Continuity of the adjoint map in various operator topologies

Uniform topology is determined by single operator norm $\Vert\cdot\Vert$ given by $\Vert T\Vert=\sup\{\Vert Tx\Vert: x\in\operatorname{B}_X\}$. Let $(T_i:i\in I)$ be any net convergent to $T$ in the uniform topology, then we have $$ \lim_{i}\Vert T_i^*-T^*\Vert =\lim_{i}\Vert (T_i-T)^*\Vert =\lim_{i}\Vert T_i-T\Vert=0 $$ Hence $^*:\mathcal{B}(H)\to\mathcal{B}(H)$ is continuous in the uniform topology.

Weak topology is determined by family of seminorms $\{\Vert\cdot\Vert_{x,y}:x,y\in H\}$ given by $\Vert T\Vert_{x,y}=|\langle Tx,y\rangle|$. Let $(T_i:i\in I)$ be any net convergent to $T$ in the weak topology. For any $x,y\in H$ we have $$ \lim_{i}\Vert T_i^*-T^*\Vert_{x,y} =\lim_{i}|\langle(T_i^*-T^*)x,y\rangle| =\lim_{i}|\langle x,(T_i-T)y\rangle|\\ =\lim_{i}\Vert T_i-T\Vert_{y,x} =0 $$ Hence $^*:\mathcal{B}(H)\to\mathcal{B}(H)$ is continuous in the weeak topology.

Strong topology is determined by family of seminorms $\{\Vert\cdot\Vert_x:x\in H\}$ given by $\Vert T\Vert_x=\Vert Tx\Vert$. For a given $x,y\in H$ we define the operator $x\bigcirc y:H\to H:z\mapsto \langle z,y\rangle x$. One can easily check that $(x\bigcirc y)^*=y\bigcirc x$. Let $\{e_n:n\in\mathbb{N}\}\subset H$ be an orthonormal family. For any $x\in H$ we get $$ \lim_{n}\Vert(e_1\bigcirc e_n)\Vert_{x} =\lim_{n}|\langle x, e_n\rangle| =0 $$ Note that the last equality is the consequence of Bessel's inequality. On the other hand $$ \lim_{n}\Vert(e_1\bigcirc e_n)^*\Vert_{x} =\lim_{n}\Vert(e_n\bigcirc e_1)\Vert_{x} =|\langle x, e_1\rangle| $$ which is not $0$ in general. Thus the map $^*:\mathcal{B}(H)\to\mathcal{B}(H)$ is not even sequentially strongly continuous.

If $T:X\to Y$ is continuous, then $T$ remains continuous if $X$ is given a finer topology or $Y$ is given a coarser topology. But if both topologies are made coarser or both finer, nothing can be said in general. In particular, if $T:X\to X$ is continuous with respect to a given topology on $X$ in both domain and codomain, you cannot generally conclude anything about continuity of $T$ when $X$ is given a finer or coarser topology on both domain and codomain. Your example illustrates this.

Here is another example to see that the adjoint is not (sequentially) continuous in the SOT: Let $(e_n)_{n=0}^\infty$ be an orthonormal basis for a Hilbert space $H$, and let $S$ be the unique bounded operator on $H$ such that $Se_n=e_{n+1}$ for all $n$. Then $S$ is an isometry sometimes called a unilateral shift. The sequence $({S^*}^n)$ converges in the SOT to the zero operator, but the sequence $(S^n)=(({{S^*}^n})^*)$ does not converge in the SOT.

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.