My question is the following:
Let $A$ be a bounded linear operator on a Hilbert space $\mathcal{H}$, and $x_n$ be a bounded sequence in $\mathcal{H}$ such that $(Ax_n)$ converges strongly in $\mathcal{H}$ to some element $y$ in $\mathcal{H}$. Suppose also that $A$ is injective, show that there exists an $x \in \mathcal{H}$ such that $(x_n)$ converges weakly to $x$ and $Ax = y$.
I tried the Riesz representation and found that we can conclude that $Ax=y$ from the fact that $A$ is injective. The only issue here is to prove that there exists an $x \in \mathcal{H}$ such that $(x_n)$ converges weakly to $x$. Need any hint or help.
By Banach Alaouglu, there exists a weakly convergent subsequence $\{x_{n_k}\}$ and an $x$, with $x_{n_k}\to x$ weakly. Then, for any $z\in H$, $$ \langle Ax_{n_k}-Ax,z\rangle=\langle A(x_{n_k}-x),z\rangle=\langle x_{n_k}-x,A^*z\rangle\xrightarrow[k]{}0. $$ So $Ax_{n_k}\to Ax$ weakly. We also have $Ax_{n_k}\to y$, so $y=Ax$.
As $A$ is injective, $$ \{0\}=\ker A=(\operatorname{ran} A^*)^\perp. $$ So $A^*$ has dense range. For $z\in H$, $$\tag1 |\langle x_n-x,A^*z\rangle|=|\langle Ax_n-Ax,z\rangle|=|\langle Ax_n-y,z\rangle|\leq \|Ax_n-y\|\,\|z\|\xrightarrow[n]{}0. $$ Given any $w\in H$ and $\varepsilon>0$, there exists $z\in H$ with $\|w-A^*z\|<\varepsilon$. Let $c=\sup\|x_n\|$. Then $$ |\langle x_n-x,w\rangle|\leq|\langle x_n-x,A^*z\rangle|+|\langle x_n-x,w-A^*z\rangle| \leq |\langle x_n-x,A^*z\rangle|+2c\varepsilon. $$ So, using $(1)$, $$ \limsup_n|\langle x_n-x,w\rangle|\leq2c\varepsilon. $$ As $\varepsilon$ was arbitrary, we get that $\lim_n\langle x_n-x,w\rangle=0$, so $x_n\to x$ weakly.