$ L^p $ and inequality

Question

I am trying to solve the following problem in Measure Theory.

I assume that I have to use Hölder's Inequality but I don't see how.

Let $ E $ measurable, $m(E)<+\infty$, $1<p<+\infty$ and $ f\in L^p(E)$.

Prove that, if $ 1\leq r \leq p $, then $ f\in L^r(E)$ and $\|f\|_r \leq \|f\|_p m(E)^{\frac{1}{r} - \frac{1}{p}} $

Answer 1

You are definitely on the right track, indeed Holder inequality is all you need to apply: $$ \|f\|_r^r = \int_E |f|^r = \int_E |f|^r\cdot 1 \le \Big(\int_E |f|^{r\frac pr}\Big)^{\frac rp}\Big(\int_E 1\Big)^{1 - \frac rp} = \|f\|_p^{\frac rp}m(E)^{1 - \frac rp}. $$ Raising both sides to the power $\frac{1}{r}$ we obtain the desired inequality. Notice that here we used Holder's inequality with exponents $\frac pr$ and conjugate exponent $\frac{p}{p - r}$ (I have tacitly assumed that $r < p$, since otherwise there is nothing to prove).

Answer 2

Observe that

$$ \int_E |f|^p =\int_E 1|f|^{p} $$

Now what you want to do is to use the Holder inequality, so that, using as exponent for f $ \quad \frac{p}{r} \ \ $

$$ \int_E |f|^r \leq \left( \int_E 1^{} \right)^\frac{p-r}{r} ||f ||_p ^{p}$$