Let $(x_n)$ be a sequences in a Hilbert space $H$, with $\|x\|\leq 1$, where $x_n \rightharpoonup 0$. Let $A: H\rightarrow H$ be a linear, compact, self-adjoint projection i.e. $A^2=A$ and $A=A^*$. Then $(A x_n)$ is relativally compact i.e. any sequence from $\overline{\lbrace Ax_1, Ax_2,\ldots\rbrace}$ has a convergent subsequence.
How can I prove $\|A x_n\|\rightarrow 0$?
Note that $x_n \rightharpoonup 0$ means $x_n$ converges weakly to $0$. In a Hilbert space this is equivalent in saying $|\langle x_n, y\rangle|\rightarrow 0$ for all $y\in H$.
I thought this way:
$\|A x_n\|^2= \langle Ax_n,Ax_n\rangle=\langle x_n,Ax_n\rangle$
Since $A$ is a selfadjoint projection.
And since $\langle x_n,Ax_k\rangle\xrightarrow{n\rightarrow \infty} 0$ for all $k\in \mathbb{N}$, one could conclude that the diagonalsequence $\langle x_n,Ax_n\rangle$ also converge to $0$ for $n\rightarrow \infty$.
But do I then really need the compactness of $A$?
There are in total $5$ needless assumptions. First $||x_{n}||\leq 1$ is not needed. Firstly, A weakly convergent seuqence is always bounded and it is the boundedness that you'll end up using. Secondly, as everyone mentioned in the comments, self adjointness. And the third is that this result is true in Banach spaces. It might not be proper to call the third needless as one may end up in a Hilbert space. Fourth, The operator need not be a projection. And lastly, the sequence may converge to any point $x$.
Here's the most general form and proof.
Let $X,Y$ be normed linear spaces such that $X$ is Banach and $A:X\to Y$ be a compact operator. Then for any $x_{n}\xrightarrow{w} x$, you have that $Ax_{n}\xrightarrow{||\cdot||} Ax$.
First you show that $Ax_{n}\xrightarrow{w} Ax$. This is easy as for any $g\in Y^{*}$ , you have that $g\circ A\in X^{*}$ . Thus $g(Ax_{n})\to g(Ax)$ for all $g\in Y^{*}$.
Now consider $Ax_{n_{k}}$ to be an arbitrary subsequence of $Ax_{n}$.
Weakly convergent sequences are bounded by an use of Uniform Boundedness Principle and here's where I require $X$ to be Banach.
So by compactness, $Ax_{n_{k}}$ will have a convergent subsequence $Ax_{n_{k_l}}$ (By definition of compact operators) . But where does it converge to? You see that this sub-subsequence already converges weakly to $Ax$ and hence, if it converged in norm, it must do so to the same point $Ax$ again.
Thus given any subsequence $Ax_{n_{k}}$ there exists a further subsequence $Ax_{n_{k_{l}}}$ which converges to $Ax$. This means that the whole sequence $Ax_{n}$ converges to $Ax$.
(A sequence $x_{n}$ in a metric space converges to $x$ if and only if given any subsequence, there exists a further subsequence which converges to $x$.)