y}\frac{\partial f}{\partial x}(x,y) = \frac{1+x}{\sqrt{(1+x)^2+y^2}} - \frac{1-x}{\sqrt{(1-x)^2+y^2}} = 0\]

if and only if

\[(1+x) \sqrt{(1-x)^2+y^2} = (1-x)\sqrt{(1+x)^2+y^2}.\]

The left hand side of the derivative equation is strictly increasing in $ x $, so it has at most one zero. Squaring the equation reduces it to $ (1+x)^2 = (1-x)^2 $, and hence the zero is $ x=0 $. Since $ f(x,y)\to\infty $ as $ |x|\to\infty $, this is the unique minimum for fixed $ y $. Thus, fix $ x = 0 $. Then

\[\frac{d}{dy} f(0, y) = \frac{y^2 - \sqrt{y^2+1}-1}{\sqrt{y^5+y^3}} = 0\]

if and only if $ y^2 - 1 = \sqrt{y^2+1} $, which implies $ y = \sqrt{3} $. Since $ f(0,y)\to\infty $ as $ y\to0^+ $ or $ y\to\infty $, this is the unique global minimum. Thus, our minimum is achieved at $ (0, \sqrt{3}) $. $ \blacksquare $

2022 F P7.   Let $ f_n \colon \mathbb{R}^2 \to \mathbb{R} $ be a sequence of continuously differentiable functions which converge pointwise to a continuously differentiable function $ f $. Suppose that for each $ n $ that $ (0, 0) $ is a local minimum for $ f_n $. Is it a local minimum for $ f $?

Proof. Not necessarily. Let $ \chi \colon \mathbb{R}\to[0,1] $ be a continuously differentiable bump function such that $ \chi(t)=1 $ for $ |t|\leq1/2 $ and $ \chi(t)=0 $ for $ |t|\geq1 $. Define

\[f_n(x,y) =-x^2-y^2+2(x^2+y^2)\chi\left(n\sqrt{x^2+y^2}\right).\]

Near the origin, $ f_n(x,y)=x^2+y^2 $, so $ (0,0) $ is a local minimum. For every fixed $ (x,y)\neq(0,0) $, the bump term eventually vanishes, and hence $ f_n $ converges pointwise to

\[f(x,y)=-x^2-y^2.\]

The origin is not a local minimum of $ f $. $ \blacksquare $

2021 F P2.   Find all $ x, y > 0 $ which minimize $ f(x, y) = x/y + y/x $ on the curve $ x^2 + 2y^2 = 3 $.

Proof. The AM-GM inequality implies

\[\frac{x}{y} + \frac{y}{x} = \frac{1}{2} \left( \frac{2x}{y} + \frac{2y}{x} \right) \geq \sqrt{4} = 2,\]

and that this minimum is achieved only at $ (1, 1) $. We see that $ 1 + 2 = 3 $, and hence is our minimum on the curve. $ \blacksquare $

2020 F P4.   Find the absolute minimum of $ x^2 y + y^2 z + z^2 w + w^2 x $ for $ xyzw = 1 $ and $ x, y, z, w > 0 $.

Proof. The $ AM-GM $ inequality implies

\[\frac{1}{4} \left( 4x^2 y + 4y^2 z + 4z^2 w + 4w^2 x \right) \geq (4^4 x^3 y^3 z^3 w^3)^{1/4} = 4,\]

and this minimum is obtained when $ x = y = z = w = 1 $. $ \blacksquare $

2018 W P7.   Find the absolute minimum of

\[f(x, y, z) = xy + yz + zx\]

on

\[g(x, y, z) = x^2 + y^2 + z^2 = 12.\]

Proof. We observe

\[2 f(x, y, z) = (x+y+z)^2 - g(x, y, z) \geq -12\]

implies $ f(x, y, z) \geq -6 $. Since $ f(-\sqrt{6}, \sqrt{6}, 0) = -6 $ we conclude that this is the minimum. $ \blacksquare $

7. Convergence of Integrals

Fubini-Tonelli theorem.   Let $ f \colon [a, b] \times [c, d] \to \mathbb{R} $ be continuous. Then

\[\int_a^b \int_c^d f(x, y) \,dy\,dx = \int_c^d \int_a^b f(x, y) \,dx\,dy.\]

2022 F P8.   Does

\[I = \int_3^{\infty} \frac{\ln(x)}{x^p \ln(ln(x))} dx\]

converge for $ p \geq 0 $?

