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Spring 2020 - Problem 2

Helly's selection theorem, Lp spaces

Assume $\func{f_n}{\R}{\R}$ is a sequence of differentiable functions satisfying

\int_\R \abs{f_n\p{x}} \,\diff{x} \leq 1 \quad\text{and}\quad \int_\R \abs{f_n'\p{x}} \,\diff{x} \leq 1.

Assume also that for any $\epsilon > 0$ there exists $R\p{\epsilon} > 0$ such that

\sup_n \int_{\set{\abs{x} \geq R\p{\epsilon}}} \abs{f_n\p{x}} \,\diff{x} < \epsilon.

Show that there exists a subsequence of $\set{f_n}_n$ that converges in $L^1\p{\R}$ .

Solution.

Observe that $\abs{f_n\p{x}} \leq 1$ everywhere for every $n \geq 1$ . Indeed, because $f_n' \in L^1\p{\R}$ , we may apply the fundamental theorem of calculus to get for any $\br{a, b}$ that

\begin{aligned} V_a^b\p{f_n} &= \sup_{a \leq t_1 < \cdots < t_n \leq b} \sum_{k=1}^{m-1} \abs{f_n\p{t_{k+1}} - f_n\p{t_k}} \\ &\leq \sup_{0 \leq t_1 < \cdots < t_n \leq 1} \sum_{k=1}^{m-1} \int_{t_k}^{t_{k+1}} \abs{f_n'\p{t}} \,\diff{t} \\ &\leq \norm{f_n'}_{L^1} \\ &\leq 1. \end{aligned}

Hence, if $f_n\p{x} = 1 + \epsilon$ for some $\epsilon > 0$ , then

\abs{f_n\p{x} - f_n\p{y}} \leq V_x^y\p{f_n} \leq 1 \implies f_n\p{y} \geq \epsilon > 0

for any $y \in \R$ , which contradicts integrability of $f_n$ . Let $\phi \in C_{\mathrm{c}}^\infty\p{\R}$ be the standard mollifier, i.e.,

\phi\p{x} = \begin{cases} C\exp\p{\frac{1}{\abs{x}^2 - 1}} & \text{if } x < 1, \\ 0 & \text{otherwise}, \end{cases}

where $C$ is chosen so that $\int_\R \phi\p{x} \,\diff{x} = 1$ , and set $\phi_\epsilon\p{x} = \frac{1}{\epsilon} \phi\p{\frac{x}{\epsilon}}$ . Observe that

\begin{aligned} \abs{\p{\phi_\epsilon * f_n}\p{x}} &\leq \int_\R \abs{f_n\p{x - y}\phi_\epsilon\p{y}} \,\diff{y} \\ &\leq \int_\R \abs{f_n\p{x - \epsilon y}\phi\p{y}} \,\diff{y} && \p{y \mapsto \frac{y}{\epsilon}} \\ &\leq \norm{f_n}_{L^1} \norm{\phi}_{L^\infty} \\ &\leq \norm{\phi}_{L^\infty}, \end{aligned}

so the $\phi_\epsilon * f_n$ are uniformly bounded. Similarly,

\begin{aligned} \abs{\p{\phi_\epsilon * f_n}'\p{x}} &\leq \int_\R \abs{f_n\p{x - y} \phi_\epsilon'\p{y}} \,\diff{y} \\ &= \frac{1}{\epsilon} \int_\R \abs{f_n\p{x - \epsilon y}\phi'\p{y}} \,\diff{y} && \p{y \mapsto \frac{y}{\epsilon}} \\ &\leq \frac{\norm{f_n}_{L^1} \norm{\phi'}_{L^\infty}}{\epsilon} \\ &\leq \frac{\norm{\phi'}_{L^\infty}}{\epsilon}, \end{aligned}

so the $\phi_\epsilon * f_n$ are also equicontinuous for each fixed $\epsilon > 0$ . Thus, for every $m \geq 1$ , Arzelà-Ascoli gives a subsequence $\set{\phi_{1/m} * f_{n_k}}_k$ which converges locally uniformly to some $g_m$ . Now fix $\epsilon > 0$ . By assumption, there exists $R\p{\epsilon} > 0$ such that

\sup_n \int_{\set{\abs{x} > R\p{\epsilon}}} \abs{f_n\p{x}} \,\diff{x} < \epsilon.

Also,

\begin{aligned} \norm{\phi_{1/m} * f_{n_k} - f_{n_k}}_{L^1} &\leq \int_\R \int_\R \abs{f_{n_k}\p{x - y} - f_{n_k}\p{x}}\phi_{1/m}\p{y} \,\diff{y} \,\diff{x} \\ &= \int_\R \int_\R \abs{f_{n_k}\p{x - \frac{y}{m}} - f_{n_k}\p{x}}\phi\p{y} \,\diff{y} \,\diff{x} \\ &= \int_\R \norm{f_{n_k}\p{\cdot - \frac{y}{m}} - f_{n_k}}_{L^1} \phi\p{y} \,\diff{y} && \p{\text{Fubini-Tonelli}}. \end{aligned}

Since translation is continuous in $L^1\p{\R}$ , it follows that the integrand is continuous, and it is bounded by $2\norm{f_{n_k}}_{L^1} \leq 2$ . Hence, by dominated convergence, we get

\lim_{m\to\infty} \norm{\phi_{1/m} * f_{n_k} - f_{n_k}}_{L^1} = 0.

Hence,

\begin{aligned} \norm{f_{n_k} - f_{n_\ell}}_{L^1} &\leq \norm{f_{n_k} - f_{n_\ell}}_{L^1\p{\br{-R\p{\epsilon}, R\epsilon}}} + 2\epsilon \\ &\leq \norm{f_{n_k} - \phi_{1/m} * f_{n_k}}_{L^1} + \norm{f_{n_\ell} - \phi_{1/m} * f_{n_\ell}}_{L^1} + \norm{\phi_{1/m} * f_{n_k} - g_m}_{L^1\p{\br{-R\p{\epsilon}, R\p{\epsilon}}}} + \norm{\phi_{1/m} * f_{n_\ell} - g_m}_{L^1\p{\br{-R\p{\epsilon}, R\p{\epsilon}}}} + 2\epsilon. \end{aligned}

By $L^1$ convergence, if $m$ is large enough, then

\norm{f_{n_k} - \phi_{1/m} * f_{n_k}}_{L^1} + \norm{f_{n_\ell} - \phi_{1/m} * f_{n_\ell}}_{L^1} \leq \epsilon.

Since $\br{-R\p{\epsilon}, R\p{\epsilon}}$ is compact, we have for large $k, \ell$ that

\norm{\phi_{1/m} * f_{n_k} - g_m}_{L^1\p{\br{-R\p{\epsilon}, R\p{\epsilon}}}} + \norm{\phi_{1/m} * f_{n_\ell} - g_m}_{L^1\p{\br{-R\p{\epsilon}, R\p{\epsilon}}}} \leq \epsilon.

Overall, we get

\norm{f_{n_k} - f_{n_\ell}}_{L^1} \leq 4\epsilon,

for large $k, \ell$ (and how large only depends on $\epsilon$ , e.g., it does not depend on $m$ ).