Abstract. These are solutions to some of the Indiana University analysis qualifying problems. They are indexed by "[Year] [Semester] [Number]". I also include in each section some theorems which might be useful. I would like to thank Chia-Tz Liang and Anuvertika Pandey for pointing out errors or suggesting solutions. Please email me if you find any errors or have any solutions you would like to add. I also omitted adding a section on integration.

Table of Contents

1. Compact spaces
2. Sequences and series
3. Uniform continuity
4. Uniform convergence
5. Derivatives
6. Optimization
7. Convergence of integrals
8. Polynomials and Stone-Weierstrass
9. Stoke's theorem
10. Inverse and implicit function theorems

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 of radius $ 1/n $ at $ x_n $, i.e. $ B(x_n, 1/n) $ is not contained in any $ U_{\alpha} $. By theorem 1.2, there exists a convergent subsequence $ {x_{n_i}} $. Consequently, for $ N \gg 0 $ large, we have $ i > N $ implies $ x_{n_i} $ is contained in some $ U_{\alpha} $. We conclude that for $ i $ large, we have $ B(x_i, 1/n_i) $ is contained in $ U_{\alpha} $ as well, 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 $. Without loss of generality, there exists infinitely many $ m $ such that $ a_{1, m} = 0 $. Deleting every term with $ a_{1, m} = 1 $ yields us an infinite subsequence. Repeating this deletion process inductively gives us a convergent subsequence.

(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 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 continuity, there exists a $ \delta > 0 $ such that $ |x-y| < \delta $ implies $ |f(x) - f(y)| \leq \varepsilon $; consequently, we can assume $ |x-y| \geq \delta $. Then the Heinel-Borel theorem and 2019 F P5 show that $ K $ and $ f(K) $ are bounded, i.e. $ |f(x) - f(y)| < 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 convergences, i.e. it is a complete metric space.

Ratio Test.   Let $ \sum a_n $ be a series in $ \mathbb{R} $. Write $ L = \lim_n |a_{n+1}/a_n| $. 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 $ < 1 $ implies absolute convergence; $ r > 1 $ implies divergence; and $ r = 1 $ is inconclusive.

Integral Test.   Let $ f \colon \mathbb{R} \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 sequences in $ \mathbb{R} $. Suppose $ L = \lim a_n / b_n $ exists. 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 let $ f \colon X \to X $ be contractive. Then $ X $ has a unique fixed point $ f(x_\ast) = x_\ast $, which is given by taking n arbitrary $ x_0 \in X $, setting $ x_{n+1} = f(x_n) $, and evaluating $ \lim_n f(x_n) = x_\ast $.

Problems

2023 W P2.   Consider the series $ \sum_{a, b \geq 0} p^{-a} p^{-b} $ for fixed $ p, q $ prime. 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. We see

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

Integrating yields

\[\frac{2}{3} - \frac{2}{3n} \leq \sum_{k=1}^{n} \frac{\sqrt{k}}{n} \leq \frac{2}{3} \sqrt{n},\]

or equivalently

\[-\frac{2}{3n} \leq \sum_{k=1}^{n} \frac{\sqrt{k}}{n} - \frac{2}{3} \sqrt{n} \leq 0.\]

Letting $ n $ approach infinity 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} | < \varepsilon $, then the sequence is convergent.

Proof. Fix $ \varepsilon > 0 $. Then there exists an $ N \gg 0 $ such that $ m \geq n \geq N $ implies

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

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

\[\sum_n \frac{a_{n+1}-a_n}{a_n} = \sum_n \frac{a_{n+1}}{a_n} - 1 \geq \sum_n \log(a_{n+1}/a_n) = \lim_n a_n - a_1,\]

which diverges 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. Consider $ f(x) = \sin(\cos(x))) $ as a function $ f \colon [0, 2 \pi] \to [0, 2 \pi] $. Set $ g(x) = f(f(x)) $. Computing the derivative, we observe that $ |g'(x)| < 1 $ for each $ x \in [0, 2 \pi] $, so $ g $ is a contraction on $ [0, 2 \pi] $, and consequently $ g $ admits a unique fixed point by the Banach fixed point theorem. We conclude that our desired sequence has a unique limit independent of $ t_0 $, because $ f $ does not oscillate between 2 distinct points anywhere. $ \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. The limit is $ 0 $, but I am not sure how to show this directly. $ \blacksquare $

2021 F P8.   Is the series

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

convergent?