Proof. The answer depends on the value of $ p $. Substitute $ u = \ln x $ and $ du = x^{-1} dx $. Then

\[I = \int_{\ln 3}^{\infty} \frac{u e^{-u(p-1)}}{\ln(u)} du.\]

If $ p < 1 $, then $ -(p-1) > 0 $ implies for $ u $ large that $ u e^{-u(p-1)} > \ln(u) $ and $ I $ diverges. If $ p = 1 $ then $ u > \ln(u) $ for $ u $ large and $ I $ diverges. Thus, we are reduced to the case when $ p > 1 $. The integrand is continuous on the finite interval $ [\ln3,e] $, while $ 1/\ln u\leq1 $ for $ u\geq e $. Therefore,

\[\int_e^\infty\frac{u e^{-u(p-1)}}{\ln u}du \leq\int_e^\infty u e^{-u(p-1)}du <\infty.\]

Thus $ I $ converges when $ p > 1 $. $ \blacksquare $

2022 W P5.   Does $ I = \int_0^{\infty} \cos(x^{2/3}) dx $ converge?

Proof. Set $ u = x^{2/3} $ and $ dx = \frac{3}{2} \sqrt{u} du $. Then

\[I = \frac{3}{2} \int_0^{\infty} \sqrt{u} \cos(u) du.\]

For $ R\geq1 $, integration by parts gives

\[\begin{aligned} \int_1^R\sqrt{u}\cos(u)du &=\sqrt{R}\sin(R)-\sin(1) -\frac12\int_1^R u^{-1/2}\sin(u)du. \end{aligned}\]

The last integral converges as $ R\to\infty $ by the Dirichlet test. Along $ R_k=\pi/2+2\pi k $, however, the first term equals $ \sqrt{R_k} $ and is unbounded. Therefore, the improper integral does not converge. $ \blacksquare $

2022 W P8.   Let $ f \colon \mathbb{R}^2 \to \mathbb{R} $ be a continuous compactly support function. Define $ g \colon \mathbb{R}^2 \to \mathbb{R} $ by

\[g(x) = \int_{\mathbb{R}^2} \frac{f(y)}{\|x-y\|} dy\]

Prove that this integral converges (and $ g $ is continuous).

Proof. Let $ K $ be a compact set supporting $ f $, and choose $ M $ such that $ |f|\leq M $. For any $ x\in\mathbb{R}^2 $, split $ K $ into the parts where $ |x-y|<1 $ and $ |x-y|\geq1 $. Then

\[\int_K\frac{\|f(y)\|}{\|x-y\|}dy \leq M\int_{B(x,1)}\frac{1}{\|x-y\|}dy+M\,\operatorname{Area}(K).\]

The first integral is $ 2\pi $ by polar coordinates, so the defining integral converges absolutely.

To prove continuity, fix $ x_0 $ and $ \varepsilon>0 $. Choose $ r>0 $ so small that $ 12\pi Mr<\varepsilon/2 $. If $ |x-x_0|<r $, then the contribution to $ |g(x)-g(x_0)| $ from $ K\cap B(x_0,2r) $ is bounded by

\[M\int_{B(x,3r)}\frac{1}{\|x-y\|}dy +M\int_{B(x_0,2r)}\frac{1}{\|x_0-y\|}dy \leq 10\pi Mr <\frac{\varepsilon}{2}.\]

On the compact set $ K\setminus B(x_0,2r) $, the functions $ y\mapsto1/|x-y| $ converge uniformly to $ y\mapsto1/|x_0-y| $ as $ x\to x_0 $. Therefore, for $ x $ sufficiently close to $ x_0 $, the remaining contribution is less than $ \varepsilon/2 $. Thus, $ g $ is continuous. $ \blacksquare $

2021 F P5.   Let $ f \colon \mathbb{R}^2 \to \mathbb{R} $ be continuous. Suppose $ \int_0^{\infty} f(x, y) dy $ exists for every $ x \in [0, 1] $. Assume there exists a $ C \geq 0 $ such that for $ z > 0 $,

\[\left\| \int_z^{\infty} f(x, y) dy \right\| \leq \frac{C}{\log(2+z)}\]

for every fixed $ x $. Show

\[\int_0^1 \left[ \int_0^{\infty} f(x, y) dy \right] dx = \int_0^{\infty} \left[ \int_0^1 f(x, y) dx \right] dy.\]

