I have the following problem:
let $1 < p< 4$ and let $f \in L_p((1, \infty))$.
We consider $g: (1, \infty ) \rightarrow [- \infty, \infty]$ , defined by $g(x):=\frac{1}{x} \int_{[x,10x]} \frac{f(t)}{t^{\frac{1}{4}}}dλ_1(t).$
I must show that $g \in L_2((1, \infty))$
I have alredy proven this theorem:
Let $1<p<r<q< \infty$ with $\frac{1}{r}=\frac{m}{p}+\frac{1-m}{q}$, $m \in (0,1)$ and $n \geq 1$. Than $||f||_{L_r} \leq ||f||^{m}_{L_p}||f||^{1-m}_{L_q} $ for all $f \in L_p(\mathbb{R}^n) \bigcap L_q(\mathbb{R}^n)$
And
Theorem: Let $1<p<q< \infty$ and $U ⊂ \mathbb{R}^n$ open with $λ_n(U) < \infty$. Then $L^q(U) ⊂ L^p(U)$.
So what I thought for solving this problem was to show that $(\int_{[x,10x]}|g(x)|^2)^{1/2}< \infty$ for doing it I thought to maybe use the first theorem but here I am having some problems
what I have done is the following:
$(\frac{1}{x} \int_{[x,10x]} |\frac{f(t)}{t^{\frac{1}{4}}}dλ_1(t)|^2)^{1/2} \leq (\int_{[x,10x]}|f(t)|^{p})^{m/p}*(\int_{[x,10x]}|t^{-1/4}|^q)^{1-m/q}$
Doing this I can easily calculate the second integral and showing that is finite. For the first Integral I can maybe recall the second theorem and using the fact that $f \in L_p((1, \infty))$ I can say that is finite.
The problem that I am having is that I cannot choose $m,p,q$. Can someone help me?
If my calculations are wrong please comment, I'll delete this answer. Thanks.
Here $\mu(t)=\lambda_1(t)$.
Given $$g=\frac1x\int_x^{10x}\frac{f(t)}{t^{\frac14}}\,d\mu(t)$$.
\begin{align}||g||^2_{L^2(1,\infty)}&\leq\int_1^{\infty}\left[\int_x^{10x}\frac{(f(t))^2}{t^{\frac12}x^2}\,d\mu(t)\right]\,d\nu(x)\\&=\int_1^{\infty}\left[\int_{t/10}^t\frac{(f(t))^2}{t^{\frac12}x^2}\,d\nu(x)\right]\,d\mu(t)\\&=9\int_1^\infty\frac{(f(t))^2}{t^{\frac32}}\,d\mu(t)\\&\leq 9 ||(f(t))^2||_{L^{\frac32}(1,\infty)}\left|\left|\frac1{t^{\frac32}}\right|\right|_{L^3(1,\infty)}<\infty\quad (\because f\in L^3(1,\infty))\end{align}