TITLE: Continuity of the adjoint map in various operator topologies
QUESTION [11 upvotes]: I am currently reading about operator topologies in the book "Methods of Modern Mathematical Physics: Functional Analysis" by Reed and Simon. 
In their treatment of the Hilbert space adjoint, a theorem (Theorem VI.3) states that the adjoint map $T \mapsto T^*$ is always continuous in the weak and uniform topology, but is continuous in the strong operator topology only if the Hilbert space is finite-dimensional.
However, a previous comment (p.184) states that the weak operator topology is weaker than the strong operator topology which in turn is weaker than the uniform topology. Doesn't this mean that the continuity of the adjoint map in the uniform operator topology implies its continuity in the strong operator topology?
Unfortunately the proof for the theorem above is not really helpful for the beginner because it states that continuity in the uniform and weak topology are trivial. An accessible counterexample, however, is provided to demonstrate the case where continuity in the strong operator topology fails.
Thanks for helping me out of my confusion!

REPLY [2 votes]: Although this question was sufficiently answered long ago let me slightly generalize the problem (for whomever may come across this in the future):

When is the map ${}^*:\mathcal B(H,G)\to\mathcal B(G,H)$ continuous (in the respective operator topologies) when considering different (but of course non-trivial) Hilbert spaces $G,H$?


As Norbert showed in his answer ${}^*$ is continuous on $\mathcal B(H)$ in the operator norm as well as the weak operator topology---this continues to hold for arbitrary Hilbert spaces $H,G$ as is easily seen.
Interestingly enough ${}^*:\mathcal B(H,G)\to\mathcal B(G,H)$ is continuous in the strong operator
topology if and only if $H$ is finite-dimensional (so there are
no assumptions on $G$). Indeed if $H$ is infinite-dimensional then the counterexample Norbert gave is still valid. An easy way to see the converse is via the following facts:


Given non-trivial normed spaces $X,Y$
(a) the strong operator topology coincides with the norm topology on $\mathcal B(X,Y)$ if and only if $\operatorname{dim}(X)<\infty$
(b) the weak operator topology coincides with the strong operator topology on $\mathcal B(X,Y)$ if and only if $\operatorname{dim}(Y)<\infty$.

Thus asking about "strong continuity" of ${}^*$ for $\operatorname{dim}(H)<\infty$ is the same as asking about continuity of $${}^*:(\mathcal B(H,G),\|\cdot\|)\to(\mathcal B(G,H),\text{weak op.})\,.$$
But this map is a composition of ${}^*:(\mathcal B(H,G),\|\cdot\|)\to(\mathcal B(G,H),\|\cdot\|)$ (* is always norm-continuous) and $\operatorname{id}:(\mathcal B(G,H),\|\cdot\|)\to(\mathcal B(G,H),\text{weak op.})$ (continuous because weak op. is well-known to be weaker than norm top.), hence continuous itself.