Analysis Problem of the Day 71

Today’s problem appeared as Problem 2 on UCLA’s Spring 2023 Analysis Qual:

Problem 2. Let $C$ be the Banach space of continuous functions on $[0,1]$ with the supremum norm, and let $L$ be the Banach space of Lipschitz functions with norm

$\|f\|_L :=|f(0)|+\sup_{x \not =y, x,y \in [0,1]} \frac{|f(x)-f(y)|}{|x-y|}.$

Let $B_C$ and $B_L$ the closed unit balls in each space, respectively.

a) Show that $B_L$ is closed in $C$ in the norm topology.

b) Let

$\Phi(f) := \int_0^1 f(x)^4 dx - \left(\int_0^1 f(x)dx\right)^4.$

Prove that $\Phi$ attains its maximum on $B_L,$ i.e. for some $g \in B_L,$ $\sup_{f \in B_L} \Phi(f)=\Phi(g).$

c) Show that $\Phi$ does not attain its maximum on $B_C.$

Solution: a) Let $u_n \in B_L, u_n \to u$ in $C([0,1]),$ and for sake of contradiction, suppose that $\|u\|_L > 1+\epsilon$ for some $\epsilon>0,$ i.e. for some $x_0,y_0 \in [0,1],$ $|u(0)|+\frac{|u(x_0)-u(y_0)|}{|x_0-y_0|} > 1+\epsilon.$ But since $u_n \to u$ uniformly, one has that $|u_n(0)|+\frac{|u_n(x_0)-u_n(y_0)|}{|x_0-y_0|} \to |u(0)|+\frac{|u(x_0)-u(y_0)|}{|x_0-y_0|}$ as $n \to \infty,$ and the left hand side is bounded above by $1$ by assumption. This is a contradiction. Thus, $\|u\|_L \leq 1,$ i.e. $B_L$ is closed in $C.$

b) Note that $B_L$ is contained in the set of 1-Lipschitz functions with $|f(0)| \leq 1,$ for which therefore $\sup_{[0,1]} |f| \leq |f(0)|+|x| \leq 1+1=2.$ Thus, $\Phi(f) \leq \int_0^1 f(x)^4 dx \leq 2^4 = 16$ on $B_L,$ so $\Phi$ is bounded above on $B_L$ and the supremum of $\Phi$ on $B_L$ thus exists (in fact, note that $\Phi$ is always non-negative by Jensen’s formula). Let $u_n \in B_L$ be such that $\Phi(u_n) \to \sup_{u \in B_L} \Phi(u).$ Our goal is to extract some kind of convergence from a subsequence $u_{n_k}$ to ultimately show $u_{n_k} \to u,$ where $u \in B_L$ is the function where $\Phi$ achieves its desired maximum. Since we are dealing with continuous functions, we are immediately reminded of the compactness criterion for continuous functions, namely, Arzela-Ascoli’s theorem. Indeed, functions in $B_L$ are uniformly bounded (since $\|f\|_\infty \leq 2$ for $f \in B_L$ ) and uniformly equicontinuous (since $|f(x)-f(y)| \leq |x-y|<\epsilon$ for $|x-y|<\delta=\epsilon$ for all $f \in B_L$ ). Thus, there does exist a uniformly convergent subsequence $u_{n_k} \to u,$ and since $u_{n_k} \in B_L,$ by (a) one has $u \in B_L.$ Finally, it is clear that if $u_n \to u$ uniformly, then $\Phi(u_n) \to \Phi(u),$ so $\Phi(u_{n_k}) \to \Phi(u)=\sup_{f \in B_L} \Phi(f),$ so that the maximum of $\Phi$ is achieved on $B_L.$

c) Note that for $\|f\|_\infty \leq 1,$ $\Phi(f) \leq 1.$ We now show that $\Phi$ gets arbitrarily close to $1$ on $B_C$ but never exactly attains it. Indeed, for $f_n \in C$ defined as

$f_n(x)=1, \quad x \in \left[0,\frac{1}{2}-\frac{1}{2n}\right],$

$f_n(x)=-n(x-\frac12), \quad x \in \left[\frac{1}{2}-\frac{1}{2n},\frac{1}{2}+\frac{1}{2n}\right]$

$f_n(x)=-1, \quad x \in \left[\frac{1}{2}+\frac{1}{2n},1\right],$

by symmetry one has that $\int_0^1 f(x)dx=0,$ yet $f_n^4 \to 1$ pointwise a.e. and is uniformly bounded by 1, so by Dominated Convergence Theorem, $\Phi(f_n) = \int_0^1 f_n(x)^4 dx \to 1.$ However, if $\Phi(f)=1,$ then $\int_0^1 f(x)=0$ but $\int_0^1 f^4(x)dx=1.$ This is impossible for $\|f\|_\infty \leq 1$ and $f$ continuous, since $\int_0^1 f^4(x)dx =1$ requires $f \equiv 1$ or $f \equiv -1,$ neither of which satisfy $\int_0^1 f(x)dx=0.$ Thus, $\Phi$ achieves its maximum on $B_L$ but not $B_C.$

Remark: Note that $|f(0)|+\frac{|f(x)|}{|x|} \leq 1$ implies $|f(x)|^4 \leq (|x|(1-|f(0)|)^4,$ i.e. $\int_0^1 |f(x)|^4 dx \leq \frac{(1-|f(0)|)^4}{5} \leq \frac15.$ It is then easy to check that $f(x) = x-\frac12$ achieves the desired maximum of $\Phi(f)=\frac15$ on $B_L$ if $f$ is evaluated at $x=\frac12$ instead of $x=0$ in the Lipschitz norm.

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