If $\int_{1}^{\infty}f\left(x\right)dx$ converges absolutely then $\int_{1}^{\infty}f^{2}\left(x\right)dx$ converges? Let $f:[0,\infty)\to\mathbb{R}$ be a continuous function. I was asked to prove/disprove the following statement:
If $\int_{1}^{\infty}f\left(x\right)dx$ converges absolutely then $\int_{1}^{\infty}f^{2}\left(x\right)dx$ converges as well. 
I figure that my only way of proving this is by direct comparison. However, for that to work I need $f(x)\leq1$  for sufficiently large $x$, but $\int_{1}^{\infty}f\left(x\right)dx$ converging (Even absolutely) does not guarantee this.
If the statement is false I'd appreciate a hint on how to construct a counter example, rather then one pulled out of thin air.
 A: If
$f(x)$
has very narrow peaks
that grow in size
so that the sum of the area of
the $n$-th peak
converges,
the sum of
the peaks of
$f^2(x)$
can diverge.
For example,
if
$f(x)
= n^a
$
for
$n \le x \le n+n^{-b}
$,
then
$\int_n^{n+1} f(x) dx
=n^{a-b}
$
and
$\int_n^{n+1} f^2(x) dx
=n^{2a-b}
$.
To get the first sum to converge,
we need
$a-b < -1$
and to get the
second sum to diverge
we need
$2a-b > -1$.
If we choose,
for some $c > 0$,
$a-b=-1-c$
and
$2a-b=-1+c$,
then
$a = 2c$
and
$b=1+3c$
will work.
Choosing
$c=1/4$,
we get
$a = 1/2$
and
$b = 7/4$.
For this,
$n^{a-b}
=n^{-5/4}
$
and the sum of these converges,
and
$n^{2a-b}
=n^{-3/4}
$
and the sum of these diverges.
Round off the corners
if you don't like boxes.
A: Sketch: The answer is yes if $f$ is bounded, as you noticed. But $f$ need not be bounded: One way towards a counterexample is to think  of tall thin triangles marching out to $\infty.$ Similar to the triangle example is an infinite seres that's fairly easy to write down:
$$f(x) = \sum_{n=1}^{\infty}n\sin^2(n^3\pi x)\mathbb \chi_{[n,n+n^{-3}]}(x).$$
A: Idea: $\ f$ is the zero function except tiny neighbourhoods around positive integers.

Above is an idea for such a function. Observe that $f(n)=2^{n-1}$.
Let's call the trapezoids $T_1,T_2,T_3$ and so on.
Make sure that the area below the upper base of each trapezoid $T_n$ is $2^{-n}$. (by manipulating its endpoints)
Also make sure that the sum of the area under the left and right sides of each trapezoid is $2^{-n}$.
So, the area under each trapezoid is $2.2^{-n}$.
So, the integral $\displaystyle \int_0^\infty f(x) \ \mathrm dx = \sum_{n=1}^\infty 2. 2^{-n} = 2$. $\implies f$ is absolutely convergent.
Now, when you square $f$, the height of the upper base of each $T_n$ will be squared.
For example, the height of $T_1$ will be $1$, the height of $T_2$ will be $4$ etc.
Also, the shape of the left and right sides of each trapezoid will be different, but we're not interested in that.
Anyways, the area under the upper base of $T_n$ will be $(2^{n-1})^2 \cdot 2^{-n}= 2^{n-2}$.
$\implies$ areas under upper bases of "squared trapezoids" will be divergent.
Hence $\displaystyle \int_0^\infty f^2(x) \ \mathrm dx$ is divergent.