Proof. Set

\[h_z(x)=\int_0^z f(x,y)dy \qquad\text{and}\qquad h(x)=\int_0^\infty f(x,y)dy.\]

The tail estimate gives

\[\sup_{x\in[0,1]}\|h(x)-h_z(x)\| \leq\frac{C}{\log(2+z)} \longrightarrow0.\]

Thus, theorem 4.1 and finite-rectangle Fubini give

\[\begin{aligned} \int_0^1\left[\int_0^\infty f(x,y)dy\right]dx &=\lim_{z\to\infty}\int_0^1\int_0^z f(x,y)dy\,dx \\ &=\lim_{z\to\infty}\int_0^z\int_0^1 f(x,y)dx\,dy \\ &=\int_0^\infty\left[\int_0^1 f(x,y)dx\right]dy. \end{aligned}\]

The same calculation also proves that the improper integral on the right exists. $ \blacksquare $

2020 F P5.   Set $ a _0 = 0 $ and for $ k \geq 1 $

\[a_k = \sqrt{1 + \frac{1}{2} + \cdots + \frac{1}{k}}.\]

Assume $ {b_k} $ is a sequence in $ \mathbb{R}^+ $ such that $ \sum_k b_k^2 < \infty $, and that $ f \colon \mathbb{R}^2 \to \mathbb{R}^+ $ is a continuous function such that $ f(x) \leq b_k $ when $ a_{k-1} \leq |x| \leq a_k $. Show that $ \int_{\mathbb{R}^2} f(x) dA $ exists.

Proof. We have

\[\begin{aligned} \int_{\mathbb{R}^2} f(x) dx & = \sum_{k=0}^{\infty} \int_0^{2 \pi} \int_{a_k}^{a_{k+1}} f(r \cos \theta, r \sin \theta) r dr d \theta \\ & \leq \sum_{k=0}^{\infty} \int_0^{2 \pi} \int_{a_k}^{a_{k+1}} b_{k+1} r dr d \theta \\ & \leq \sum_{k=0}^{\infty} \pi b_{k+1} [a_{k+1}^2 - a_k^2] \\ & = \sum_{k=0}^{\infty} \pi b_{k+1} \frac{1}{k+1} \\ & \leq \sum_{k=1}^{\infty} \pi \text{max} \{b_k^2, \frac{1}{k^2}\} \\ & \leq \pi \sum_{k=1}^{\infty} b_k^2 + \pi \sum_{k=1}^{\infty} \frac{1}{k^2} \\ & < \infty. \end{aligned}\]

We conclude that the limit exists. $ \blacksquare $

2020 W P5.   Let $ f \colon \mathbb{R}^2 \to \mathbb{R} $ be a differentiable function such that $ f(0, 0) = 0 $. Show that

\[I = \int \int_{x^2+y^2 \leq 1} \frac{f(x, y)}{(x^2+y^2)^{4/3}} dx dy\]

converges.

Proof. Differentiability at the origin and $ f(0,0)=0 $ imply that there exist $ 0<\delta\leq1 $ and $ C>0 $ such that

\[\|f(x,y)\|\leq C\sqrt{x^2+y^2}\]

whenever $ x^2+y^2<\delta^2 $. Thus, the absolute value of the integral over the disk of radius $ \delta $ is at most

\[\int_0^{2\pi}\int_0^\delta\frac{Cr}{r^{8/3}}r\,dr\,d\theta =2\pi C\int_0^\delta r^{-2/3}dr =6\pi C\delta^{1/3} <\infty.\]

On the compact annulus $ \delta\leq\sqrt{x^2+y^2}\leq1 $, the integrand is continuous and hence integrable. Therefore, the original integral converges absolutely. $ \blacksquare $

8. Polynomials and Stone Weierstrass

Weierstrass approximation theorem.   Let $ f \colon [a, b] \to \mathbb{R} $ be continuous. Then there exists a sequence $ p_n \in \mathbb{R}[x] $ that converges uniformly to $ f $.

