Lp spaces (Hölder, Minkovski)

Question

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?

Answer 1

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*}