Bounded operators: $\langle T_n (x),x \rangle \longrightarrow 0 \implies T_n (x) \rightarrow 0$? Suppose $(T_n)_{n=1}^\infty$ is a sequence of bounded operators on a Hilbert space $\mathcal{H}$ and that for every $x \in \mathcal{H}$
$$
\langle T_n (x),x \rangle \longrightarrow 0 \quad \text{as } n \to \infty.
$$
Is it then true that $T_n (x) \rightarrow 0$ as $n \to \infty$ for every $x \in \mathcal{H}$?

I am not sure whether or not this is true. One the one hand it could be that $T_n (x)$ converges to something in the orthogonal complement of a subsapce containing $x$ but also this seems odd if this holds for any $x \in \mathcal{H}$. We know that the inner-product is continuous in the first entry but I can't see that this is useful here to show the claim.
Any help towards a proof, counter-example or reference is much appreciated!

For seperable real Hilbert spaces there is a counter-example: Let $T_n \equiv T$ such that $Te_{2i}=e_{2i+1} ,Te_{2i+1}=-e_{2i}$ for $i \ge 0$, where $\{ e_i \}_{i \ge 0}$ is an orthonormal base. Then clearly $T_n(x) \equiv T(x) \ne 0$ but $\langle T_n(x),x \rangle \equiv \langle T(x),x \rangle = 0$.
For inseperable real Hilbert spaces similar construction can be made if there is a convolution with no fixed point on the orthonormal base, but I'm afraid some set-theoretic methods will be involved.
Another Construction: Choose a two-dimensional subspace $V$ of the ambient Hilbert space. Suppose $a,b$ is an orthonormal base of $V$. Define the bounded linear map $T$ to be $Ta=b,Tb=-a,T|_{V^{\perp}}=0$ and take $T_n \equiv T$. This works for every real Hilbert space.
For the complex case, as @user3002473 pointed out, a counter-example can be taken to be a WOT-null operator sequence that is not SOT-null. So in the seperable case a counter-example is $T_n=R^n$, where $Re_i=e_{i+1}$ is the right-shift operator. (This apparently also works for the real case.)