Proof. 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 = \frac{1}{2}(x + \frac{4}{x}) $. Then $ f'(x) = \frac{1}{2} (1 - \frac{4}{x^2}) $ is strictly less than $ 1 $ on our domain, so $ f $ is a contraction. Applying the Banach fixed point theorem to $ [1.5, \infty) $ tells us our sequence has a unique limit. 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 in $ \mathbb{R}^+ $ with $ \lim a_n = 0 $. Show there exists infinitely many $ N \in \mathbb{N} $ such that $ n \geq N $ implies $ a_n \leq a_N $.

Proof. Let

\[N = \sup\{n \mid a_n \geq a_0 \},\]

which is finite since our sequence converges to 0. Then $ n \geq N $ implies $ a_n \leq a_N $. Now delete the first $ n $ terms and repeat to get infinitely many such $ N $. $ \blacksquare $

2020 w P2.   Write $ {a_n} $ for a sequence in $ \mathbb{R}^+ $ 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 will show the partial sums converge. Fix $ \varepsilon > 0 $, and choose $ N \gg 0 $ such that $ a_n < \varepsilon/2 $ for $ n \geq N $ and

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

(This is possible by the p-series, or integral, test.) Then for each $ M > N $, we see that if $ N $ is odd

\[|\sum_{n=N}^M (-1)^{n-1} a_n| \leq \sum_{n=N/2}^{\lceil M/2 \rceil} |a_n - a_{n+1}| < \frac{\varepsilon}{2}\]

and if $ N $ is even

\[|\sum_{n=N}^M (-1)^{n-1} a_n| \leq |a_N| + |\sum_{n=N+1}^M (-1)^{n-1} a_n | < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}.\]

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

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

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

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

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

\[\lim_{n \to \infty} \ln(a_n) \leq \sum_{k=1}^{\infty} \frac{1}{n^2} \sqrt{n} \leq \sum_{k=1}^{\infty} \frac{1}{n^{3/2}},\]

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 \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) \leq \delta $ implies $ | f(x)-f(y) | < 1 $. Set $ M = \lceil 1 / \delta \rceil $. Now fix $ x \in X $, and define $ x_n = \frac{n}{M} x $ for $ 0 \leq n \leq M $. Then, using that $ d(x_{n+1}, x_n) < \delta $, we have

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

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 $. Without loss of generality, we suppose $ f'(x_n) > C $. Since $ f' $ is uniformly continuous, there exists a $ \delta > 0 $ so that $ | x-y | < \delta $ implies $ | f'(x) - f'(y) | < C/2 $. Therefore,

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

which implies $ f(x_n + \delta) \geq f(x_n - \delta) + \delta C $. 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) Suppose not. Then there exists sequences $ {x_n} $ and $ {y_n} $ in $ (0, 1] $ with $ \lim_n x_n = 0 $ (resp. $ \lim y_n = 0 $), $ f(x_n) $ monotonically increasing (resp. $ f(y_n) $ monotonically decreasing), and $ \lim_n f(x_n) = \sup f(x) $ (resp. $ \lim_n f(y_n) = \inf f(x) $). Since $ f $ is uniformly continuous, there exists a $ \delta > 0 $ such that $ |x-y| < \delta $.

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. Using the same proof technique as 2020 W P3, we are reduced to showing $ \rho $ is a bounded function on $ B $. Clearly $ \rho(x, y) \leq 2 $. $ \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

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

(This need not be a metric if $ d_Y $ is unbounded on $ f(X) $.) We say a sequence of functions $ f_n \colon X \to Y $ converges uniformly to $ f_* $ if it does with respect to $ \rho $. 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}^n $. 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}^n $ be a sequence of functions. 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}^n $ be a sequence of functions defined on a set $ X $. 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 monotonically increasing with 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. Recall that for the $ \ell^p $ norm on sequences, we have \(\lim_p \| (a_n) \|_p = \| (a_n) \|_{\infty}\). Hence,

