0

Let $f$ be continuously differentiable on $[0,\infty)$,$\int_0^{\infty}f^2(x)dx<\infty$ and $|f'(x)|\leq C$, then prove $\lim_{x\to \infty}f(x)=0$.

I think $\int_0^{\infty}f^2(x)dx<\infty\implies f^2(x)\to 0$, so $f(x)\to 0$. But my professor gave a complicated answer. Am I right? Or what details have I overlooked? Thanks in advance!

Ychen
  • 565
  • 6
    $\int_0^{\infty}f^2(x)dx<\infty\implies f^2(x)\to 0$ is not true in general. See e.g. https://math.stackexchange.com/questions/3696388/ You need to use the properties given for $f$. – Gary Nov 17 '23 at 12:01
  • In the cited solution $f(x)$,is not nonnegative but here $f^2(x)$ is. – kmitov Nov 17 '23 at 12:07
  • @kmitov That's not a problem, see this: https://math.stackexchange.com/q/1231906/1104384 – Bruno B Nov 17 '23 at 12:13
  • 4
    @notpron Your mistake here is that if a function does not tend to $0$, then it doesn't necessarily tend to another number, it might have no limit at infinity just like in the case of sequences, for example think of a sine or cosine curve. The difference in behaviour between series and integrals at infinity on whether the sequence/function converges to $0$ is due to the "non-discrete" aspect of integrals, which allows you to create a function with spikes which increase in height but decrease in width fast enough as you go towards infinity which can allow your function to be integrable. – Bruno B Nov 17 '23 at 12:44
  • You're absolutely right, this was a very silly mistake of mine. I will delete the comment because it might be misleading. Thank you. – notpron Nov 17 '23 at 12:46
  • You don't really have to, it's constructive for OP and other people to be honest. But it's fine either way. – Bruno B Nov 17 '23 at 12:48
  • 2
    See this post: Square integrable function with bounded derivative goes to zero pointwise? – Mengchun Zhang Nov 17 '23 at 16:25

1 Answers1

4

As it's said in comments, it's not true that if $\int_0^\infty f^2(x)\, dx$ converges then $\lim\limits_{x \to \infty} f^2(x) = 0$. Counterexample: let $w(x)$ be bump function: $w(0) = 1$, $w(x) = 0$ if $|x| > 1$, $w(x) \in [0, 1]$, $w$ continuously differentiable. Now, let $f(x) = \sum\limits_{n=0}^\infty w(2^n (x - n))$ - narrow bumps around integer numbers.

To prove the result, we need the restriction on derivative. Note that if $|f(x_0)| = \epsilon > 0$, then (assuming wlof $f(x_0) > 0$), $f(x_0 + x) \geq \epsilon / 2$ for $x \in \left[0, \frac{\epsilon}{2C}\right]$ (here we use $|f'| \leq C$), and thus $\int_{x_0}^x f^2(t)\, dx \geq \frac{\epsilon^3}{8C}$. However, by Cauchy criteria, for all large enough $x_0$ and $x > x_0$, we have $\int_{x_0}^x f^2(t)\, dx <\frac{\epsilon^3}{8C}$, and thus $|f(x)| < \epsilon$.

mihaild
  • 15,368