Convergence of functions in $L^p$ Let $\{f_k\} \subset L^2(\Omega)$, where $\Omega \subset \mathbb{R}^n$ is a bounded domain  and suppose that $f_k \to f$ in $L^2(\Omega)$. Now if $a \geq 1$ is some constant, is it possible to say that $|f_k|^a \to |f|^a$ in $L^p$ for some $p$ (depending on $a$ and also possibly depending on $n$)? 
Showing the statement is true would probably require a smart way of bounding $\left| |f_k|^a - |f|^a \right|$ by a term including the factor $|f_k - f|^2$. However, I don't really know what to do with the fact that $a$ doesn't have to be an integer...
 A: Some answers for $a = 1, 2$:
If $a = 1$, then $\big||f_k| - |f|\big| \leq |f_k - f|$, so you have convergence in $L^2$. And you'd also have convergence in $L^p$ for $p < 2$ since $L^p$ norms increase with $p$ on bounded domains. But never for $p > 2$ since $|f|^{p}$ needs to be integrable.
If $a = 2$, then you can use $|f_k^2 - f^2| = |f_k - f| |f_k + f|$. So by Cauchy Schwarz 
$$\int_{\Omega}|f_k^2 - f^2| \leq (\int_{\Omega}|f_k - f|^2)^{1 \over 2} (\int_{\Omega}|f_k + f|^2)^{1 \over 2}$$
The left factor goes to zero by assumption, and the right factor is bounded by a constant since $|f_k + f|^2 \leq 2|f_k|^2 + 2|f|^2$. Thus the product on the right goes to zero, and the same therefore must be true for the product on the left. So $f_k^2$ goes to $f^2$ in $L^1$. It can't coverge in any higher $L^p$ since $f$ is not assumed to be in $L^{2p}$ for any $p > 1$.
A: First, one cannot expect better results than $p=\frac 2 a$ because we only know $f\in L^2(\Omega)$. And I do think it is true for $p=\frac 2 a$. Proof is as follows. 
Note that $x^a$ is a convex increasing function for $x\ge 0$, hence (draw a picture and you can see this)
$$0\le \frac{u^a-v^a}{u-v}\le a\max\{u^{a-1},v^{a-1}\}, \forall u, v\ge 0, u\neq v.$$
Plugging in $u=|f_k|, v=|f|$ and noticing that $||f_k|-|f||\le |f_k-f|$, we have
$$||f_k|^a-|f|^a|\le a|f_k-f|\max\{|f_k|^{a-1},|f|^{a-1}\}.$$
Then, raising the last inequality to the power $p=\frac 2 a$, by Hölder's inequality, 
$$\||f_k|^a-|f|^a\|_{L^p}^p\le a^p \|f_k-f\|_{L^2}^{2/p}\|\max\{|f_k|^{a-1},|f|^{a-1}\}\|_{L^{\frac{2}{a-1}}}^{\frac a {a-1}}.$$
(When we apply Hölder, $|f_k-f|^p\in L^a,$ and $\max\{|f_k|^{a-1},|f|^{a-1}\}^p\in L^r, r=a^*=\frac a {a-1}.$ You have to check the exponents to see if I made any mistake.)
Since $f_k$ have bounded $L^2$-norm (they converge),$$\|\max\{|f_k|^{a-1},|f|^{a-1}\}\|_{L^{\frac{2}{a-1}}}$$
is bounded by some constant. Sending $k\to\infty$, we have $|f_k|^a\to|f|^a$ in $L^{2/a}$.

A:  A partial answer . 
Fix $a\geq 1$ and define $\varphi(x)=x^a$. By the Mean value Theorem we have 
$$
|\varphi(|f_k|)-\varphi(|f|)|\leq 
a \sup_{t\in[0,1]} \big| t|f_k|+(1-t)|f|\big |^{a-1}\cdot \big| |f_k|- |f|\big|
$$
Assume that $f_k$ and $f$ are bounded. From the previous inequality we get that 
$$
|\varphi(|f_k|)-\varphi(|f|)|\leq 
M \big| |f_k|-|f|\big|\leq M|f_k-f|
$$
the last inequality is obtained by the second triangular inequality.
Taking square in both sides we end up with
$$
||f_k|^a-|f|^a|^2\leq \tilde{M} |f_k-f|^2
$$
therefore you can chose $p=2$ and you have the convergence you are looking for.
Right now I don't know how to remove the strong assumption I made about the boundedness of $f_k$ and $f$. 
Perhaps other participants of this forum can show us how to remove this assumption or give us a better argument. 
