Let $f$ be a function such as, for every $\alpha > 0$,
$$\omega (\alpha) = | \lbrace x \in \mathbb{R}^n : |f(x)| > \alpha \rbrace | \leq c (1 + \alpha)^{-p}$$
Prove that $f \in L^{r}(\mathbb{R}^n)$, $\;$ if $\;$ $0 < r < p$.
This is what I have done:
Knowing that: $\displaystyle || f ||^p_p = p \int^{\infty}_0 t^{r-1} \omega (t) dt$ , for $p > 0$.
We can make: $\displaystyle ||f||^{r}_{r} = r \int^{\infty}_{0} t^{r-1} \omega (t) dt$.
But for every $\;$ $t > 0$ $\;$ we got: $\displaystyle ||f||^{r}_{r} \leq r \int^{\infty}_{0} t^{r-1} \frac{c}{(1+t)^{p}} dt$.
Then: $\displaystyle ||f||^{r}_{r} \leq cr \int^{\infty}_{0} \frac{t^{r-1}}{(1+t)^{p}} dt$.
And working the expression:
$\displaystyle \frac{t^{r-1}}{(1+t)^{p}} \leq \frac{t^{r-1}}{t^{p}} = \frac{1}{t^{p-(r-1)}} \leq \frac{1}{t^s}$.
Where $s = p-(r-1) > 1$ , cause $r < p \Rightarrow r-1 < p-1 \Rightarrow 1 < p-(r-1)$.
Finally, $\displaystyle ||f||^{r}_{r} \leq \int^{\infty}_{0} \frac{1}{t^s} < \infty $.
Now, here is my doubt... is this ok? jaja