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Show that if $f\in L^2 (\mathbb{R})$ then $$\lim_{x \rightarrow \infty} g(x)=0,$$ where $$g(x)=\int_x^{x+1}f(t)\,dt$$

Since $f \in L^2 (\mathbb{R})$ then $f \in L^2[x,x+1]$ for every $x$. Then $$|g(x)| \leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}} \int_x^{x+1} 1 dt$$ $$|g(x)|\leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}}$$

6 Answers 6

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Using Jensen's Inequality, the sum $$ \begin{align} \sum_{n=0}^\infty\left(\int_n^{n+1}|f(x)|\,\mathrm{d}x\right)^2 &\le\sum_{n=0}^\infty\int_n^{n+1}|f(x)|^2\,\mathrm{d}x\\ &=\int_0^\infty|f(x)|^2\,\mathrm{d}x\\ &\le\|f\|_{L^2(\mathbb{R})}^2 \end{align} $$ converges, which means the terms must go to $0$. That is, $$ \lim_{n\to\infty}\int_n^{n+1}f(x)\,\mathrm{d}x=0 $$

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    Is using $\sum_{n=0}^\infty$ rigorous? $g(x)=\int_x^{x+1}f(t)dt$ can take any $x\in \mathbb{R}$. – 2018-03-24
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    @user398843: $\left|\,\int_x^{x+1}f(x)\,\mathrm{d}x\,\right|\le\int_{\lfloor x\rfloor}^{\lfloor x\rfloor+2}|\,f(x)\,|\,\mathrm{d}x$, and we've just shown that $\int_{\lfloor x\rfloor}^{\lfloor x\rfloor+1}|\,f(x)\,|\,\mathrm{d}x\to0$ and $\int_{\lfloor x\rfloor+1}^{\lfloor x\rfloor+2}|\,f(x)\,|\,\mathrm{d}x\to0$ – 2018-03-25
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Continuing from your last step, you have

$$|g(x)|\leq\left(\int_\mathbb{R} f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}$$ where $\chi_{[x,x+1]}(t)$ is the indicator function on the interval $[x,x+1]$.

So $$\lim_{x\to\infty}|g(x)|\leq\lim_{x\to\infty}\left(\int_\mathbb{R} f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}$$

Since $|f(t)^2\chi_{[x,x+1]}|\leq|f(t)^2|$ which is integrable, by Lebesgue's Dominated Convergence Theorem, we can move the limit inside the integral, so

$$\lim_{x\to\infty}\left(\int_\mathbb{R} f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}=\left(\int_\mathbb{R} \lim_{x\to\infty}f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}=0$$

since $\lim_{x\to\infty}\chi_{[x,x+1]}=0$

Hence $\lim_{x\to\infty}|g(x)|\leq 0$, and we can conclude from there.

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    Could you please elaborate in what sense does $\lim_{x\to\infty}\chi_{[x,x+1]} = 0$. – 2017-01-03
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    In the sense of convergence as a function of $t\in\mathbb{R}$: $\chi_{[x,x+1]}(t)$ converges to the zero function for each $t\in\mathbb{R}$. – 2017-01-03
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    Oh, right, thank you. I was misinterpreting it as a function $x\mapsto \chi_{[x,x+1]}\colon \Bbb R\to L^2(\Bbb R)$ instead of $x\mapsto \chi_{[x,x+1]}(t)\colon \Bbb R\to \Bbb R$ for fixed $t$. – 2017-01-03
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The result is valid for functions in $C_0^\infty(\mathbb R)$. So, it is true for functions in $L^2(\mathbb R)$ (as well as for functions in $L^p(\mathbb R)$ with $1\leq p<\infty$) by density.

