If $f_m \rightarrow f$ in $L^p$, does $|f_m|^r \rightarrow |f|^r$ in $L^{\frac{p}{r}}$? Suppose that $(f_m)_m, f \subset L^p(\Omega)$ are nonnegative, such that $f_m \rightarrow f$ in $L^p$; that is:
$$
\|f_m - f\|_p \rightarrow 0
$$
It is clear that $({f_m}^r)_m, f^r \subset L^{\frac{p}{r}}(\Omega)$, since $\int |f|^p = \int|f^r|^{\frac{p}{r}}$. Can we conclude that ${f_m}^r \rightarrow f^r$ in $L^{\frac{p}{r}}$? That is:
$$
\left\|
{f_m}^r - f^r
\right\|_{\frac{p}{r}} \rightarrow 0
$$
Note: we assume $1<r<p<\infty$, and $\Omega \subset \mathbb{R}^n$ is measurable.
 A: We can use the following result: if a sequence $\left(g_n\right)_{n\geqslant 1}$  is such that $g_n\to g$ almost everywhere and $\int\left\lvert g_n\right\rvert^q\to \int\left\lvert g\right\rvert^q$ then $\int\left\lvert g_n-g\right\rvert^q\to 0$. This follows from an application of Fatou's lemma to the sequence $\left(h_n\right)_{n\geqslant 1}$ defined by $h_n=2^{q-1}\left(\left\lvert g_n\right\rvert^q +\left\lvert g\right\rvert^q\right)-\left\lvert g_n-g\right\rvert^q$. 
Extract an almost everywhere convergent subsequence $\left(f_{n_k}\right)$ of $\left(f_n\right)$, $q=p/r$, $g_k:=\left\lvert f_{n_k}\right\rvert^r$  to get that $\left\|
{f_{n_k}}^r - f^r
\right\|_{\frac{p}{r}} \rightarrow 0$. Apply this reasoning to an arbitrary subsequence of $\left(f_n\right)$ instead of of $\left(f_n\right)$ to get the wanted result. This shows that for all subsequence, you can extract a further subsequence which converges to $f^r$ in $L^{p/r}$, which gives the wanted result (note that in the link, convergence of sequences is treated but it works with the same proof for normed spaces). 
A: The answer is yes. The proof of this fact relies on the following lemma:

Let $E \subset \mathbb{R}^n$ be measurable, and $1<p<\infty$. Then the function $\phi_p(u) = |u|^{p-1} u$ defines a homeomorphism from $L^p(E)$ to $L^1(E)$. Moreover, $\forall u,v \in L^p(E)$:
$$
2^{1-p} \|u-v\|_p^p \leq \|\phi_p(u) - \phi_p(v)\|_1 \leq p (\|u\|_p+\|v\|_p)^{p-1} \|u-v\|_p
$$

This lemma follows from exercises 46-48 of Royden & Fitzpatrick's Real Analysis (4th edition), p.154 (end of chapter 7).
The result follows easily from here. Since the ${f_m}^r$ and $f^r$ are in $L^{\frac{p}{r}}$, we use the left inequality with the value $p/r$ to get:
$$
2^{1-\frac{p}{r}} {\|{f_m}^r - f^r \|_{\frac{p}{r}}}^{\frac{p}{r}}
\leq
\| \phi_{\frac{p}{r}}({f_m}^r) - \phi_{\frac{p}{r}}(f^r) \|_1
=
\| {f_m}^p - f^p \|_1
$$
Then we use the right inequality with the value $p$ to get:
$$
\|{f_m}^p - f^p\|_1
=
\| \phi_p(f_m) - \phi_p(f) \|_1
\leq
p (\|f_m\|_p+\|f\|_p)^{p-1} \|f_m-f\|_p
$$
Finally, since $f_m \rightarrow f$ in $L^p$, we can bound $\|f_m\|_p$ uniformly. Putting this together with the inequalities above, we conclude that there are constants $C_1, C_2 > 0$ such that:
$$
C_1 {\|{f_m}^r - f^r \|_{\frac{p}{r}}}^{\frac{p}{r}}
\leq
C_2 \|f_m-f\|_p
$$
So, as $m \rightarrow \infty$, we know by assumption that the right-hand side goes to $0$, and it is clear that this implies that the left-hand side goes to $0$. Hence ${f_m}^r \rightarrow f^r$ in $L^{\frac{p}{r}}$.