Problems

2023 W P9.   For $ n \geq 2 $ let $ p \colon \mathbb{R}^n \to \mathbb{R} $ be

\[p(x_1, \dots, x_n) = \sum_{j=1}^n x_j^{2j+1}.\]

Suppose $ f = (f_1, \dots, f_n) \colon \mathbb{R}^n \to \mathbb{R}^n $ is continuously differentiable function with $ p (f(x)) = 0 $ for all $ x \in \mathbb{R}^n $. Prove $ \det f' = 0 $.

Proof. Suppose $ \det f'(x_0)\neq0 $ at some $ x_0 $. By the inverse function theorem, there are neighborhoods $ U $ of $ x_0 $ and $ V $ of $ f(x_0) $ such that $ f(U)=V $, where $ V $ is open. But $ p(f(x))=0 $ for every $ x\in U $, so $ p $ vanishes on the open set $ V $. A nonzero real polynomial cannot vanish on an open set, contradicting the definition of $ p $. Therefore, $ \det f'(x)=0 $ for every $ x $. $ \blacksquare $

2020 F P8.   Let $ f \colon [0, 1] \to \mathbb{R} $ be continuous such that $ \int_0^1 f(x) x^n dx = 0 $ for $ n = 3, 4, \dots $. Prove $ f(x) = 0 $.

Proof. The Weierstrass approximation theorem implies we can approximate $ f(x) x^4 $ by polynomials $ p_n \in \mathbb{R}[x] $. Since this sequence converges uniformly, theorem 4.1 then implies

\[\lim_{n \to \infty} \int_0^1 [f(x) x^4] p_n(x) dx = \int_0^1 f(x)^2 x^8 dx.\]

Each integral on the left hand side is $ 0 $, and hence the right hand side is too. Thus, the continuous nonnegative function $ f(x)^2x^8 $ is identically zero, so $ f(x)=0 $ for $ x>0 $. Continuity then gives $ f(0)=0 $. $ \blacksquare $

2018 W P10.   Let $ f \colon \mathbb{R}^2 \to \mathbb{R} $ be a function such that $ f(x_0, y) $ is a polynomial in $ y $ for every fixed $ x_0 $, and $ f(x, y_0) $ is a polynomial in $ x $ for every fixed $ y_0 $. Show that $ f $ is polynomial.

Proof. For $ N\geq0 $, let

\[Y_N=\{y\in\mathbb{R}\mid \deg_x f(x,y)\leq N\},\]

where the zero polynomial is included. Since $ \mathbb{R}=\bigcup_{N\geq0}Y_N $ is uncountable, some $ Y_N $ is infinite. Choose $ N+1 $ distinct real numbers $ x_0,\dots,x_N $, and let $ L_0,\dots,L_N $ be their Lagrange interpolation polynomials. For every $ y\in Y_N $,

\[f(x,y)=\sum_{j=0}^N f(x_j,y)L_j(x).\]

Define the right hand side to be $ P(x,y) $. Each function $ y\mapsto f(x_j,y) $ is a polynomial, so $ P $ is a polynomial in both variables. For each fixed $ x $, the polynomial

\[y\longmapsto f(x,y)-P(x,y)\]

vanishes on the infinite set $ Y_N $, and hence is identically zero. Therefore, $ f=P $ is a polynomial in both variables. $ \blacksquare $

2010 W P5.   Let $ g_n \colon [0, 1] \to \mathbb{R} $ be a sequence of continuous functions. Suppose the $ g_n $ are uniformly bounded by $ M $, and that there exists a continuous $ g \colon [0, 1] \to \mathbb{R} $ such that

\[\lim_n \int_0^1 g_n(x) p(x) dx = \int_0^1 g(x) p(x) dx\]

for every $ p \in \mathbb{R}[x] $. Show that $ | g(x) | \leq M $, and that

\[\lim_n \int_0^1 g_n(x) f(x) dx = \int_0^1 g(x) f(x) dx\]

for every continuous $ f \colon [0, 1] \to \mathbb{R} $.

Proof. First let $ f $ be continuous and choose polynomials $ p_k $ converging uniformly to $ f $. Since $ |g_n|_\infty\leq M $ and $ g $ is bounded,

\[\left\|\int_0^1g_n(f-p_k)\right\|\leq M\|f-p_k\|_\infty, \qquad \left\|\int_0^1g(f-p_k)\right\|\leq\|g\|_\infty\|f-p_k\|_\infty.\]

For a fixed sufficiently large $ k $, apply the hypothesis to $ p_k $ and then let $ k\to\infty $. This proves the asserted convergence for every continuous $ f $.

