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.



Why is $q(f,g) = (f-g,0)$ not adjointable?
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 $s\_2\in J$.


