weak convergence implies boundedness.

I have these in books without proof, mostly as a corollary. I was wondering if I could get a proof.

Suppose $$\lim_{n\to \infty} \int_0^1 f_ng dx = \int_0^1 fg dx$$ for all $g\in L^2(0,1)$, where $f_n, f \in L^2(0,1)$. Then there exists a constant $K$ such that $\|f_n\|_{L^2} \leq K \lt \infty$ for all $n$.

Banach-Steinhaus tells us that a family of linear functionals on a Banach space is either unbounded on a dense $G_\delta$ set or is uniformly bounded in norm. Since this family converges weakly we have for any $g \in L^2$ that $\sup_n \langle f_n , g \rangle < \infty$ and therefore the family of functionals is uniformly bounded in norm.
• @Josh Banach-Steinhaus (aka the uniform boundedness principle) says that exactly one of two things is true: either we can find a lot of functions (in particular, one exists) in $L^2$ for which $\sup_n \langle f_n , g\rangle = \infty$ or $\|f_n\| \leq C$ for some uniform constant $C$. But the $f_n$ sequence converges weakly, so that for any $g$ $\lim_{n \to \infty} \langle f_n , g \rangle$ exists. A convergent sequence of reals is bounded and so in particular we know that the set of $g$ for which $\sup_n \langle f_n , g \rangle = \infty$ is empty, so we must have a uniform norm bound. – Chris Janjigian May 6 '12 at 21:06
• What does the symbol $\langle f_n , g\rangle$ mean? – Josh May 7 '12 at 21:35