It remains to prove the bound on $ g $. Suppose, for example, that $ g(x_0)>M $. By continuity, there are an $ \eta>0 $ and a neighborhood $ U $ of $ x_0 $ on which $ g>M+\eta $. Choose a nonzero continuous function $ f\geq0 $ supported in $ U $. Then

\[\int_0^1g(x)f(x)dx>(M+\eta)\int_0^1f(x)dx,\]

while $ \int_0^1g_n(x)f(x)dx\leq M\int_0^1f(x)dx $ for every $ n $, contradicting the convergence just proved. Applying the same argument to $ -g $ rules out $ g(x_0)<-M $. Thus, $ |g(x)|\leq M $ for every $ x $. $ \blacksquare $

9. Stokes' theorem

Stokes' theorem.   Let $ S $ be a smooth oriented surface in $ \mathbb{R}^3 $ with boundary $ \partial S $. If $ F \colon \mathbb{R}^3 \to \mathbb{R}^3 $ is a vector field with continuous first order partials, then

\[\int \int_S (\text{Curl} F \cdot n) dA = \int_{\partial S} F \cdot d\mathbf{r},\]

where $ n $ is the oriented unit normal and $ \partial S $ has the induced orientation.

Divergence theorem.   Let $ M $ be a bounded region in $ \mathbb{R}^n $ with piecewise smooth boundary $ \partial M $, and let $ F $ be continuously differentiable on a neighborhood of $ M $. Then

\[\int_M \text{Div } F dV = \int_{\partial M} (F \cdot n) dS,\]

where $ n $ is the outward unit normal.

Problems

2022 F P3.   Find the value of $ \int \int_E F \cdot n dS $ where $ F(x, y, z) = (x, ze^x, y^2) $,

\[E = \{(x, y, z) \in \mathbb{R}^3 \mid \|(x, y, z) \| = 1, z \geq 0\},\]

and $ n $ is the outward unit normal.

Proof. Write

\[H = \{(x, y, z) \in \mathbb{R}^3 \mid \| (x, y, z) \| \leq 1, z \geq 0\}.\]

The divergence theorem tells us

\[\int_H \text{Div} F dV = \int_E (F \cdot n) dS + \int_{D^2} (F \cdot n) dS,\]

where $ D^2 $ denotes the unit disk with $ z=0 $. We see

\[\int_H \text{Div} F dV = \int_H 1 dV = \frac{2}{3} \pi.\]

Next our unit normal on $ D^2 $ is $ (0, 0, -1) $, so

\[\int_{D^2} (F \cdot n) dS = \int_{D^2} -y^2 dS = - \frac{\pi}{4}.\]

Thus, $ \int_E F \cdot n dS = \frac{2 \pi}{3} + \frac{\pi}{4} $. $ \blacksquare $

2023 W P6.   Let $ C $ be a simple closed curve that lies in the plane $ x + y + z = 1 $, oriented positively with respect to $n=(1,1,1)/\sqrt3$. Show

\[\int_C z dx - 2xdy + 3y dz\]

only depends on the region $ R $ enclosed by $ C $.

Proof. The curl of $ (z, -2x, 3y) $ is $ (3, 1, -2) $. We use $ n = (1, 1, 1)/\sqrt{3} $ to specify the positive orientation. By Stokes' theorem,

\[\int_C z dx - 2xdy + 3y dz = \frac{2}{\sqrt{3}}\operatorname{Area}(R),\]

where the area is measured in the plane $ x+y+z=1 $. Equivalently, the integral is twice the area of the projection of $ R $ onto the $ xy $-plane. Thus, it depends only on the oriented region $ R $. $ \blacksquare $

2022 W P9.   Let $ F = (6yz, 2xz, 4xy) $, and define $ \alpha, \gamma \colon [-\pi, \pi] \to \mathbb{R}^3 $ by

\[\alpha(t) = (\cos(t), \sin(t), 0)\]

and

\[\gamma(t) = (\cos(t), \sin(t), 4+\sin(t)\cos(t^3)).\]

(a) Apply Stokes' theorem on the ruled surface joining $ \alpha $ to $ \gamma $ to express $ \int_{\gamma} F\cdot d\mathbf r $ in terms of $ \int_{\alpha}F\cdot d\mathbf r $. (b) Use (a) to evaluate the first integral.

