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!
 A: The polar identity states
$$
\langle Tx, y\rangle = \frac14\sum_{k=0}^3 i^k \langle T(x + i^ky), x + i^ky\rangle
$$
This is a well-known identity, which can be derived from two simpler observations:
$$
\begin{aligned}
\langle Tx, y\rangle + \langle Ty, x\rangle & = \frac12\left(\langle T(x + y), x + y\rangle - \langle T(x - y), x - y\rangle\right) \\
\langle Tx, y\rangle - \langle Ty, x\rangle & = \frac1{2i}\left(\langle T(x - iy), x - iy\rangle - \langle T(x + iy), x + iy\rangle\right)
\end{aligned}
$$
Thus, if $\langle T_n x, x\rangle\to 0$ for all $x\in\mathcal{H}$, then $\langle T_n x, y\rangle$, being a sum of four terms of the form $\langle T_n z, z\rangle$ for some $z\in\mathcal{H}$, also tends to zero. So $T_n$ converges to zero in the weak operator topology (or WOT).
On the other hand, $T_nx\to 0$ for all $x\in\mathcal{H}$ is just saying $T_n\to 0$ in the strong operator topology (or SOT).
In general, these topologies are not equivalent; the SOT is strictly finer than the WOT, which is to say that there are more SOT-open sets than WOT-open sets, or equivalently every SOT-convergent net of operators is WOT-convergent (via the Cauchy-Schwarz identity).
Consider the right-shift operator $R \in\mathcal{B}(\ell_2)$ taking $(a_1, a_2, ...)$ to $(0, a_1, a_2, ...)$. It's not too hard to check that $R^n$ converges weakly to zero, but each $R^n$ is an isometry, and so cannot converge strongly to zero.

Clarifying the discussion in the comments, I had mistakenly claimed that the WOT and SOT coincide for certain types of sets. This isn't true, and it's an easy trap to fall into (I've made this mistake before)!
What is true:

*

*If $U\subseteq\mathcal{B}(\mathcal{H})$ is bounded, then the strong and ultra-strong (resp. weak and ultra-weak) topologies on $U$ are identical: they have the same open sets.

*If $C\subseteq\mathcal{B}(\mathcal{H})$ is convex, then $\overline{C}^{WOT} = \overline{C}^{SOT}$. This is because the WOT and SOT topologies have precisely the same continuous linear functionals (in fact, any two locally convex topologies on a vector space with the same continuous linear functionals enjoy this property: see this question).

This second fact does not mean that the WOT-open and SOT-open subsets of $C$ are the same.
A: 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.)
