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I read in a book that the condition $\int |f(x)|^2 dx <\infty$ is less restrictive than $\int |f(x)| dx <\infty$. That means whenever $\int |f(x)| dx$ is finite, $\int |f(x)|^2 dx$ is also finite, right?

My understanding is that $|f(x)|$ may have a thick tail to make the integral blow up, but $|f(x)|^2$ may decay quickly enough to have a finite integral. Can someone give me an example that $\int |f(x)| dx=\infty$ but $\int |f(x)|^2 dx <\infty$. Suppose $f(x)$ is an absolutely continous function and bounded on $(-\infty, \infty)$.

Did
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JACKY88
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    It is not the case that $\int |f(x)|,dx < \infty \implies \int |f(x)|^2,dx < \infty$. Consider $f(x) = \frac1{\sqrt x}$ on $(0,1)$. – kahen Nov 01 '12 at 14:24
  • @kahen Yes, you are right. But I am interested in those $f$ which are bounded on $\mathbb{R}$. Maybe I should state that in the problem. Thank you for the good point. – JACKY88 Nov 02 '12 at 00:03

7 Answers7

19

$$f(x)=\frac1{\sqrt{1+x^2}}{}$$

Did
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You can think of the harmonic series:

$$\sum_{n \geq 1} \frac{1}{n}=\infty$$

but

$$ \sum_{n\geq 1}\frac{1}{n^2}<\infty$$.

Therefore you can choose $$f(x) = \sum_{n \geq 1} \frac{1}{n} \chi_{[n,n+1)}$$ where $\chi_X$ is the characteristic function of the set $X$.

Beni Bogosel
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For example take $f(x)=1/\lfloor x\rfloor$. Then $$\int_1^\infty|f(x)|dx=\sum_{n=1}^\infty \frac1n=\infty$$ But $$\int_1^\infty|f(x)|^2dx=\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}<\infty$$

Dennis Gulko
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How about: $$f(x) = \left\{ \begin{array}{ccc} \frac{1}{|x|} && |x| > 1 \\ 1 && |x| \leq 1 \end{array} \right.$$

Nathaniel Bubis
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Let $$f(x)=\begin{cases}\frac1{|x|} & |x|\geq 1\\1 & |x|<1.\end{cases}$$

Cameron Buie
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The most noticeable one I think is the sinc function $$\mathrm{sinc}(x)=\frac{\sin(\pi x)}{\pi x}$$

chaohuang
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Firstly, i'm presuming you mean an integral over all space?

The first answer that comes to mind, for a continuous function is going to be annoying simple;

$$ \int_{-\infty}^{\infty} x dx = 0 $$

while

$$ \int_{-\infty}^{\infty} x^2 dx = 0 $$

So any odd function will always satisfy $\int f(x) dx < \infty$, but not necessarily $\int f^2(x) dx < \infty$. That's a whole class of functions that always fit one of the conditions but not always the other.

will
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