Proof. Set $ h(t)=4+\sin(t)\cos(t^3) $ and parameterize the joining surface by

\[X(t,u)=(\cos t,\sin t,u h(t)), \qquad -\pi\leq t\leq\pi,\quad0\leq u\leq1.\]

Since $3\leq h(t)\leq5$, this parametrizes the joining surface. Apply Stokes' theorem with the orientation for which the boundary is $ \gamma-\alpha $. Since

\[\operatorname{Curl}F=(2x,2y,-4z)\]

and the oriented vector area element is $ (-h(t)\cos t,-h(t)\sin t,0)\,dt\,du $, we obtain

\[\int_\gamma F\cdot d\mathbf r-\int_\alpha F\cdot d\mathbf r =-2\int_{-\pi}^{\pi}h(t)dt =-16\pi,\]

because $\sin(t)\cos(t^3)$ is odd. Finally, $ F(\alpha(t))=(0,0,4\sin t\cos t) $ is orthogonal to $ \alpha'(t)=(-\sin t,\cos t,0) $, so $ \int_\alpha F\cdot d\mathbf r=0 $. Therefore,

\[\int_\gamma F\cdot d\mathbf r =-16\pi.\]

$ \blacksquare $

2021 F P4.   Let $ E $ be the square-based pyramid in $ \mathbb{R}^3 $ with top vertex $ (1, 2, 5) $ and base $ (x, y, 0) $ with $ 0 \leq x, y, \leq 3 $, and let $ S_1, S_2, S_3, S_4 $ be the triangular sides of $ E $. Define $ F \colon \mathbb{R}^3 \to \mathbb{R}^3 $ by

\[F(x, y, z) = (3x-y+4z, x+5y-2z, x^2+y^2-z).\]

If $ n $ is the outward unit normal, find

\[\sum_{i=1}^4 \int \int_{S_i} F \cdot n d A.\]

Proof. Let $ B = [0, 3] \times [0, 3] \times 0 $ denote the base of our pyramid. The divergence theorem tells us

\[\sum_{i=1}^4 \int \int_{S_i} F \cdot n d A = \int_E \text{Div} F dV - \int_B (F \cdot n) dA.\]

We see $ \text{Div} F = 7 $ and the volume of any pyramid is $ \frac{1}{3} b h $ where $ b = 9 $ and $ h = 5 $, so

\[\int_E \text{Div} F dV = 7 \text{Vol}(E) = 7 (\frac{1}{3} 9 \cdot 5) = 105.\]

Likewise

\[\int_B (F \cdot n) dA = \int_0^3 \int_0^3 (3x-y, x+5y, x^2+y^2) \cdot (0, 0, -1) = -54.\]

We conclude that our desired value is $ 105+54 = 159 $. $ \blacksquare $

2020 F P2.   Find the value of $ \int \int_E F \cdot n dS $, where $ F(x, y, z) = (yz^2, \sin x, x^2) $,

\[E = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + 4z^2 = 1, z \geq 0\},\]

and $ n $ is the outward unit normal.

Proof. Write

\[V = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + 4z^2 \leq 1, z \geq 0\}.\]

The divergence theorem tells us

\[\int \int_E (F \cdot n) dS = \int_V \text{Div} F dV - \int_{D^2} (F \cdot n) dA,\]

where $ D^2 $ is the unit disk with $ z = 0 $. We see that $ \text{Div} F = 0 $ implies $ \int_V \text{Div} F dV = 0 $. On $ D^2 $ the unit normal is $ n = (0, 0, -1) $, and we compute

\[\begin{aligned} \int_{D^2} (F \cdot n) dA & = \int_{D^2} - x^2 dA \\ & = \int_0^{2 \pi} \int_0^1 -r^3 \cos^2(\theta) dr d \theta \\ & = - \frac{\pi}{4}. \end{aligned}\]

Thus, $ \int \int_E (F \cdot n) dS = \pi/4 $. $ \blacksquare $

10. Inverse and implicit function theorems

Inverse function theorem.   Let $ f \colon \mathbb{R}^n \to \mathbb{R}^n $ be continuously differentiable. If $ \det f'(x) \neq 0 $, then there are neighborhoods $ U $ of $ x $ and $ V $ of $ f(x) $ such that $ f\vert_U \colon U\to V $ is a continuously differentiable diffeomorphism. Conversely, the derivative of a local continuously differentiable diffeomorphism is invertible.

