Characterizing when composition by a power function lies in an $L^p$-space This is a past qual question that I have been struggling with: let $p,q,r \geq 1$. One would like to characterize the constants $q$ such that $f(x^r) \in L^q ((0,1))$ for all $f \in L^p((0,1))$, that is, find all such $q$ for fixed constants $p,r$.
My current attempt is the following: we can apply Holder's inequality to get range of $q$'s that work: 
$$
\int_0^1 f(x^r)^q dx = \frac{1}{r} \int_0^1 \frac{f(u)^q}{u^{(r-1)/r}} du \leq \frac{1}{r} \| u^{-(r-1)/r} \|_{L^a} \| f(u)^q \|_{L^b},
$$
and this is certainly finite if  $\frac{a(r-1)}{r} < 1$ and if $qb = p$, and $1 = a^{-1} + b^{-1}$. As $r,q \geq 1$, it follows that $b \in (r,p]$ and therefore that $q \in [1,\frac{p}{r})$. 
What is unclear is whether the interval $[1,\frac{p}{r})$ consists of $\textit{all}$ such $q$. If there are indeed values of $q$ that satisfy the above condition and lie outside of this interval, how can we find them? If there are not, can we construct a function in $L^p$ to demonstrate that such $q$'s do not work?
 A: You can rule out every $q > p/r$.
For, suppose $q > p/r$, i.e., $qr > p$. Then, let $\epsilon > 0$ such that $qr > p+\epsilon$, let $f(x) = x^{-\frac{1}{p+\epsilon}}$; we then have
$$\int_0^1 f(x)^p\,dx = \int_0^1 x^{-\frac{p}{p+\epsilon}}\,dx < \infty$$
since $p/(p+\epsilon) < 1$, so $f(x) \in L^p$. On the other hand,
$$\int_0^1 f(x^r)^q\,dx = \int_0^1 x^{-\frac{qr}{p+\epsilon}}\,dx = \infty$$
since $qr > p+\epsilon$ implies $qr/(p+\epsilon) > 1$.
Now for $q = p/r$, the answer depends on $r$. If $r = 1$, then clearly $q = p/r = p$ is an allowable value of $q$. On the other hand, if $r > 1$, then I'm not quite sure…
EDIT: Here is the last case. Suppose $r > 1$. Then, consider the function
$$f(x) = \frac{\chi_{(0,1/e)}}{x^{1/p}(-\log x)^{r/p}}.$$
$f(x) \in L^p$ since
$$\int_0^1 f(x)^p\,dx = \int_0^{1/e} \frac{dx}{x(-\log x)^r} = \int_1^\infty \frac{du}{u^r} < \infty$$
by substitution. On the other hand, $f(x^r) \notin L^{p/r}$ since
$$\int_0^1 f(x^r)^{p/r}\,dx = \int_0^{1/e} \frac{dx}{x(-r\log x)} = \frac{1}{r} \int_1^\infty \frac{du}{u}$$
diverges.
