Convergence in norm and in mean Here is my problem: If $\{f_k\}$ is a sequence in $L^2$ and $f_k\to f$ in mean, show that $\{||f_k||_2\}$ is a bounded sequence of real numbers.
Before I start doing the problem, I would like to know the difference between convergence in mean and convergence in norm.
To me, they are very similar.
Convergence in mean: $\displaystyle \lim_{n\to\infty} \int (f-f_n)^2 \,\mathrm{d}x = 0$
Convergence in norm: $\displaystyle \lim_{n\to\infty} ||f-f_n||_2 = 0$, which means $\displaystyle \lim_{n\to\infty} \left( \int |f-f_n|^2 \,\mathrm{d}x \right)^\frac{1}{2} = 0$
Are convergence in mean and in norm different? What are some counter-examples?
Thank you.
 A: $\quad$ I understand that I am too late to help you solve your problem. But just in case someone comes here and see this post, I want to rectify a mistake that has been made all along. 
$\quad$ The definition you give for convergence in mean is wrong. $\left( f_k \right)$ converges in mean to $f$ signifies that
\begin{equation*}
\lim_{k\to+\infty} \int \left\vert f-f_k \right\vert dx \;=\; 0
\end{equation*}
This is equivalent to convergence in $L^1$. 
$\quad$ You are right about convergence in $L^2$.
A: For real valued functions, both notions you write are the same, because if we set $x_n := \int |f-f_n|^2 \, dx$, then convergence in mean means $x_n \to 0$, while convergence in ($L^2$)-norm means $\sqrt{x_n}\to 0$.
Since $x_n\geq 0$, these statements are equivalent (use continuity of the square root and of $x \mapsto x^2$).
Finally, continuity of the norm yields $\Vert f_n\Vert_2 \to \Vert f\Vert_2$, where $\Vert f \Vert_2 \leq \Vert f-f_n\Vert_2 + \Vert f_n\Vert_2<\infty$, by the triangle inequality. But convergent sequences of real numbers are bounded.
Alternatively, you can use that $\Vert f_n \Vert_2 \leq \Vert f-f_n \Vert_2 + \Vert f \Vert_2$, and $(\Vert f- f_n\Vert_2 )_n$ is a null-sequence, hence bounded.
