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Spring 2021 - Problem 4

Arzelà-Ascoli

Define

I\p{f} = \int_0^1 \p{\frac{1}{2} \p{f'\p{x}}^2 + \sin\p{f\p{x}} + f^4\p{x}} \,\diff{x}

for any $f \in C^1\p{\br{0,1}; \R}$ . Let $f_n \in C^1\p{\br{0,1}; \R}$ be such that

I\p{f_n} \to \int_{f \in C^1\p{\br{0,1}; \R}} I\p{f}.

Show that the sequence $\set{f_n}$ has a limit point in the space $C\p{\br{0,1}; \R}$ .

Solution.

First, observe that $\set{I\p{f_n}}_n$ is bounded by some $M > 0$ , and so

\int_0^1 \frac{1}{2}\p{f'\p{x}}^2 - 1 + f^4\p{x} \,\diff{x} \leq I\p{f} \leq M \implies \frac{1}{2} \norm{f'}_{L^2}^2 + \norm{f}_{L^4}^4 \leq M + 1,

so the $\set{f_n'}_n$ are uniformly bounded in $L^2\p{\br{0,1}}$ and the $\set{f_n}_n$ are uniformly bounded in $L^4\p{\br{0,1}}$ . Hence, given any $x, y \in \br{0, 1}$ , we have

\abs{f_k\p{x} - f_k\p{y}} \leq \int_{\br{x,y}} \abs{f_k'\p{t}} \,\diff{t} \leq \norm{f_k'}_{L^2} \abs{x - y}^{1/2} \leq \p{\sup_n \,\norm{f_n'}_{L^2}} \abs{x - y}^{1/2},

and the constant is independent of $n$ by our first estimate. In particular, $\set{f_n}_n$ is equicontinuous. Similarly, by Hölder's inequality,

\norm{f_k}_{L^1} \leq \norm{f_k}_{L^4} \leq \sup_n \,\norm{f_n}_{L^4} \quad\text{and}\quad \norm{f_k'}_{L^1} \leq \norm{f_k'}_{L^2} \leq \sup_n \,\norm{f_n'}_{L^2}.

Hence,

\begin{gathered} \abs{f_k\p{x}} \leq \abs{f_k\p{y}} + \int_{\br{x,y}} \abs{f_k'\p{t}} \,\diff{t} \\ \implies \abs{f_k\p{x}} \leq \norm{f_k}_{L^1} + \norm{f_k'}_{L^1} \leq \sup_n \,\p{\norm{f_n}_{L^1} + \norm{f_n'}_{L^1}} \end{gathered}

by integrating in $y$ . Thus, the $\set{f_n}_n$ are pointwise bounded, so because $\br{0, 1}$ is compact, we may apply Arzelà-Ascoli. Thus, $\set{f_n}_n$ admits a limit point in $C\p{\br{0,1}; \R}$ , which was what we wanted to show.