Convergence in $L^p$ by using Holder's inequality

Question

Let $1\lt p \lt \infty$ and $f\in L_p[0,\infty )$. Show that

a) $$\left\vert\int_0^x f(t)\,dt\right\vert\le\|f\|_px^{1-\frac{1}{p}},$$ for $x\gt 0$.

b) $$\lim_{x\to \infty} \frac{1}{x^{1-\frac1{p}}}\int_0^xf(t)\,dt=0.$$

For a) just I used Holder's inequality and got the result but for b) I'm not able to figure out. I tried same Holder here too but I got $\le |f(t)|$ instead of $0$. Could you please just give me some hints for b)? Thanks in advance.

Answer 1

Use an approximation argument: it is true when $f$ is a simple function (linear combination of characteristic function), and for any $s$ simple and each positive $x$, $$x^{-(1-1/p)}\int_0^x|f(t)|\mathrm dt\leqslant \lVert f-s\rVert_p+x^{-(1-1/p)}\int_0^x|s(t)|\mathrm dt.$$

Answer 2

And another proof:

Let $\epsilon>0$ and choose $a$ large enough so that $\|f 1_{[a,\infty)} \|_p < {\epsilon \over 2}$.

Now choose $L \ge a$ large enough so that $|\frac{1}{x^{1-\frac1{p}}}\int_0^a f(t)\,dt | < {\epsilon \over 2}$ whenever $x \ge L$.

Then we have $|\frac{1}{x^{1-\frac1{p}}}\int_0^x f(t)\,dt | = |\frac{1}{x^{1-\frac1{p}}}\int_0^a f(t)\,dt | + |\frac{1}{x^{1-\frac1{p}}}\int_a^x f(t)\,dt | < \epsilon$, from which the result follows (since $|\frac{1}{x^{1-\frac1{p}}}\int_a^x f(t)\,dt | \le \|f 1_{[a,\infty)} \|_p$).