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

uniformly. Our result then follows from theorem 4.1. $ \blacksquare $

Anuvertika's proof. Set

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

Fix $ x \in [0, 1] $; then

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

combined with $ n^{1/n} \to 1 $ implies $ a_n(x) \to f(x) $ pointwise. Since $ { a_n(x) } $ is monotonically increasing, we apply 2022 F P2 to show the convergence is uniform. $ \blacksquare $

2023 W P3.   Let $ f_n \colon [0, 1] \to \mathbb{R} $ be a sequence of function. 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 sequence $ {x_n} $ in $ [0, 1] $. Show $ {f_n} $ converges uniformly to $ f $ or provide a counterexample.

Proof. Suppose not. Then there exists sequences $ a_n \in \mathbb{N} $, $ a_n $ increasing and $ x_n \in [0, 1] $ such that

\[|f_{a_{2n}}(x_n) - f_{a_{2n+1}}(x_n) | > \varepsilon\]

for some fixed $ \varepsilon > 0 $. Since $ [0, 1] $ is compact $ x_n $ admits a convergent subsequence $ x_{n_k} $ with limit $ x_* $; let us assume without loss of generality $ x_n $ has limit $ x_* $. Define a sequence $ {y_n} $ by

\[y_{a_{2n}} = y_{a_{2n+1}} = x_n\]

and $ y_n = x_* $ elsewise. Then $ f_n(y_n) $ is not a Cauchy sequence by assumption, but $ \lim_n f_n(y_n) = f(x_*), $ a contradiction.

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

2023 W P4.   Does there exists a sequence of continuously differentiable functions that converge uniformly to a non-differentiable function?

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 $ 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.   Define $ F \colon \mathbb{R} \to \mathbb{R} $ by

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

(a) Prove that $ F $ converges uniformly on $ [1+\delta, \infty) $ for any $ \delta > 0 $ . Explain why $ f $ is continuous on $ (1, \infty) $. Is $ f $ continuous 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 Weierstrauss M-test implies $ F $ converges uniformly. $ F $ is continuous by the uniform limit theorem. $ F $ is not continuous at $ 1 $ since it approaches infinity as $ x $ approaches $ 1 $.

(b) We see that

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

Again, in order to apply the Weierstrass M-Test for $ \delta > 0 $, we observe

\[\begin{aligned} F'(1+\delta) & = -\sum_{n=1}^{\infty} \ln(n) \frac{1}{n^{-(1+\delta)}} \\ & \leq - \int_1^{\infty} \ln(x) x^{-(1+\delta)} dx; \end{aligned}\]

setting $ u = \ln(x) $ and $ du = x^{-1} $,

\[\begin{aligned} F'(1+\delta) & = \int_0^{\infty} e^{-\delta u} du < \infty. \end{aligned}\]

We conclude that $ F'(x) $ converges uniformly, and that it is indeed the derivative of $ F $ by theorem 4.2. $ \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

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

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 \in \mathbb{R} $ such that

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

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 $ with loss of generality, we see for each $ n \in \mathbb{N} $

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

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

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

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

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

Each $ f_n $ is compact (its image contains finitely many points). Although, the image of $ x_n = 2^{-n} $ by $ f $ is not compact: it contains $ 1 $ and $ 2^{-n} $ for each $ n $, but not $ 0 $. $ \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} $. $ \blacksquare $

Proof. Since $ f $ is periodic we can consider it as a bounded and uniformly continuous function on $ [0, 1] $. Let $ \varepsilon > 0 $, and choose an $ n \gg 0 $ such that $ | x-y | < 2^{-n} $ implies $ |f(x) - f(y)| < \varepsilon $. We observe that we can write $ f_n(x) $ as a sum of $ 2^n $ functions of the form $ f(x+a_i) $. In particular, we can write $ f_{n+k}(x) $ as a sum of $ 2^{n+k} $ functions of the form $ f(x+b_i) $, such that for each $ a_i $ the nearest $ 2^k $ of the $ b_i $ satisfy

\[\|a_i - b_i\| \leq \sum_{j=1}^{k} \frac{1}{2^{-n-j}} < \sum_{j=1}^{\infty} \frac{1}{2^{-n-j}} = 2^{-n}.\]

