Suffice condition for Weak convergence in $L^2$ Define 
$$ \langle f,g\rangle=\int_{\mathbb{R}}f(x)g(x)dx.$$
I know that in order to show that a sequence $\{f_n\}\in L^2(\mathbb{R})$ converge weakly to $f\in L^2(\mathbb{R})$, its suffice to show (because $L^2(\mathbb{R})$ is a Hilbert space) that 
$$\langle \phi,f_n\rangle \to \langle\phi,f\rangle$$
for all $\phi\in L^2(\mathbb{R})$.
And I also know that $C^{\infty}_{c}(\mathbb{R})$ is dense in $L^2(\mathbb{R})$.
So, if I show that 
$$\langle \varphi,f_n\rangle \to \langle\varphi,f\rangle$$
for all $\varphi \in C^{\infty}_{c}(\mathbb{R})$,
then  $\{f_n\}\in L^2(\mathbb{R})$ converge weakly to $f\in L^2(\mathbb{R})$?
 A: No
Let $f_n = n 1_{[n-1, n]}$ and $f = 0$.  If $\varphi \in C^\infty_c(\mathbb{R})$, then it is supported inside some interval $[-M, M]$.  Hence for $n \ge M+1$ we have $\langle \varphi, f_n \rangle = 0$, so $\langle \varphi, f_n \rangle \to 0 = \langle \varphi, f \rangle$.
But we do not have $f_n \to f$ weakly.  For instance, since $f_n$ are unbounded in norm, the uniform boundedness principle says they cannot converge weakly.  Or, let $g(x) = 1/x$ for $x > 1$ and $g(x) = 0$ for $x \le 1$, so that $g \in L^2$.  Then it is easy to see that $\langle g, f_n \rangle \ge 1$ for every $n > 1$, yet $\langle g, f \rangle = 0$.
However, if you add the hypothesis that $\sup_n \|f_n\| < \infty$, then your claim is true.  Suppose without loss of generality that $f=0$.  Set $M = \sup_n \|f_n\|$, let $g \in L^2$, and fix $\epsilon > 0$.  By density there exists $\varphi \in C^\infty_c$ with $\|g - \varphi\| < \epsilon$.  Now
$$\begin{align*}|\langle g, f_n \rangle| &\le |\langle g - \varphi, f_n \rangle| + |\langle \varphi, f_n \rangle| \\ &\le \|g - \varphi\| \|f_n\| + |\langle \varphi, f_n \rangle| && \text{(Cauchy-Schwarz)} \\ &\le M \epsilon + |\langle \varphi, f_n \rangle|.\end{align*}$$
Now since $\langle \varphi, f_n \rangle \to 0$ by assumption, passing to the limit as $n \to \infty$ we have
$\limsup_{n \to \infty} |\langle g, f_n \rangle| \le M \epsilon$.  But $\epsilon$ was arbitrary so $\limsup_{n \to \infty} |\langle g, f_n \rangle| = 0$.  Since $g$ was arbitrary we have shown $f_n \to 0$ weakly.