Details: As $C_0^\infty(\mathbb R)$ is dense in $L^2(\mathbb R)$, there is $(f_n)$ in $C_0^\infty(\mathbb R)$ such that $$\|f_n-f\|_{L^2(\mathbb R)}\xrightarrow{n\to\infty} 0.\tag{1}$$ Take $\varepsilon>0$. We want to show that there exist $M>0$ such that $$x>M\quad\Longrightarrow \quad |g(x)|<\varepsilon.$$

From $(1)$ there is $n_0\in\mathbb N$ such that $$\|f_{n_0}-f\|_{L^2(\mathbb R)}<\varepsilon.\tag{2}$$

As $f_{n_0}$ has compact support, there is $M>0$ such that $$|x|>M\quad\Longrightarrow \quad|f_{n_0}(x)|=0.\tag{3}$$

From $(2)$, $(3)$ and your estimate, $$\begin{align} x>M\quad\Longrightarrow \quad |g(x)|\leq \|f\|_{L^2(x,x+1)}&\leq\|f_{n_0}-f\|_{L^2(x,x+1)}+\|f_{n_0}\|_{L^2(x,x+1)}\\ &\leq\|f_{n_0}-f\|_{L^2(\mathbb R)}+\|f_{n_0}\|_{L^2(x,x+1)}\\ &<\varepsilon+0=\varepsilon. \end{align}$$

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    Hi, I am solving a similar homework problem. May I know in which book can I find the result in your answer? – 2018-03-20
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    I don't understand this "As $f_{n_0}$ has compact support, there is an $M>0$ such that ...". Would you mind explaining why there is such an $M$? – 2018-03-20
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    @user398843 Unfortunately, I can not remember any book that presents this result. Thus, let me know if you find any flaw. With respect to the second question, a compactly supported function is zero outside a compact. And a compact subset of the line is bounded. So, it is zero outside a bounded set which is a subset of $[-M,M]$ provided that we take $M$ large enough. – 2018-03-20
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We have that $\int_{-\infty}^{+\infty}f(u)^2du<+\infty \Rightarrow \lim_{x\rightarrow +\infty} \int_{-\infty}^{x}f(u)^2du=\int_{-\infty}^{+\infty}f(u)^2du$ and so for $ε>0,\ \exists M>0\ :\ \forall x$ with $x>M$, we have:

$|\int_{-\infty}^{x}f(u)^2du -\int_{-\infty}^{+\infty}f(u)^2du |<\frac{ε^2}{2}\ (1)$

When $x>M$ we also have $x>M+1$. Thus $(1)$ and the triangle inequality yield that:

$\int_{x}^{x+1}f(u)^2du=|\int_{-\infty}^{x+1}f(u)^2du -\int_{-\infty}^{x}f(u)^2du|=\\ |(\int_{-\infty}^{x+1}f(u)^2du -\int_{-\infty}^{+\infty}f(u)^2du )-(\int_{-\infty}^{x}f(u)^2du -\int_{-\infty}^{+\infty}f(u)^2du )|\leq \\ |\int_{-\infty}^{x+1}f(u)^2du -\int_{-\infty}^{+\infty}f(u)^2du |+|\int_{-\infty}^{x}f(u)^2du -\int_{-\infty}^{+\infty}f(u)^2du |<\frac{ε^2}{2}+\frac{ε^2}{2}=ε^2$

Therefore for $ε>0,\ \exists M>0 : \forall x$ with $x>M$, we have:

${(\int_{x}^{x+1}f(u)^2du)}^{\frac{1}{2}}<ε \Rightarrow |g(x)|<ε$, by your attempt.

From this point the definition of the limit to infinity finishes the job.

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You have a good start. To finish, use the folling result: If $g\in L^1(\mathbb R),$ then $\int_x^{x+1} |g| \to 0$ as $x\to \infty.$

Proof: Let $g_n = g\chi_{[n,n+1]}.$ Then $g_n \to 0$ pointwise everywhere, and $|g_n| \le |g|$ on $\mathbb R.$ By DCT, $\int_{\mathbb R} |g_n| \to 0.$ That proves the result for $x= 1,2,\dots,$ and that will give the result for arbitrary $x\to \infty.$

In your problem, use $g=f^2.$

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$d\mu= f^2 dt$ defines a positive finite measure on $\mathbb{R}$ (this is just another way of writing $\mu (A) = \int_{A} f(t)^2 dt$ if it's not clear), since $f$ is square-integrable.
Then, $\mu(x,x+1)= \mu(-\infty,x+1) - \mu(-\infty,x)$ (this is allowed since both sets have finite measure).
Then $$ \lim_{x \to +\infty} \int_{x}^{x+1} f(t)^2 dt = \lim_{x \to +\infty} \mu(x,x+1) = \mu(\mathbb{R}) - \mu(\mathbb{R}) = 0$$