Example for continuous $f$ such that $\int \limits_{0}^\infty |f(x)|dx < \infty$ and $\int \limits_{0}^\infty f^2(x)dx=\infty$.

Question

Does $f = \dfrac{\sin(x)}{\sqrt{x}}$ make it?

I know that $\int \limits_{0}^\infty \dfrac{\sin^2(x)}{x}dx = \infty$, but does $\int \limits_{0}^\infty \dfrac{|\sin(x)|}{|x|}dx < \infty$?

What would be a good example?

Answer 1

$\int \limits_{0}^\infty \dfrac{|\sin(x)|}{|x|}dx = \infty$

and

$\int \limits_{0}^\infty \dfrac{|\sin(x)|}{\sqrt x}dx = \infty$ (which is more relevant here).

$f(x)=\frac 1 {\sqrt x}$ for $0<x<1$, $f(x)=2-x$ for $1\leq x <2$ and and $f(x)=0$ for $x\geq 2$ gives such a function.

Answer 2

Consider

$$f_n(x)=\begin{cases}n(1-n^3|x-n|) &\mathrm{if}\,|x-n|<\frac1{n^3}\\0& \mathrm{otherwise}\end{cases}$$

That is, $f_n$ is continuous, has a triangular peak at $x$ of height $n$, and is $0$ outside of $[x-\frac{1}{n^3},x-\frac{1}{n^3}]$.

You can compute $\int_0^\infty f_n(x)\,\mathrm{d}x=\dfrac{1}{n^2}$ and $\int_0^\infty (f_n(x))^2\,\mathrm{d}x=\dfrac{2}{3n}$.

Let $f(x)=\sum_{n=2}^\infty f_n(x)$. Then $f$ is continuous on $[0,+\infty)$, $\int_0^\infty f(x)\,\mathrm{d}x$ is bounded, while $\int_0^\infty (f(x))^2\,\mathrm{d}x$ is not.

Here is a plot of $f$ on $[0,10]$:

Answer 3

Consider the following fact.

Let $$f(x) = \begin{cases} \frac{1}{|x|^{a}} & |x|\leq1, \\ 0 & \text{else}. \end{cases}$$Then $f\in L^{1}(\mathbb{R}^{n})$ $\iff$ $a<n.$