summable square function implies...?

Question

I have difficulty to demonstrate this:

$$ \int_{-\infty}^{\infty}|f(x)|^2dx<\infty~~~\text{(summable square function)}$$ then,

$$\lim_{|x|\rightarrow\infty }f(x)=0$$

thank you.

Answer 1

You are probably having difficulties because this fact is false: The function $$f(x)=\begin{cases} 1 & x \in \mathbb Z \\0 & x \notin \mathbb Z \end{cases}$$ is a counterexample.

Answer 2

Just take a function $$f(x)=\frac{1}{n^2+1}(x-n)^{-1/3}, \quad x\in (n,n+1]$$ for $n\in \Bbb Z$.

Answer 3

Define ${\rm J}\left(x\right) \equiv \int_{-\infty}^{x}\left\vert{\rm f}\left(x'\right)\right\vert^{2}\,{\rm d}x'$. Then, $\forall \epsilon > 0, \exists\ N, N'$ such that $\left(x + h\right) > N$ and $x > N'\quad\Longrightarrow$

$$ \left\vert{\rm J}\left(x + h\right) - {\rm J}\left(x\right)\right\vert \leq \underbrace{\quad\left\vert{\rm J}\left(x + h\right) - {\rm J}\left(\infty\right)\right\vert\quad} _{<\ \epsilon\,/\,2} + \underbrace{\quad\left\vert{\rm J}\left(x\right) - {\rm J}\left(\infty\right)\right\vert\quad} _{<\ \epsilon\,/\,2} < \epsilon $$

Then, $\forall\ \epsilon >0$; when $x > {\rm max}\left\lbrace N - h, N'\right\rbrace$, we have $\left\vert{\rm J}\left(x + h\right) - {\rm J}\left(x\right)\right\vert < \epsilon$. That means

$$ \lim_{x \to \infty}\left\vert{\rm J}\left(x + h\right) - {\rm J}\left(x\right)\right\vert = 0 $$

Also

$$ 0 = \lim_{h \to 0}\lim_{x \to \infty} \left\vert{{\rm J}\left(x + h\right) - {\rm J}\left(x\right) \over h}\right\vert = \lim_{x \to \infty}\lim_{h \to 0} \left\vert{{\rm J}\left(x + h\right) - {\rm J}\left(x\right) \over h}\right\vert = \lim_{x \to \infty}\ \left\vert{\rm f}\left(x\right)\right\vert^{2} $$

We have to be sure we can exchange the limits. That puts an extra condition on ${\rm J}\left(x\right)$ or/and ${\rm f}\left(x\right)$.