Existence of the limit of the norm of a function $h_n = \sqrt{n} \chi_{(0,1/n)}$ in $L^2(0,1)$? Consider the sequence $h_n = \sqrt{n} \chi_{(0,1/n)}$ in $L^2(0,1)$. Then $||h_n|| = 1$ for each $n\in \mathbb{N}$. But what can we say about the limit as $n\to \infty$ of the norm of $h_n$? Does it or does it not exist? What are criteria for the existence of the limit of a norm of a function in general?
 A: The Riesz-Fischer theorem is a useful tool to identify (candidates for) $L^2$-limits:

Let $h_n$ be a sequence such that $h_n \to h$ in $L^2$. Then there exists a subsequennce $h_{n_k}$ such that $h_{n_k} \to h$ almost everywhere as $k \to \infty$.

Since $h_n = \sqrt{n} \chi_{(0,/1n)}$ converges pointwise to $0$ on the interval $(0,1)$, we find that $h := 0$ is the only candidate for the $L^2$-limit. However,
$$\|h\|_{L^2} = 0 \neq 1 = \lim_{n \to \infty} \|h_n\|_{L^2},$$
and therefore $h_n$ does not converge to $h$ in $L^2$.
This shows that the sequence $(h_n)_{n \in \mathbb{N}}$ does not converge in $L^2$.
A: If you have a bounded sequence $\{h_n\}$ in $L^2$, since $L^2$ is reflexive, you can find a subsequence $\{h_{n_k}\}$ which converges weakly to some function $h$, that is
$$\int_0^1 h_{n_k}(x)g(x)\,dx\to \int_0^1 h(x)g(x)\,dx$$ for every $g\in L^2$. 
Since $\{h_{n_k}\}$ converges weakly to $h$ in $L^2$, you always have that
$$\Vert h\Vert_{L^2}\le\liminf_{k\to\infty}\Vert h_{n_k}\Vert_{L^2}$$ but in general you have strict inequality as your example shows. 
However, since $L^2$ is uniformly convex (see uniform convexity), if it happens that
$$\Vert h\Vert_{L^2}=\lim_{k\to\infty}\Vert h_{n_k}\Vert_{L^2},$$
then actually $$\lim_{k\to\infty}\Vert h_{n_k}-h\Vert_{L^2}.$$
So weak convergence in $L^2$ and existence of the limit of the norm is equivalent to strong convergence. 
