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Spring 2016 - Problem 2

Lp spaces

Let $f \in L^1\p{\R}$ . Show that the series

\sum_{n=1}^\infty \frac{1}{\sqrt{n}} f\p{x - \sqrt{n}}

converges absolutely for almost all $x \in \R$ .

Solution.

Let $k \in \N$ . Since everything is non-negative in the following, we may apply Fubini-Tonelli to obtain

\begin{aligned} \int_k^{k+1} \sum_{n=1}^\infty \frac{1}{\sqrt{n}} \abs{f\p{x - \sqrt{n}}} \,\diff{x} &= \sum_{n=1}^\infty \frac{1}{\sqrt{n}} \int_k^{k+1} \abs{f\p{x - \sqrt{n}}} \,\diff{x} \\ &= \sum_{n=1}^\infty \frac{1}{\sqrt{n}} \int_k^{k+1} \abs{f\p{x - \sqrt{n}}} \,\diff{x} && \p{x \mapsto x - \sqrt{n}} \\ &= \sum_{n=1}^\infty \frac{1}{\sqrt{n}} \int_{k-\sqrt{n}}^{k+1-\sqrt{n}} \abs{f\p{x}} \,\diff{x} \\ &= \sum_{n=1}^\infty \frac{1}{\sqrt{n}} \int_\R \chi_{\set{k-\sqrt{n} \leq x \leq k+1-\sqrt{n}}} \abs{f\p{x}} \,\diff{x} \\ &= \int_\R \abs{f\p{x}} \sum_{n=1}^\infty \frac{1}{\sqrt{n}} \chi_{\set{k-x \leq \sqrt{n} \leq k+1-x}} \,\diff{x}. \end{aligned}

To complete the proof, we will estimate the sum. Notice that because $n \geq 1$ ,

\begin{aligned} k - x \leq \sqrt{n} \leq k + 1 - x &\implies \max\,\set{1, k - x} \leq \sqrt{n} \leq 1 + \max\,\set{1, k - x} \\ &\implies \p{\max\,\set{1, k - x}}^2 \leq n \leq \p{1 + \max\,\set{1, k - x}}^2. \end{aligned}

Thus, for each $x$ , there are at most

\p{1 + \max\,\set{1, k - x}}^2 - \p{\max\,\set{1, k - x}}^2 + 1 = 2\max\,\set{1, k - x} + 2

non-zero terms in the sum. Hence,

\int_k^{k+1} \sum_{n=1}^\infty \frac{1}{\sqrt{n}} \abs{f\p{x - \sqrt{n}}} \,\diff{x} \leq \int_\R \abs{f\p{x}} \frac{2\max\,\set{1, k - x} + 2}{\max\,\set{1, k - x}} \,\diff{x} \leq 4\norm{f}_{L^1},

so the sum converges absolutely for almost every $x \in \br{k, k+1}$ . Since $k$ is arbitrary, it follows that the sum converges absolutely almost everywhere.