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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.

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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).

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