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Let $f:[1,\infty)\rightarrow \mathbb{R}$ be continuous and absolutely integrable. Show that $f^2$ is absolute integrable on $[1,\infty)$.

My solution: Because $f$ is contiuous and absolutely integrable, $f$ will be eventually zero: $\lim_{x\rightarrow \infty}f(x)=0$. So $|f|^2<|f|$ is true for sufficient large values of $x$. Other $x$ are inside a compact set and due to continuity $f$ is square absolutely integrable there as well.

Is this correct reasoning?

fasdgr
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    If by absolutely integrable you mean that $\int_1^{\infty} |f|<\infty$ then this doesn't imply $\lim_{x\to\infty}f(x)=0$, even if $f$ is continuous. Think of a piecewise linear function whose graph contains lots of "triangles", which eventually become very thin. – Mark Mar 27 '24 at 13:14

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This is unfortunately not true. Let $f_n$ be a “hat” function (its graph is an isosceles triangle) supported in $[n,n+n^{-a}]$ and with height $n^b$, and let $f = \sum_n f_n$. Clearly $f$ is continuous. We have

$$\int_1^\infty f(x) dx = \sum_{n=1}^\infty n^{b-a}$$ $$\int_1^\infty f(x)^2 dx = \frac23 \sum_{n=1}^\infty n^{2b-a}$$

To make $f$ integrable, we just need $a-b > 1$, and to make $f^2$ not integrable, we need $a-2b\leq 1$, so we can take $a=3$ and $b=1$. In particular, $f$ is unbounded, despite being integrable and continuous. What you really need is for $f$ to be uniformly continuous, so that $f$ does indeed decay to zero (see here) and in particular is bounded. Then you can argue that $f^2 \leq M |f|$ to prove that $f^2$ is integrable.

kieransquared
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  • Interesting. There is an inclusion (in one direction) between $L^1[0,1]$ and $L^2[0,1]$, but there is no inclusion (in either direction) between $L^1(1,\infty)$ and $L^2(1,\infty)$. – GEdgar Mar 27 '24 at 13:36
  • So intervals of domain for hat function are $[1,1+1^{-3}], [2,2+2^{-3}],...$. Do you mean to fill in with the zero function? – fasdgr Mar 27 '24 at 15:32
  • Yes, they are supported in those intervals, so they are zero outside of those intervals – kieransquared Mar 27 '24 at 17:39