Consequently,

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

We conclude that $ {f_n} $ is Cauchy with respect to the uniform metric, and thus 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] $, chose an $ N_x \gg 0 $ such that $ g_n(x) < \varepsilon R $ for all $ n > N_x $. Since our $ g_n $ are continuous, for each $ x $ there exisrts 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_0}, \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

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

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 $,

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

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) Consider bump functions of radius $ 2/n $ and equal volume:

\[f_n(x) = n \exp \left( - \frac{1}{1-(nx-1)^2} \right).\]

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

(b) It is not hard to see that if the left limit is nonzero and the right limit is $ 0 $, then for each $ n $ there exists a set $ A_n $ of measure greater than some fixed constant $ C $, i.e. $ \mu(A_n) > C $, such that $ |f_n(A_n)| > D $ for some fixed constant $ D $. This contradicts pointwise convergence. $ \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. Suppose not. Then $ f'(x) < C $ for some $ C > 0 $ for all $ x $. Hence, $ f(x) \leq - Cx + f(0 ) $ implies $ \displaystyle \lim_{x \to \infty} f(x) = -\infty $, a contradiction. $ \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

\[f(r \cos \theta, r \sin \theta) = r^2 \cos^3 \theta,\]

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

(b) Set $ v = (a, b) $. 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^2 \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,

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

so $ f $ is differentiable $ (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} f(x) > 1 $, and hence $ \ln f(x) > x $ for $ x \gg 0 $. Consequently, $ f(x) > e^x $ for $ x \gg 0 $, yielding $ \displaystyle \lim_{x \to \infty} f(x) e^{-k} = \infty $ for $ 0 < k < 1 $. We see our least upper bound on such $ k $ is $ 1 $. $ \blacksquare $

2020 F P7.   Let $ f \colon \mathbb{R}^2 \to \mathbb{R}^2 $ be a differentiable map with $ f = 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

\[Q(a,b) \geq 1.5(a^2 + b^2) + \left( \frac{\partial f_1}{\partial x_2} + \frac{\partial f_2}{\partial x_1} \right).\]

In order to minimize the $ ab $ term we assume $ ab \geq 0 $ and set $ \frac{\partial f_1}{\partial x_2} + \frac{\partial f_2}{\partial x_1} = -1 $. Thus,

\[Q(a, b) \geq 1.5(a^2 + b^2) - ab.\]

If $ a^2 \geq b^2 $,

\[Q(a, b) \geq 1.5(a^2 + b^2) - a^2 = 0.5 a^2 + 1.5 b^2.\]

Thus, $ 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) Set $ v = (a, b) $. 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}) $ 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 $ | 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 $ \displaystyle_{x \to 0} f(x) $ does not exist.

(b) By definition $ f'(x) = g(x) $. Let $ \varepsilon > 0 $ and choose $ \delta > 0 $ as stated in the problem. Then for any $ x, y \in \mathbb{R} $ with $ | 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} \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. We proceed by minimizing in each variable independently. Fix $ y = 1 $. Then

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

if and only if

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

Squaring both sides reduces the equation to $ (1+x)^2 = (1-x)^2 $, and hence our minimum is achieved when $ x = 0 $. 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} $. 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. Since the projection $ \pi_1 \colon \mathbb{R}^2 \to \mathbb{R} $ is continuously differentiable, we can reduce ourselves to the case $ f_n \colon \mathbb{R} \to \mathbb{R} $. Set

\[f_n(x) = -x^2 + B_1(x, n) + B_2(x, n),\]

where $ B_1(x, n) $ and $ B_2(x, n) $ are bump functions of height $ 1 $ from $ -1/n $ to $ 0 $ and from $ 0 $ to $ 1/n $, respectively. Then $ {f_n} $ converges to $ -x^2 $ pointwise and $ 0 $ is a local minimum for each $ n $ but not of the limit. $ \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

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

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 $. Then

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

Substitute $ y = u(p-1) $ and $ dy = (p-1) du $ we get

\[I \leq \frac{1}{(p-1)^2} \int_{a}^{\infty} y e^{-y} dy,\]

where $ a = \ln(3) (p-1) $. Integrating by parts shows

\[I \leq \frac{1}{(p-1)^2} [-e^{-y} y - e^{-y}]_{a}^{\infty} = \frac{1}{(p-1)^2} e^{-a}(a-1).\]

