Vrftjkry

Let $f:[0,+infty)to mathbb{R}$ be a continuous function satisfying $f(x)to 0$ for $xto +infty$. Prove that if $fin L^1([0,+infty))$ then $fin L^p([0,+infty))$ for all $pge1$.

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
$$intlimits_0^{+infty}|f(x)|dx=intlimits_0^{+infty}|e^{-x}g(x)|dx=intlimits_0^{+infty}e^{-x}|g(x)|dxle||e^{-x}||_{L^infty(0,+infty)}cdot||g(x)||_{L^1}=1cdot||g(x)||_{L^1}<+infty$$
$$intlimits_0^{+infty}|f(x)|^pdx=intlimits_0^{+infty}|e^{-x}g(x)|^pdx=intlimits_0^{+infty}e^{-px}|g(x)|^pdxle||e^{-px}||_{L^q}cdot||g(x)||_{L^p}$$

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 $Min[0,+infty)$ such that for all $xge M$, $lvert f(x)rvertle1$. So for this $M$, $$int_0^{+infty}lvert f(x)rvert^p dx=int_0^Mlvert f(x)rvert^p dx+int_M^{+infty}lvert f(x)rvert^p dx.$$ Observe how the number $int_0^Mlvert f(x)rvert^p dx$ is finite. You only worry whether $int_M^{+infty}lvert f(x)rvert^p dx$ is infinite. But $lvert f(x)rvert^plelvert f(x)rvert$ in the region $[M,+infty)$. So $$int_M^{+infty}lvert f(x)rvert^pdxleint_M^{+infty}lvert f(x)rvert dxlt+infty.$$