Analysis Problem of the Day 86

Today’s problem appeared as Problem 4 on the UCLA Fall 2023 Analysis Qual:

Problem 4. For $f \in C^1([0,1]),$ define

$E(f) = \int_0^1 |f'(x)|^2 +|f(x)|^6-|f(x)|^4 dx.$

i) Show that $E$ is bounded below on $C^1([0,1]).$

ii) Show that for a minimizing sequence $f_n, E(f_n) \to \inf_{C^1([0,1])} E,$ there exists a uniformly convergent subsequence $f_{n_k} \to f.$

Solution. i) The only term we need to bound is $\int |f(x)|^4 dx,$ since all the other terms are non-negative. Now, recall that Hölder’s inequality implies that $\|f\|_{L^p([0,1])} \leq \|f\|_{L^q([0,1])}$ for $p \leq q.$ In particular,

$\int_0^1 |f(x)|^4 dx \leq \left(\int_0^1 |f(x)|^6\right)^{\frac{2}{3}}.$

If $\int_0^1 |f(x)|^6 dx \geq 1,$ then the above is bounded by $\int_0^1 |f(x)|^6 dx$ (since $a^{\frac23} \leq a$ for $a \geq 1$ ), and otherwise, it is bounded by $1.$ Either way,

$E(f) \geq \int_0^1 |f(x)|^6 dx - \max(1, \int_0^1 |f(x)|^6 dx) \geq -1,$

so that $E$ is bounded below.

ii) Since $E$ is bounded below, $m := \inf_{C^1([0,1])} E$ exists, and thus there exists a sequence $f_n \in C^1([0,1])$ such that $E(f_n) \to m.$ Our goal is to construct a uniformly convergent subsequence, which immediately reminds us of the Arzela-Ascoli theorem. Thus, it suffices to show $\{f_n\}$ is uniformly bounded and equicontinuous. For now, let’s suppose $\{f_n(0)\}$ is bounded. Then, by the Fundamental Theorem of Calculus and Cauchy-Schwarz,

$|f_n(t)-f_n(t')| = \left|\int_{t'}^t f'(x) dx\right| \leq \left(\int_{t'}^t |f'(x)|^2 dx \right)^{\frac12}|t-t'|^{\frac12}$

$\leq \left(\int_0^1 |f'(x)|^2dx\right)^{\frac12} |t-t'|^{\frac12}.$

Since

$\int_0^1 |f'(x)|^2 dt \leq m+\epsilon + \int_0^1 |f(x)|^4 dx - \int_0^1 |f(x)|^6 dx \leq m+\epsilon+1:=C^2$

for any $\epsilon>0$ for large enough $n,$ by the discussion above, we conclude that $|f_n(t)-f_n(t')| \leq C|t-t'|^{\frac12} <\delta$ for $|t-t'|<\sqrt{\frac{\delta}{C}}$ for all $n,$ implying that $\{f_n\}$ is uniformly equicontinuous, and in fact uniformly bounded by $\sup_n |f_n(0)|+C.$

It remains to show that $\{f_n(0)\}$ is bounded. Without loss of generality, suppose that for some subsequence, $f_{n_k}(0) \to \infty.$ Then, by the uniform equicontinuity condition, we have that $f_{n_k}(x) \geq f_{n_k}(0) - C \geq 100$ for large enough $k.$ But this implies that $E(f_{n_k}) \to \infty$ since

$\int_0^1 |f_{n_k}(x)|^6 dx - \int_0^1 |f_{n_k}(x)|^4 dx \geq C'\left(\int_0^1 |f_{n_k}(x)|^6\right)^{\frac{2}{3}} \to \infty$

as $k \to \infty,$ which contradicts the fact that $f_{n_k}$ is a minimizing sequence (since $f\ equiv 0$ clearly shows that $m \leq 0$ ). It follows that $\{f_n(0)\}$ is bounded, so $\{f_n\}$ is uniformly bounded and equicontinuous, so by the Arzela-Ascoli theorem, there exists a uniformly convergent subsequence $f_{n_k} \to f,$ as claimed.

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