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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.

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1 Answer 1

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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 $s_2\in J$.

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

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