Weak convergence in $L^2$ implies boundness Let $f$, $f_n \in L^2$, we say that $f_n \rightharpoonup f$ converges weakly in $L^2$ if
$$\lim_{n \to \infty} \langle (f_n-f),g \rangle_{L^2}=0\,\,\,\forall\,\,\, g \in L^2. $$
I know that using the Uniform boundedness principle ( Banach-Steinhaus Theorem) the sequence $(\lVert f_n \rVert_2)$ is bounded.
But I wish to know a way to prove this boundness without using the uniform boundness principle. Given the particularity of the situation, I think that is not that hard, but nevertheless I still need help, thank you in advance!
I was able to prove, using Riesz- Fréchet representation theorem that for each linear functional $\varphi:L^2 \to \mathbb{R}$, the sequence $(\varphi(f_n))_n$ is bounded. There exist some particular $\varphi \in L^2$ which implies the boundness of the sequence $(\lVert f_n \rVert_2)$ ?
 A: I found the solution of my problem in the book Concentration Compactness by Kyril Tintarev and Karl - Heinz Fieseler. The Theorem that solves the problem is the following:
Theorem: A sequence $(u_n)$ in a Hilbert space $H$ is bounded if and only if for any $w \in H$ the sequence $\langle w,u_n \rangle$ is bounded.
Proof: ($\implies$) This is an immediate consequence of the Cauchy-Schwarz inequality $|\langle w,u_n \rangle| \leq \lVert w \rVert \lVert u_n \rVert.$
($\impliedby$) For an unbounded sequence $(u_n)$, we construct a vector $w \in H$, such that the sequence $\langle w,u_n \rangle$ is unbounded.
We may assume $\lim_{n \to \infty} \lVert u_n\rVert=+\infty$  or even $\lVert u_n\rVert=4^n$ resp. $u_n=4^nv_n$ with vectors $v_n$ of lenght $1$. Namely, for any $n \in \mathbb{N}$ there is a $k_n$ such that $\lVert u_{k_n} \rVert \geq 4^n$, then replace $u_n$ with the sequence $û_n := 4^n \lVert u_{k_n} \rVert^{-1}u_{k_n}.$ We define
$$w:=\sum_{k=1}^{\infty} \sigma_k 3^{-k}v_k$$
where $\sigma_k=\pm 1$ is inductively chosen: We take $\sigma_1:=1$, and given $\sigma_1,\dots, \sigma_{k-1}$ we let
$$\sigma_k := \text{sign} \left ( \sum_{i=1}^{k-1} \sigma_i 3^{-i} \langle v_i, v_k \rangle\right) $$
using the convention $\text{sign(0)=1}$. Now
$$ \langle w,u_n  \rangle=4^n\sum_{i=1}^{\infty} \sigma_i3^{-i}\langle v_i,v_n \rangle$$
and thus
\begin{align*}
|\langle w,u_n \rangle| &\geq 4^n \left(\left| \sum_{i=1}^{n} \sigma_i 3^{-i} \langle v_i,v_n \rangle \right|-\sum_{i=n+1}^{\infty} 3^{-i}|\langle v_i,v_n \rangle| \right)\\
&\geq \left( \frac{4}{3}\right)^n-4^n\left( \sum_{i=n+1}^{\infty} 3^{-i}\right)\\
&= \left( \frac{4}{3} \right)^n \left( 1-\frac{1}{3}\frac{1}{1-\frac{1}{3}}\right)=\frac{1}{2}\left(\frac{4}{3}\right)^n. \square
\end{align*}
Corollary: Every weakly convergent sequence in a Hilbert space is bounded.
