$f\in L^{2} \implies \int_{n}^{n+1} f(x) dx \to 0$?

Question

Let $f\in L^{2}(\mathbb R)$, that is, $\int_{\mathbb R} |f(x)|^{2} dx < \infty.$

My Question: (1)Can we say $\int_{n}^{n+1} f(x) dx \to 0$ as $n\to \infty$? (2) Let $\phi \in L^{2}(\mathbb R).$ Can we say $\int_{\mathbb R } \phi(x-x_n)f(x) dx \to 0$ as $x_n\to \infty$?

Answer 1

Note that $\int |f|^2 = \sum_n \int_n^{n+1} |f|^2 $ so $\int_n^{n+1} |f|^2 \to 0$.

Cauchy Schwarz gives $\int_n^{n+1} |f| \le \sqrt{\int_n^{n+1} |f|^2}$, hence the result.

This is similar in spirit to Joey's answer:

Choose $\epsilon>0$ and let $\tilde{\phi}, \tilde{f}$ be compactly supported function such that $\|\tilde{\phi} \|_2 \le \| \phi \|_2 $, $\|\tilde{f} \|_2 \le \| f \|_2 $, $\|\tilde{\phi} - \phi \|_2 < \epsilon$, $\|\tilde{f} - f \|_2 < \epsilon$.

Let $(T_y g)(x) = g(x-y)$, and $(a \square b)(y) = \int (T_y a) (x) b(x) dx$

Choose $N$ such that the support of $T_{x_n}\tilde{\phi}$ and $\tilde{f}$ are disjoint for $n \ge N$. Then \begin{eqnarray} \|T_{x_n} {\phi} \square {f} \|_1 &\le& \| T_{x_n} \tilde{\phi} \square \tilde{f} \|_1 + \| T_{x_n} \phi \square f - T_{x_n} \tilde{\phi} \square \tilde{f} \|_1 \\ &\le& \| T_{x_n} \phi \square f - T_{x_n} \tilde{\phi} \square {f} \|_1 + \| T_{x_n} \tilde{\phi} \square f - T_{x_n} \tilde{\phi} \square \tilde{f} \|_1 \\ &\le & \| T_{x_n} \phi - T_{x_n} \tilde{\phi} \|_2 \| f\|_2 + \| T_{x_n} \tilde{\phi} \|_2 \|f-\tilde{f}\|_2 \\ &=& \|\phi - \tilde{\phi} \|_2 \| f\|_2 + \| \tilde{\phi} \|_2 \|f-\tilde{f}\|_2 \\ &\le& (\|f\|_2 + \|\phi\|_2) \epsilon \end{eqnarray}

Answer 2

(1) Yes, we can. Denote $a_n$ as $\int_n^{n+1}f(x)\mathrm{d}x$. If $\{a_n\}$ doesn't converge to 0, then there's an $\epsilon>0$ such that there are infinite entries in the sequence satisfying $|a_n|\geq \epsilon$ (which implies $|a_n|^2\geq \epsilon$). Thus $f \notin \mathrm{L}^2(\mathbb{R})$, a contradiction.

(2) I think it can be solved by using Cauchy-Schwarz inequality, since $\phi(x-x_n) \in \mathrm{L}^2(\mathbb{R})$.

EDIT: Wow, someone already answered while I was typing :D

Answer 3

For (2), we first note that for any $f\in L^2(\mathbb{R})$ we have $$\lim\limits_{y\rightarrow\infty}{\int\limits_{|x|\ge y}{|f|^2\text{ d}x}} = 0.$$ This follows by applying the Dominated Convergence Theorem on $|f|^2\mathbb{1}_{|x|\ge y}$.

Without loss of generality, take $\int\limits_{\mathbb{R}}{|\phi|^2} = \int\limits_{\mathbb{R}^2}{f^2}=1$. Notice that \begin{align} \left|\int\limits_{\mathbb{R}}{\phi(x-y)f(x)\text{ d}x}\right| &\le \int\limits_{x<y/2}{|\phi(x-y)f(x)|\text{ d}x} + \int\limits_{x\ge y/2}{|\phi(x-y)f(x)|\text{ d}x}\\ &=\int\limits_{x<-y/2}{|\phi(x)f(x+y)|\text{ d}x} + \int\limits_{x\ge y/2}{|\phi(x-y)f(x)|\text{ d}x} \\ &\le\left(\int\limits_{x<-y/2}{|\phi(x)|^2\text{ d}x}\right)^{1/2}\left(\int\limits_{x<-y/2}{|f(x+y)|^2\text{ d}x}\right)^{1/2} + \left(\int\limits_{x\ge y/2}{|\phi(x-y)|^2\text{ d}x}\right)^{1/2}\left(\int\limits_{x\ge y/2}{|f(x)|^2\text{ d}x}\right)^{1/2} \\ &\le\left(\int\limits_{x<-y/2}{|\phi(x)|^2\text{ d}x}\right)^{1/2} + \left(\int\limits_{x\ge y/2}{|f(x)|^2\text{ d}x}\right)^{1/2} \\ &\le\left(\int\limits_{|x|\ge y/2}{|\phi(x)|^2\text{ d}x}\right)^{1/2} + \left(\int\limits_{|x|\ge y/2}{|f(x)|^2\text{ d}x}\right)^{1/2}. \end{align} It follows that $$\limsup\limits_{y\rightarrow\infty}{\left|\int\limits_{\mathbb{R}}{\phi(x-y)f(x)\text{ d}x}\right|}\le\limsup\limits_{y\rightarrow\infty}{\left(\left(\int\limits_{|x|\ge y/2}{|\phi(x)|^2\text{ d}x}\right)^{1/2} + \left(\int\limits_{|x|\ge y/2}{|f(x)|^2\text{ d}x}\right)^{1/2}\right)} = 0$$ and hence $$\lim\limits_{y\rightarrow\infty}{\int\limits_{\mathbb{R}}{\phi(x-y)f(x)\text{ d}x}} = 0$$ as desired.