Definition of weak operator topology

Question

Let $\mathcal{L}(H)$ be the space of bounded linear operators on a Hilbert space $H$ with inner product $\langle \cdot, \cdot \rangle$. For $\{T_n\}_n \in \mathcal{L}(H)$, we say that $T_n$ converges to $T \in \mathcal{L}(H)$ in the weak operator topology if

$$\langle T_n x , y \rangle \to \langle Tx,y \rangle \, , \quad \forall x, y \in H \, .$$

On the other hand, in some lecture notes, I have seen it defined by

$$\langle T_n x , x \rangle \to \langle Tx,x \rangle \, , \quad \forall x \in H \, .$$

Are these equivalent? Are these equivalent if the $T_n$'s are self-adjoint? Can you provide a proof?

Answer 1

I think I found a proof that both topologies are equivalent, even if the operators are not self-adjoint. Let me know if there is a mistake.

Assume that $\langle T_n x, x \rangle \to 0$, $\forall x \in H$ (we can take $T=0$ w.l.o.g. because the ${T_n}'s$ are linear). Then we have

• $\langle T_n(x+y), x+y \rangle = \langle T_n x ,x \rangle + \langle T_n x ,y \rangle +\langle T_n y ,x \rangle +\langle T_n y ,y \rangle \to 0$

• $\langle T_n(x-y), x-y \rangle = \langle T_n x ,x \rangle - \langle T_n x ,y \rangle -\langle T_n y ,x \rangle +\langle T_n y ,y \rangle \to 0 .$

Substracting both we have

$$\langle T_n x ,y \rangle +\langle T_n y ,x \rangle \to 0. \qquad (1) $$

Similarly,

• $\langle T_n(x+iy), x+iy\rangle = \langle T_n x ,x \rangle - i \langle T_n x ,y \rangle + i \langle T_n y ,x \rangle +\langle T_n y ,y \rangle \to 0$

• $\langle T_n(x-iy), x-iy \rangle = \langle T_n x ,x \rangle +i \langle T_n x ,y \rangle - i\langle T_n y ,x \rangle +\langle T_n y ,y \rangle \to 0 .$

Substracting both we have

$$\langle T_n x ,y \rangle -\langle T_n y ,x \rangle \to 0. \qquad (2) $$

Now add $(1)$ and $(2)$, which gives

$$\langle T_n x , y \rangle \to 0 \, , \quad \forall x, y \in H.$$

Answer 2

I know that both definitions are equivalent whenever $T_n$ and $T$ are all self-adjoint operators.

Suppose first that $H$ is a complex Hilbert space (the proofs when it is a real Hilbert space are analogous). There is one implication that is immediate, so we only need to prove that the lecture notes' definition implies the first definition given in the post.

Recall that for the inner product one has the polarization identity, which is given by: $$\left\langle u, v\right\rangle=\frac{1}{4}\sum_{k=0}^{3} i^{k}\cdot||u+i^{k}\cdot v||^2$$

Now consider the function $\beta:H\times H\to \mathbb{C}$ given by $\beta(u,v)=\langle Tu,v\rangle$. We have that $\beta$ is linear in the first coordinate and antilinear in the second coordinate (not necessarily anti-symmetric). When $T$ is self-adjoint, then it is anti-symmetric, and we can also compute $\beta$ in a similar fashion to how we can compute the inner product using the polarization identity. If $Q(u)=\beta(u,u)$ for every $u\in H$ then we have that: $$\beta(u,v) = \frac{1}{4}\sum_{i=0}^{3} i^{k} \cdot Q(u+i^{k}\cdot v)$$ which can be proved analogously to the polarization identity.

Now note that if $\beta_n(x, y)=\langle T_n x, y\rangle$ and $Q_n(x)=\beta_n(x,x)$ then we will have that $Q_n(x)\to Q(x)$, so if $\beta_n$, $\beta$ are symmetric, for example if $T_n$, $T$ are self-adjoint, we will have that: $$ \langle T_n x, y\rangle = \beta_n(x,y) = \frac{1}{4}\sum_{i=0}^{3} i^{k} \cdot Q_n(x+i^{k}\cdot y) \to \frac{1}{4}\sum_{i=0}^{3} i^{k} \cdot Q(x+i^{k}\cdot y) = \beta(x,y) = \langle Tx, y\rangle$$ which completes the proof.