Let $A= C([0,1])$ and $J= \{f \in A: f(0) = 0\}$. Consider the Hilbert $C^*$-module $E:= A \oplus J$ (with the obvious right $A$-action and inner product). I want to prove that $$q: E \to E: (f,g) \mapsto (f-g, 0)$$ is not adjointable. This is claimed in Lance's book on Hilbert $C^*$-modules, p22.

Here is what I tried. Assume to the contrary that $q$ is adjointable. Then there is $q^*: E \to E$ such that $$ (\overline{f-g})s= \langle q(f,g) , (s,t)\rangle = \langle (f,g), q^*(s,t)\rangle.$$

In particular, $q^*(s,t)$ does not depend on $t$ so we have $q^*(s,t) = q^*(s,0)$. Then I'm stuck.


1 Answer 1


This is just a calculation. Continuning your argument, $q^*(s,t) = q^*(s,0) = (s_1,-s_2)$ for some $s_1\in A, s_2\in J$ (I add the minus sign for convenience later). Then $$ \overline{f} s - \overline{g}s = \langle (f,g), (s_1,-s_2) \rangle = \overline{f} s_1 - \overline{g}s_2, $$ for all $f\in A, g\in J$. Set $f=1,g=0$ to see that $s = s_1$; set $f=0$ to see that $$ \overline{g} s = \overline{g} s_2, $$ for all $g\in J$. Letting $g$ run through an approximate identity for $J$ (so a net $(g_i)$ with $g_i(x)\rightarrow 1$ for each $x>0$) we conclude that $s(x) = s_2(x)$ for all $x>0$. If for example $s=1\in A\setminus J$ this shows that $s_2(x)=1$ for all $x>0$, contradicting that $s_2\in J$.

  • $\begingroup$ @MathQED Yep, fixed! $\endgroup$ Aug 15, 2021 at 8:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.