Implicit function theorem.   Let $ D \subseteq \mathbb{R}^{n+m} $ be open, and let $ f \colon D \to \mathbb{R}^n $ be continuously differentiable with $ f(a, b) = 0 $ for some $ (a, b) \in D $. Put $ A = f'(a, b) $ and assume $ A_x $ is invertible. Then there exist open neighborhoods $ U \subseteq \mathbb{R}^n $ of $ a $ and $ W \subseteq \mathbb{R}^m $ of $ b $ with $ U\times W\subseteq D $ such that there is a unique continuously differentiable $ g \colon W \to U $ with $ g(b) = a $, $ f(g(y), y) = 0 $, and $ g'(b) = -(A_x)^{-1} A_y $.

Problems

2023 W P8.   Define $ f \colon \mathbb{R}^3 \to \mathbb{R} $ by $ f(x, y, z) = x^2y + e^x + z $. (a) Show there exists a continuously differentiable $ \phi $ defined in a neighborhood $ U $ of $ (1, -1) $ such that $ \phi(1, -1) = 0 $ and $ f(\phi(y, z), y, z) = 0 $ for all $ (y, z) \in U $. (b) Find $ \nabla \phi(1, -1) $.

Proof. (a) Observe $ f(0, 1, -1) = 0 $ and $ \frac{\partial f}{\partial x}(0, 1, -1) = 1 $. Hence, we can apply the implicit function theorem to obtain $ U $ and $ \phi $.

(b) The implicit function theorem tells us

\[\nabla \phi(1, -1) = \phi'(1, -1)^T = - 1 \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix},\]

so $ \nabla \phi(1, -1) = (0, -1)^T $. $ \blacksquare $

2022 F P4.   (a) Let $ G \colon \mathbb{R}^3 \to \mathbb{R}^2 $ with $ G = (g_1, g_2) $ and $ G(x_0, y_0, z_0) = 0 $. When does there exist continuously differentiable $ \phi \colon I \to \mathbb{R} $ and $ \psi \colon I \to \mathbb{R} $ defined on an open interval $ x_0 \in I $ such that

\[\{(x_1, x_2, x_3) \mid G(x_1, x_2, x_3) = 0 \} = \{ (x_1, \phi(x_1), \psi(x_1)) \mid x_1 \in I \}\]

in a neighborhood of $ (x_0, y_0, z_0) $?

(b) Suppose $ f \colon \mathbb{R}^2 \to \mathbb{R} $ is continuously differentiable, that $ f(1, 1) = 1 $, and $ \frac{\partial f}{\partial x_1}(1, 1) \neq 0 $, $ \frac{\partial f}{\partial x_2}(1, 1) \neq 0 $, and $ (\frac{\partial f}{\partial x_2}(1, 1))^2 \neq 1 $. Show the system

\[f(x_3, f(x_1, x_2)) = 1, \qquad f(f(x_1, x_3), x_2) = 1\]

defines functions $ x_2 = \phi(x_1) $ and $ x_3 = \psi(x_1) $ in a neighborhood of $ 1 $ satisfying

\[f(\psi(x_1), f(x_1, \phi(x_1))) = 1, \qquad f(f(x_1, \psi(x_1)), \phi(x_1)) = 1.\]

Proof. (a) When the derivative $ A_x $ where $ x = (x_2, x_3) $ is invertible and $ G $ is continuously differentiable; i.e.

\[A_x = \begin{bmatrix} \frac{\partial g_1}{\partial x_2} & \frac{\partial g_1}{\partial x_3} \\ \frac{\partial g_2}{\partial x_2} & \frac{\partial g_2}{\partial x_3} \\ \end{bmatrix}\]

is invertible, which is equivalent to $ \det A_x\neq0 $.

