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!

  • 4
    $\begingroup$ I'm guessing the same topologies are being used on both the domain and target in each case - if this is so, then moving to a weaker topology doesn't necessarily preserve continuity. $\endgroup$ Commented Aug 23, 2013 at 14:20

4 Answers 4


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.

  • $\begingroup$ I really like your explanation, would you recommend an alteranative source on Functional Analysis to Reed and Simon's book, something similar in style to your answer? $\endgroup$
    – harlekin
    Commented Aug 24, 2013 at 9:07
  • $\begingroup$ @harlekin take a look at this question $\endgroup$
    – Norbert
    Commented Aug 24, 2013 at 9:29

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.


You could think of the following sequence of operators(it converges strongly to zero but the sequences of adjoints does not) take Sn to be the n-th power of the left shift.


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