Let $H$ be an infinite-dimensional Hilbert space and $\mathcal{B}(H)$ the set of bounded linear operators on $H$.

One way to define the strong operator topology (SOT) on $\mathcal{B}(H)$ is by specifying how nets converge: if $\{T_\lambda\}_{\lambda\in\Lambda}$ is a net in $\mathcal{B}(H)$ and $T\in\mathcal{B}(H)$, then $T_\lambda\to T$ if $$T_\lambda(x)\to T(x)\tag{1}\label{convergence}$$ in $H$ for all $x$.

This appears to be related to the general notion of "topology of pointwise convergence" for a family $F$ of functions $X\to Y$, where $X$ and $Y$ are topological spaces. This is the subspace topology that $F$ inherits from the space of all functions $Y^{|X|}$ equipped with the product topology.

The following two questions are kind of sanity checks to see if I've understood things correctly.

Question 1: Is the SOT on $\mathcal{B}(H)$ the topology of pointwise convergence on the set $\mathcal{B}(H)\subseteq H^{|H|}$, where $H$ is equipped with the norm topology?

Now let $\mathcal{L}(H)$ be the set of all linear maps $H\to H$ (not necessarily bounded). Then we could also put a "strong operator topology" on $\mathcal{L}(H)$ in the same way, using $\eqref{convergence}$. To distinguish this topology, let's call it SOT$_{\mathcal{L}(H)}$. Then it seems that the SOT on $\mathcal{B}(H)$ is really the subspace topology inherited from $(\mathcal{L}(H),\text{SOT}_{\mathcal{L}(H)})$.

Question 2: Is $\mathcal{B}(H)$ closed in $(\mathcal{L}(H),\text{SOT}_{\mathcal{L}(H)})$?

I suspect the answer is no, from looking at this answer.

Remark: I wanted to check this because it seems to be a slightly subtle fact that (by the uniform boundedness principle), pointwise convergence (at all $x\in H$) of a sequence $T_n\in\mathcal{B}(H)$ implies that the operator $T$ defined by $T(x)=\lim_{n\to\infty}T_n(x)$ is bounded, whereas this is not true for a general net $\{T_\lambda\}_{\lambda\in\Lambda}$. So one cannot really detect boundedness of the operator $T$ using nets as in \eqref{convergence}, but rather it should be thought of as a characterisation of convergence of nets when one already knows the limit $T$ is bounded.


1 Answer 1


Answer to question 1:

Yes. A net $f_\alpha$ is convergent pointwise to some $f$ iff $f_\alpha(x)\to f(x)$ for all $x$. This is precisely how you have characterised SOT.

Answer to question 2:

No. Let $L$ be an arbitrary linear map. Let $\mathcal V$ denote the set of finite dimensional subspaces of $H$. The relation $V\subseteq W$ makes $\mathcal V$ into a directed set as can be checked easily. For any $V\in\mathcal V$ let $p_V$ denote the orthogonal projection to $V$.

Then the net $L_V:= p_VLp_V$ converges pointwise to $L$, since for any $x\in H$ there is a $V\in\mathcal V$ with $x\in V$ and $L(x)\in V$, then for all $W$ with $V\subseteq W$ you have $L_W(x)=L(x)$ and $L_V\to L$ pointwise.

Further if you write $H=V\oplus V^\perp$ then $L_V=p_VL p_V$ operates only on the first factor, which is finite dimensional. So $L_V$ is continuous.

So $\mathcal B(H)$ is dense in $\mathcal L(H)$ in pointwise convergence (infact this shows already that the finite rank linear maps are dense).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .