$f \geq 0$ continuous, $\lim_{x \to \infty} f(x)$ exists, $\int_{0}^{\infty}f(x)dx< \infty$, Prove: $\int_{0}^{\infty}f^2(x)dx< \infty$

Something that bothers me with the following question: $f: [0, \infty] \to \mathbb{R}$,$f \geq 0$, $\lim_{x \to \infty} f(x)$ exists and finite, and $\int_{0}^{\infty}f(x)dx$ converges, I need to show that $\int_{0}^{\infty}f^2(x)dx$

I separated the integral in the following way: $\int_{0}^{1}f^2(x)dx$+$\int_{1}^{\infty}f^2(x)dx$, while for the second one we know that $\lim_{x \to \infty}\frac{f^2(x)}{f(x)}$ so they converges together, but what happens in the first range, does my $\lim_{x \to 0}f(x)$ has to be finite? Do you have an example for an integral between $0$ and $\infty$ which converges and the $\lim_{x \to 0}$ is not finite? What should I do in my case?

Thanks!

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The integral over the closed (finite) interval $[a,b]$ of a function continuous on $[a,b]$ always exists. –  André Nicolas Feb 10 '12 at 21:28
@DavideGiraudo: Thank you, I fixed it. –  Jozef Feb 10 '12 at 21:38
@JavaMan: Your reminder today helped! I hope it will remain tomorrow as well ;) –  Jozef Feb 10 '12 at 21:39

We first answer your specific question. That answer is only partly relevant to the existence of $\int_0^\infty f^2(x)\,dx$. So in a remark at the end, we deal with that problem.

The integral over the closed (finite) interval $[a,b]$ of a function continuous on $[a,b]$ always exists. The function $f^2$ is continuous on $[0,1]$.

Remark: The sketch you made for $[1,\infty)$ does not appear to lead to a proof. We describe an approach that does work.

We are told that the limit of $f(x)$ as $x\to\infty$ exists. That limit must be $0$, else $\int_0^\infty f(x)\,dx$ would not exist (here the fact that $f(x)\ge 0$ is relevant).

So after a while $0\le f(x)\le 1$. It follows that after a while we have $0\le f^2(x)\le f(x)$. Thus by Comparison everything is OK after a while. And everything is always OK (for continuous functions) in a finite interval.

I am confident that you will be able to turn this deliberately informal sketch into a full proof.

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Oh I see, so basically you're saying that due to $f$ being continues the integral exists, and the opposite does not true always. Thanks a lot! –  Jozef Feb 10 '12 at 21:42
@Jozef: I have amplified the remark about the result. I hope that now everything will be clear. –  André Nicolas Feb 10 '12 at 22:06

Let $E$ be the set $\{x: f(x)>1\}$. Since $f\in L^{1}$, we know that $E$ is measurable and has finite measure. Since $f\in L^{1}$ and continuous, we know that $\lim_{x\to\infty}f(x)=0$. Hence, $f$ is bounded above by a constant $M$. Thus, we have $$\int_{[0,\infty]}f^{2}\,dx=\int_{E}f^{2}\,dx+\int_{E^{c}}f^{2}\,dx\leq m(E)M^{2}+\int_{E^{c}}f\,dx<\infty.$$ This proves that $f\in L^{2}$.

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To answer your questions regarding the behaviour at $0$ first. Note that $f:[0;\infty[ \ \to \mathbb{R}$, hence $f$ is well-defined and continuous at $0$ - according to the assumptions you state. In particular this means that $\lim_{x\to 0}f(x) = f(0)$, so there is really no issue at $x=0$; your function is well-defined at this end of the interval. Had this not been the case then you would have had to consider what might go wrong here (the function $f(x) = x^{-1/2}$ could be an instructional case to consider when integrating).

I am not sure I follow your reasoning in the case when $x\to \infty$, but I think it might be a good idea to try and deduce what the possible values of $\lim_{x\to \infty}f(x)$ could be, if the integral $\int_0^{\infty}f(x)dx$ is suppose to converge, and then maybe use $0<f(x)^2\leq f(x)$ under the right circumstances...

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