I'm trying to prove the next statement:

Suppose that $(f^2)^{'}, f^{'}\in L_{2}(0,\infty).$ Then $\displaystyle\lim_{x\rightarrow\infty}f^{2}(x)$ exists. If $f\in L_{2}$ then $\displaystyle\lim_{x\rightarrow\infty}f(x)=0.$

I was trying use Cauchy inequality to bound $f^{2}$ and fundamental theorem calculus but I'm not sure if this is a good path to prove it.I'm stuck.

Even more, why does the hypothesis of belonging to $L_{2}$ imply that the limit of $f$ should be zero at infinity?

Any suggestions?

  • 2
    $\begingroup$ Isn’t $f(x)=ln(1+x)$ a counterexample? $\endgroup$ – Vogel Nov 6 '17 at 7:41

The first part would be correct if you require $(f^2)'\in L^1(0,\infty)$. Note that by $(f^2)'=2f f'$ this holds if $f,f'\in L^2(0,\infty)$. Moreover, if the limit is nonzero, then $f$ cannot be square integrable.


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