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

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

Hence, for each $ n \geq 0 $ we have

\[\begin{aligned} \int_{n \pi - \pi/2}^{2 \pi + \pi/2} \sqrt{u} \cos(u) du & \geq \sqrt{n \pi - \pi/2} \int_{n \pi - \pi/2}^{n \pi+\pi/2} \cos(u) \\ & = \sqrt{n \pi - n/2}(\sin(\pi/2) - \sin(-\pi/2)) \\ & = 2 \sqrt{n \pi - \pi/2}. \end{aligned}\]

We conclude that the 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. Without loss of generality, assume $ f $ is supported on $ D^2 $ and $ x = 0 $. Since $ f $ is compactly supported, there exists an $ M $ such that $ \sup | f | \leq M $. Thus,

\[g(0) = \int_{D^2} \frac{f(y)}{\|y\|} dy \leq M \int_{D^2} \frac{1}{\|y\|} dy.\]

Substituting polar coordinates yields

\[g(0) \leq M \int_0^{2 \pi} \int_0^1 dr d \theta = 2 \pi M.\]

Thus, $ g $ converges. (I have not shown continuity.) $ \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. We apply the Fubini-Tonelli theorem as follows. For each $ z \in (0, \infty) $, we have

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

Observing

\[\begin{aligned} \lim_{z \to \infty} \int_0^1 \int_z^{\infty} f(x, y) dy dx & \leq \lim_{z \to \infty} \int_0^1 \left\| \int_z^{\infty} f(x, y) dy \right\| dx \\ & \leq \lim_{z \to \infty} \frac{C}{\log(2+z)} \\ & = 0 \end{aligned}\]

we get our desired inequality. $ \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-q} \leq |x| \leq a_k $ . Show that $ \int_{\mathbb{R}^2} f(x) dA $ exists.

Proof. Indeed,

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

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. The following proof holds if $ f $ is $ C^1 $. Performing a polar substitution

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

yields

\[\begin{aligned} I & = \int_0^{2 \pi} \int_0^1 \frac{f(r \cos \theta, r \sin \theta)}{r^{5/3}} dr d \theta \\ & \int_0^{2 \pi} \int_0^1 \frac{g(r, \theta) - g(0, 0)}{r} \frac{1}{r^{2/3}} dr d \theta \\ \end{aligned}\]

Since $ f $ is continuously differentiable and our domain is compact we see $ f' $ is bounded by som $ M $, and so $ g' $ is bounded by some $ N $. Fixing $ \theta $, the mean value theorem implies that for each $ r $ there exists an $ r_0 $ such that

\[\left\| \frac{g(r, \theta) - g(0, 0)}{r} \right\| = \left\| g'(r_0, \theta) \right\| \leq N\]

where the derivative is taken in the first variable. Hence,

\[\begin{aligned} I & \leq \int_0^{2 \pi} \int_0^1 \left\| \frac{g(r, \theta) - g(0, 0)}{r} \right\| \frac{1}{r^{2/3}} dr d \theta \\ & \leq N \leq \int_0^{2 \pi} \int_0^1 \frac{1}{r^{2/3}} dr d \theta \\ & 6 \pi N. \end{aligned}\]

Thus, our integral converges. $ \blacksquare $

8. Polynomials and Stone Weierstrass

Stone-Weierstrass 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 not. Then there exists a point $ x = (x_1, \dots, x_n) \in \mathbb{R} $ with $ \det f'(x) \neq 0 $. Since $ \det f'(x) \neq 0 $ and the determinant and derivatives are continuous, we can assume $ f_j(x) \neq 0 $. The identity $ p(f) = 0 $ implies $ d [p(f)] = 0 $, or equivalently $ p'(f) f' = 0 $. Expanding this equation, we see that for each $ 1 \leq j \leq n $ we have

\[(2j+1) f_j^{2j}(x_1, \dots, x_n) \left( \sum_{k=1}^{n} \partial_j f_k \right) = 0.\]

