So i have a Problem with a proof.
given is a function $f:[1,\infty) \to R$, which is in every intervall $[1,b]$ with $b>1$ Riemann-Integratable.
There are also given $p,q,r ∈ R$ with $1≤p≤r≤q$.
Show that if $\int_1^\infty |f(x)| dx$ exists, then exists $\int_1^\infty |f(x)|^2 dx$.
I tried the Hölder's inequality, but i don't know how to choose $r$.
my Proof Idea:
There is a $M > 0$ so that:
$\int_1^b |f(x)|dx ≤M$ for all $1<b<\infty$ ( from Cauchy ).
let $\int_1^b |f(x)|^2dx$ be integratable then :
$\int_1^b |f(x)|^2dx = \int_1^b ||f(x)|^2 . 1|dx ≤ (\int_1^b |f(x)|^{2r}dx)^{1/r}.(\int_1^b 1dx)^{1/s} = (\int_1^b |f(x)|dx)^{1/r}.(\int_1^b 1dx)^{1/s} ≤ M^{1/r}.(b-1)^{1/s} := M'$
from that we know that $\int_1^b |f(x)|^2dx ≤ M'$
I don't know but how should i choose r and s so that $2r=1$ and $1/r+1/s=1$ with $r,s\ge 1$.
If you have an Idea on how to choose r and s or if you have a better solution let me know.
Here is a counterexample to the problem as written:
$$f(x) = \begin{cases} \lfloor x \rfloor & \text{if } x - \lfloor x \rfloor < \lfloor x \rfloor ^{-3} \\ 0 & \text{otherwise} \end{cases} $$
The function consists of a sequence of rectangular bumps of height $n$ but area $n^{-2}$. So $\int_1^\infty |f(x)|\,dx = \pi^2/6$.
However $f(x)^2$ has bumps of the same width but height $n^2$. So the area of each bump is now $n^{-1}$, and thus $\int_1^\infty |f(x)|^2\,dx$ diverges logarithmically.
Perhaps there's some condition that this $f(x)$ fails, which you forgot to copy in the question? I notice that the $p,q,r$ are not mentioned in the statement of what you want to prove at all, except that they are real numbers ...