if $f(x)$ is summable square function, then...

Question

I have a question. If a normed function, that is to say

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

$$\underset{\begin{array}{c} x\rightarrow+\infty\\ x\rightarrow-\infty \end{array}}{lim}f(x)=0$$

It is true, why?

thanks

One can surely construct a function that peaks every now and then but with peaks so narrow the square is still integrable. — Pedro, Aug 08 '13 at 15:02
Isn't standard terminology here square-integrable? Should we say summable only when talking about infinite sums, or are the two really interchangeable? — Clayton, Aug 08 '13 at 15:08

Clayton · Answer 1 · 2013-08-08T15:14:26.563

3

It isn't true, even if we restrict the function to be continuous. For example, take the function that forms the top part of a triangle of width $1/n^2$ at each integer and goes up to its peak, $1$, at the integer. Then it is easy to see the integral converges to a value less than $\pi^2/3$ but $$\lim_{x\to\pm\infty}f(x)=\text{DNE}.$$

edited Aug 08 '13 at 15:14

answered Aug 08 '13 at 15:06

Clayton

24,751

score 2 · Answer 2 · answered Aug 08 '13 at 15:06

No, this is not true even if you have a continuous function. Think of a function which is zero on a large set, has periodically bumps on smaller and smaller intervals.

For example, take a piecewise linear continuous function, which is 1 in integer points $x=1$, is linear (like a tent) on the interval $[n-\frac{1}{3n^2},n+\frac{1}{3n^2}]$ for $n\neq0$, and constant 0 otherwise.

if $f(x)$ is summable square function, then...

2 Answers2