Let $f:[0,+\infty)\to \mathbb{R}$ be a continuous function satisfying $f(x)\to 0$ for $x\to +\infty$. Prove that if $f\in L^1([0,+\infty))$ then $f\in L^p([0,+\infty))$ for all $p\ge1$.
My attemp was:
We can assume w.l.o.g. that $f(x)=e^{-x}g(x)$ with $e^{-x}\in L^p([0,+\infty))$ and $g(x)\in L^1([0,+\infty))$. Compute now the $L^p$-norms of $f(x)$, we obtain $$\int\limits_0^{+\infty}|f(x)|dx=\int\limits_0^{+\infty}|e^{-x}g(x)|dx=\int\limits_0^{+\infty}e^{-x}|g(x)|dx\le||e^{-x}||_{L^\infty(0,+\infty)}\cdot||g(x)||_{L^1}=1\cdot||g(x)||_{L^1}<+\infty$$ $$\int\limits_0^{+\infty}|f(x)|^pdx=\int\limits_0^{+\infty}|e^{-x}g(x)|^pdx=\int\limits_0^{+\infty}e^{-px}|g(x)|^pdx\le||e^{-px}||_{L^q}\cdot||g(x)||_{L^p}$$
But at this point I can't conclude nothing. I thought other ways but without any solutions. Maybe the solution is really easy but I can't see it. Some hints?
That $f$ goes to zero tells you that there exists some large $M\in[0,+\infty)$ such that for all $x\ge M$, $\lvert f(x)\rvert\le1$. So for this $M$, $$\int_0^{+\infty}\lvert f(x)\rvert^p dx=\int_0^M\lvert f(x)\rvert^p dx+\int_M^{+\infty}\lvert f(x)\rvert^p dx.$$ Observe how the number $\int_0^M\lvert f(x)\rvert^p dx$ is finite. You only worry whether $\int_M^{+\infty}\lvert f(x)\rvert^p dx$ is infinite. But $\lvert f(x)\rvert^p\le\lvert f(x)\rvert$ in the region $[M,+\infty)$. So $$\int_M^{+\infty}\lvert f(x)\rvert^pdx\le\int_M^{+\infty}\lvert f(x)\rvert dx\lt+\infty.$$