Since $ f_j(x) \neq 0 $ for each $ j $, we have $ \sum_k \partial_j f_k = 0 $. This implies the rows of $ f'(x) $ sum to $ 0 $, and thus are linearly dependent, meaning $ \det f(x) = 0 $. This contradicts the assumption that $ \det f'(x) \neq 0 $. $ \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 Stone-Weierstrass 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

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

Each integral on the left hand side is $ 0 $, and hence the right hand side is too. This can only happen when $ f(x) = 0 $. $ \blacksquare $

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

Proof. Write

\[f(x, y) = \sum_{n=0}^{\infty} c_n(y) x^n\]

and

\[Y_n = \{ y_0 \in \mathbb{R} \mid \deg f(x, y_0) = n \}.\]

Since $ \bigcup_n Y_n = \mathbb{R} $ and $ \mathbb{R} $ is uncountable while our union is indexed by a countable set, we see that for some $ N $ that $ Y_N $ is infinite. We denote the restriction of $ f $ to $ \mathbb{R} \times Y_N $ by $ Y_N $, where

\[Y_n = \sum_{n=0}^{N} c_n(y) x^n.\]

Choosing $ N $ distinct reals $ x_0, \dots, x_N \in \mathbb{R} $ allows us to solve for the $ c_n $ on this subset:

\[c_n(y) = \sum_{n=0}^{N} a_{j, n} f_N(x_j, y).\]

Using that $ f_N $ is polynomial for each fixed $ x_j $ we see that

\[f_N(x, y) = \sum_{n=0}^{N} a_{j, n} f_N(x_j, y) x^n\]

is polynomial in both variables. Finally, for each $ x \in \mathbb{R} $ we observe

\[f(x, y) - \sum_{n=0}^{N} a_{j, n} f(x_j, y) x^n\]

is $ 0 $ for infinitely many $ y $, so

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

Thus, it is 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. The second statement follows from the Stone-Weierstrass theorem in a manner similar to 2020 F P8. For the first statement, suppose $ g(x) \neq 0 $ and choose $ f(x) $ such that $ g(x) f(x) = M $. Then

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

implies $ | f(x) | \geq 1 $ and $ |g(x)| \leq M $. $ \blacksquare $

9. Stoke's theorem

Stoke's 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

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

where $ n $ is the unit normal.

Divergence theorem.   Let $ M $ be a smooth manifold in $ \mathbb{R}^n $ with boundary $ \partial M $. Then

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

where $ n $ is the 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^z) $,

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

and $ n $ is the 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 $. 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 see $ n = (1, 1, 1)/\sqrt{3} $. Hence,

\[\int_C z dx - 2xdy + 3y dz = \int \int_R 2 dA,\]

and thus our integral is $ 2 \text{Vol}(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

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

(a) Apply Stoke's theorem on

\[S = \{(\cos(t), \sin(t), z) \mid t \in [-\pi, \pi], 0 \leq z \leq 4t \sin(t) \cos(t^3)\}\]

to express $ \int_{\gamma} (F \cdot n) dA $ in terms of $ \int_{\alpha} (F \cdot n) dA$. (b) Use (a) to evaluate the first integral.

Proof. (a) Stoke's theorem tells us

\[\int_S (\text{Curl} \cdot n) dS = \int_{\alpha} (F \cdot n) dA + \int_{\gamma} (F \cdot n) dA.\]

Here, $ \text{Curl} F = (2x, 2y, -4z) $ and our unit normal is $ (\cos(t), \sin(t), 0) $. Therefore

\[\begin{aligned} \int_S (\text{Curl} F \cdot n) dS & = \int_{-\pi}^{\pi} \int_0^{4 \sin(t) \cos(t^3)} 2 \cos^2(t) + 2 \sin^2(t) dz dt \\ & = 2 \int_{- \pi}^{\pi} 4t \sin(t) \cos(t^3) dt \\ & = 16 \pi. \end{aligned}\]

Thus,

\[16 \pi = \int_{\alpha} (F \cdot n) dA + \int_{\gamma} (F \cdot n) dA.\]

(b) We see

\[\begin{aligned} \int_{\alpha} (F \cdot n) dA & = \int_{-\pi}^{\pi} F(\alpha(t)) \cdot n(t) \\ & = \int_{-\pi}^{\pi} (0, 0, 4 \cos(t) \sin(t^3)) \cdot (-\cos(t), -\sin(t), 0) dt \\ & = 0. \end{aligned}\]

We conclude that $ \int_{\gamma} (F \cdot n) dA = 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 unit normal with positive component, 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 $ \mathbb{R}^n \to \mathbb{R}^n $ be a continuously differentiable function. Then $ \det f'(x) \neq 0 $, i.e. $ f'(x) $ invertible as a linear transformation, if and only if $ f $ is invertible in a neighborhood of $ x $.

