Showing that the operator is bounded and find its norm.

I have this operator $T: L^p(0,\infty)\rightarrow L^p(0,\infty)$, $1<p<\infty$ :

$$(Tf)(x)=1/x\int_0^xf(t)dt.$$

I am supposed to show that it is bounded and fint its norm. I had an idea that almost worked for showing that it was bounded, but then I ran into a problem. And also I do not see how to show that its norm is $q$, where $q$ is given by $1/p+1/q=1$.

First we have:

$|(Tf)(x)|\le1/x\int_{[0,x]}|f|dt=1/x\int_{[0,x]}|f|\cdot1dt$, and then by Hölder:

$$\le1/x(\int_{[0,x]}|f|^pdt)^{1/p}(\int_{[0,x]}1^gdt)^{1/q}\le \|f\|_px^{-1+1/q}=\|f\|_p/x^{1/p}.$$ But this function is not in $L^p(0,\infty)$. And showing that the norm is $q$, I really don't know how to do.

Do you guys have any tips?

Since continuous functions with compact support obviously belongs to $\mathcal{L}^{p}(0,\infty)$, we shall first solve the problem for that class of functions and then try to finish it of by a standard density-argument. So suppose $f$ is non-negative as well. Then it is pretty straightforward to prove by the fundamental theorem of calculus that $$\lim_{x\rightarrow 0+} T(f)(x)=\lim_{x\rightarrow 0+}\frac{1}{x}\int_{0}^{x}f(t)\, dt=f(0+)$$ and that the operator is well-defined for every other $x>0$. Now an integration by parts gives that $$||T(f)||_{p}^{p}= x\cdot T(f)(x)^{p} \,|_{0}^{\infty} +p\int_{0}^{\infty}\left(T(f)(x)^{p} -T(f)(x)^{p-1}f(x)\right) \, dx= $$ $$p\cdot ||T(f)||_{p}^{p} -p\int_{0}^{\infty}T(f)(x)^{p-1}f(x) \, dx$$

Collecting the terms involving $||T(f)||_{p}^{p}$ and dividing and using Hölders inequality we get $$||T(f)||_{p}^{p}= \frac{p}{p-1}\int_{0}^{\infty}T(f)(x)^{p-1}f(x)\, dx\leq \frac{p}{p-1}\left(\int_{0}^{\infty}T(f)(x)^{(p-1)q}\, dx\right)^{1/q}\cdot ||f||_{p}=$$ $$\frac{p}{p-1}\left(\int_{0}^{\infty}T(f)(x)^{p}\, dx\right)^{1/q}\cdot ||f||_{p}= \frac{p}{p-1}\cdot ||T(f)||_{p}^{\frac{p}{q}}\cdot ||f||_{p}$$ where $\frac{1}{p}+\frac{1}{q}=1$. Now assuming $||T(f)||_{p}$ is non-zero otherwise the claim is trivial, we divide and obtain $$||T(f)||_{p} \leq \frac{p}{p-1}\cdot ||f||_{p.}$$ for every non-negative continuous function $f$ with compact support.

Now using the simple fact that $|T(f)(x)| \leq T(|f|)(x)$ for each $x\in (0,\infty)$,the result immediately extends to all continuous functions with compact support on $(0,\infty)$. Now to finish off the proof, let $g$ be an arbitrary $\mathcal{L}^{p}(0,\infty)$ and $\left\{f_{n}\right\}_{n=1}^{\infty}$ a sequence of continuous functions with compact support that converges to $f$ in $\mathcal{L}^{p}$-norm. We can extract a subsequence $\left\{f_{n_{j}}\right\}_{j=1}^{\infty}$ which converges to $g$ almost everywhere on $(0,\infty)$.

Now using Fatou's lemma, we finally get that $$||T(g)||_{p}\leq \liminf_{j\rightarrow \infty} \, ||T(f_{n_{j}})||_{p}\leq \liminf_{j\rightarrow \infty} \, \frac{p}{p-1}||f_{n_{j}}||_{p}\leq \limsup_{j\rightarrow \infty} \, \frac{p}{p-1}\left(||f_{n_{j}}-g||_{p} +||g||_{p}\right)= \frac{p}{p-1}||g||_{p}$$