(b) Set

\[G(x, y, z) = (f(z, f(x, y))-1, f(f(x, z), y)-1).\]

At $ (x,y,z)=(1,1,1) $, write

\[a=\frac{\partial f}{\partial x_1}(1,1) \qquad\text{and}\qquad b=\frac{\partial f}{\partial x_2}(1,1).\]

The chain rule gives

\[G'(1,1,1) = \begin{bmatrix} ab & b^2 & a \\ a^2 & b & ab \end{bmatrix}.\]

Hence, the derivative block in the dependent variables $ (y,z) $ satisfies

\[\begin{aligned} \det A_{(y,z)}(1,1,1) & = \det \begin{bmatrix} b^2 & a \\ b & ab \end{bmatrix} \\ & = ab(b^2-1) \\ & \neq 0. \end{aligned}\]

Thus, $ \phi, \psi $ exists by (a). $ \blacksquare $

2022 W P4.   Let $ G \subseteq \mathbb{R}^5 $ be the set of vectors $ A = (a_0, \dots, a_4) $ such that the monic quintic

\[P_A(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4+x^5\]

has five distinct real roots. Show that $ G $ is open.

Proof. Define $ f \colon \mathbb{R}^6 \to \mathbb{R} $ by $ f(A, x) = P_A(x) $; this is a $ C^1 $ function since it is polynomial. Fix an $ A_0 \in G $, and let $ \alpha_1,\dots,\alpha_5 $ be the five distinct real roots of $ P_{A_0} $. They are simple, so $ P_{A_0}'(\alpha_i)\neq0 $ for every $ i $. Hence, the implicit function theorem gives, after shrinking to a common neighborhood $ U $ of $ A_0 $, continuously differentiable functions $ \phi_i\colon U\to\mathbb{R} $ such that

\[P_A(\phi_i(A))=0\]

and $ \phi_i(A_0)=\alpha_i $. Shrinking $ U $ again, continuity and the distinctness of the $ \alpha_i $ ensure that the five functions $ \phi_i $ have pairwise distinct values. Thus, every $ A\in U $ lies in $ G $, so $ G $ is open. $ \blacksquare $

2019 F P8.   Write $ F(x, y, z) = xe^{2y} + ye^z - ze^x $ and

\[G(x, y, z) = \ln(1+x+2y+3z) + \sin(2x-y+z).\]

(a) Prove that in a neighborhood of $ 0 $ the intersection of $ F = 0 $ and $ G = 0 $ can be represented as a continuously differentiable curve parameterized by $ x $. (b) Find a tangent vector to this curve at $ 0 $.

Proof. Define $ f \colon \mathbb{R}^3 \to \mathbb{R}^2 $ by

\[f(x, y, z) = (F(x, y, z), G(x, y, z)).\]

Then $ f(0, 0, 0) = (0, 0) $ and

\[f'(0, 0, 0) = \begin{bmatrix} 1 & 1 & -1 \\ 3 & 1 & 4 \end{bmatrix}\]

Let $ B $ be the derivative with respect to the dependent variables $ (y,z) $. Then

\[\det B = \det \begin{bmatrix} 1 & -1 \\ 1 & 4 \end{bmatrix} = 5.\]

Thus, we can apply the implicit function theorem to get our curve $ \phi(x) $.

(b) We see

\[\phi'(0) = -B^{-1} \begin{bmatrix} 1 \\ 3 \end{bmatrix} = \frac{-1}{5} \begin{bmatrix} 4 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} -7/5 \\ -2/5 \end{bmatrix}.\]

We conclude that

\[(1,-7/5,-2/5)\]

is tangent to our curve at $ 0 $. $ \blacksquare $

2018 W P8.   Let $ f \colon \mathbb{R}^2 \to \mathbb{R}^2 $ be a continuously differentiable map such that $ f^{-1}(y) $ is a finite set for all $ y $. Show that the Jacobian determinant $ \det df $ cannot vanish on any open subset of $ \mathbb{R}^2 $.

Proof. Suppose $ \det df $ vanishes on a nonempty open set $ U $. If $ df $ has rank $ 1 $ at some point of $ U $, then some $ 1\times1 $ minor is nonzero there. On a sufficiently small neighborhood $ V\subseteq U $, that minor remains nonzero, while the determinant remains zero, so $ df $ has constant rank $ 1 $ on $ V $. The constant rank theorem then shows that a fiber of $ f\vert_V $ contains a one-dimensional curve, contradicting the assumption that every fiber is finite.

If no point of $ U $ has rank $ 1 $, then $ df=0 $ throughout $ U $. On any open ball $ B\subseteq U $, the mean value theorem applied along line segments shows that $ f $ is constant. The corresponding fiber contains all of $ B $, again a contradiction. Thus, $ \det df $ cannot vanish on a nonempty open set. $ \blacksquare $