Implicit function theorem.   Let $ f \colon D \subseteq \mathbb{R}^{n+m} \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 exists open subsets $ U \subseteq D $ and $ W \subseteq \mathbb{R}^m $ with $ (a, b) \in U $ and $ b \in W $, such that there is a unique continuously differentiable $ g \colon W \to \mathbb{R}^n $ with $ g(b) = a $, $ f(g(y), y) = 0 $, and $ g'(b) = -(A_x)^{-1} A_y $.

Problems

2023 W P8.   Define $ f^3 \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 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 our $ 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

\[f(\psi(x_1), f(x_1, \phi(x_1))) = 1, \qquad f(f(x_1, \phi(x_2)), \phi(x_1)) = 1.\]

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 its derivative being nonzero.

(b) Set

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

Then

\[G' = \begin{bmatrix} \frac{\partial f}{\partial x_1} \frac{\partial f}{\partial x_2} & \left(\frac{\partial f}{\partial x_2} \right)^2 & \frac{\partial f}{\partial x_1} \\ \left(\frac{\partial f}{\partial x_1} \right)^2 & \frac{\partial f}{\partial x_2} & \frac{\partial f}{\partial x_1} \frac{\partial f}{\partial x_2} \end{bmatrix}.\]

Hence,

\[\begin{aligned} \det A_x(1, 1) & = \det \begin{bmatrix} \left(\frac{\partial f}{\partial x_2}(1, 1) \right)^2 & \frac{\partial f}{\partial x_1}(1, 1) \\ \frac{\partial f}{\partial x_2}(1, 1) & \frac{\partial f}{\partial x_1}(1, 1) \frac{\partial f}{\partial x_2}(1, 1) \end{bmatrix} \\ & = \frac{\partial f}{\partial x_1}(1, 1) \frac{\partial f}{\partial x_2}(1, 1) \left [ \left( \frac{\partial f}{\partial x_2}(1, 1)\right)^2 - 1 \right] \\ & \neq 0. \end{aligned}\]

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

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

has $ 5 $ distinct 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 $ be root of $ P_{A_0} $. Since our roots are distinct we have $ P_{A_0}'(\alpha) \neq 0 $. Hence, we can apply the implicit function theorem to find a unique function $ \phi_{\alpha} \colon U \to \mathbb{R} $, where $ U $ is a neighborhood of $ A_0 $, such that $ f(x, \phi_{\alpha}(x)) = 0 $. Since our $ \phi_{\alpha} $ are continuously differentiable, there exists a neighborhood of $ A_0 $ such that they do not intersect, i.e. $ P_A $ has distinct roots. $ \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

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

so

\[\det A_x = \det \begin{bmatrix} 1 & -1 \\ 1 & 4 \end{bmatrix} = 6.\]

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

(b) We see

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

We conclude that $ (-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 determinant $ \det df $ of the Jacobi matrix of $ f $ cannot vanish on any open subset of $ \mathbb{R}^2 $.

Proof. Suppose $ \det df $ vanishes on $ U $. Then the inverse function theorem tells us the image $ f(U) $ cannot be locally isomorphic to $ \mathbb{R}^2 $, else it would be invertible in a neighborhood, and hence have nonvanishing determinant. Thus, the image of $ f $ is locally isomorphic to $ \mathbb{R} $. The kernel of a surjective map from a 2-manifold to a 1-manifold is a 1-manifold, and hence $ f $ has infinite kernel.

Note: It is also a well known fact that there exists no continuous bijections from $ \mathbb{R}^2 \to \mathbb{R} $ (this requires the Baire category theorem), and hence our map has infinite kernel. $ \blacksquare $