Lp spaces (Hölder, Minkovski) Let $1<p<\infty$ and $f\in L_p(0,\infty)$. Show that 
$$\lim_{x\to\infty}\frac{1}{x^{1-\frac{1}{p}}}\int_0^x f(t)dt=0$$
by assuming that f is compactly supported.
Any idea so it can help me how to show it?
 A: Let $\frac{1}{p}+\frac{1}{q}=1$. Then
\begin{eqnarray*}
\varphi _{f}(x) &=&\frac{1}{x^{1-\frac{1}{p}}}\int_{0}^{x}dtf(t)=\frac{1}{x^{%
\frac{1}{q}}}\int_{0}^{x}dtf(t)=x^{-1/q}\int_{0}^{x}dtf(t) \\
&=&x^{-1/q}\int_{0}^{\infty }dt\theta (x-t)f(t)
\end{eqnarray*}
Next
\begin{eqnarray*}
|\int_{0}^{\infty }dt\theta (x-t)f(t)| &\leqslant &\left[ \int_{0}^{\infty
}dt\theta (x-t)^{q}\right] ^{1/q}\left[ \int_{0}^{\infty }dt|f(t)|^{p}\right]
^{1/p} \\
&=&\left[ \int_{0}^{\infty }dt\theta (x-t)^{q}\right] ^{1/q}\left[
\int_{0}^{\infty }dt|f(t)|^{p}\right] ^{1/p} \\
&=&\left[ \int_{0}^{\infty }dt\theta (x-t)\right] ^{1/q}\parallel f\parallel
==\left[ \int_{0}^{x}dt\right] ^{1/q}\parallel f\parallel =x^{1/q}\parallel
f\parallel
\end{eqnarray*}
so
\begin{equation*}
|\varphi _{f}(x)|\leqslant \parallel f\parallel
\end{equation*}
Let $\{f_{n}(x)\}$ be a sequence of compactly supported functions converging
to $f$. We have
\begin{equation*}
\varphi _{f_{n}}(x)\rightarrow 0
\end{equation*}
Now
\begin{equation*}
|\varphi _{f_{n}}(x)-\varphi _{f}(x)|\leqslant \parallel f_{n}-f\parallel
\end{equation*}
where the left hand side converges to $|\varphi _{f}(x)|$ and the right hand
side to $0$. Hence
\begin{equation*}
\varphi _{f}(x)\rightarrow 0
\end{equation*}
