Bounding a linear map from $L^q$ to $C_c$ Let $C_0((0, \infty))$ be the set of all functions such that $\lim_{x \to \infty} f(x) = 0$ and $\lim_{x \to 0} f(x) = 0$, this is a Banach space under the $\sup$ norm. 
Now if we fix a $p$, $1 < p < \infty$ and define $ \displaystyle Tf(x) = x^{\frac{-1}{p}}\int_{0}^{x}{f(t) \: dt}$, for locally integrable functions on $[0, \infty)$. Show that T is a bounded linear map from $L^q((0, \infty))$ to $C_0((0, \infty))$ where $\frac{1}{p} + \frac{1}{q} = 1$.
I have shown everything except that the operator is bounded, and have tried a few methods for this. It doesn't seem easy to control the size of the sup norm of $Tf$ through the controlling of the $L^q$ norm of $f$, so showing continuity in a neighborhood of $0$ hasn't worked. There must be some way of using $x^{\frac{-1}{p}}$ to prevent having a very narrow and tall function from having arbitrarily large sup norm after the transformation. I'm not having any luck with this though.
 A: This is just Holder's inequality. By definition $$\int_0^x f=\int\chi_{(0,x)}f,$$so $$\left|\int_0^xf\right|\le||\chi_{(0,x)}||_p||f||_q=x^{1/p}||f||_q.$$Hence $|Tf(x)|\le||f||_q$.
You say you've done everything else. It's not clear to me how you showed that $Tf(x)$ tends to $0$ at infinity without first proving the inequality above. In any case, for the benefit of any readers who don't see how that goes, a nice clean way to do it is like so: For $A>0$ define $$T_Af(x)=x^{-1/p}\int_0^x\chi_{(0,A)}f.$$ Arguing as above shows that $$|T_Af(x)|\le x^{-1/p}A^{1/p}||f||_q,$$so $T_Af(x)\to0$ at $\infty$. But $$\lim_{A\to\infty}||f-\chi_{(0,A)}f||_q=0,$$ so the first inequality we proved shows that $T_af\to Tf$ uniformly as $A\to\infty$.
Finally, the fact that $Tf$ is continuous on $[0,\infty)$ follows from, I wonder what, oh right it follows from Holder's inequality. First, Holder shows that $$|Tf(x)|\le\left(\int_0^x|f|^q\right)^{1/q},$$so $\lim_{x\to0}Tf(x)=0$. Now to show $Tf$ is continuous on $(0,\infty$ it's enough to show that $Sf$ is continuous on $(0,\infty)$, where $$Sf(x)=\int_0^x f.$$But if $x,y\in(0,\infty$ Holder shows that $$|Sf(x)-Sf(y)|\le|x-y|^{1/p}||f||_q.$$
Moral Try Holder's inequality.
In fact this is one of those things where if it follows from Holder people won't even say why it's true; if it follows from Holder people will take it to be obvious without comment.
