Different answers for adjoint of $T : \ell^2 \to \ell^2$ such that $T(x_1, x_2, \dots) = (c_1 x_1, c_2 x_2, \dots),$ where's the mistake?

Question

$c = \{c_n\} \in \ell^{\infty}$ and $T : \ell^2 \to \ell^2$ is defined as $T(\{x_n\}) = \{c_n x_n\},$ all spaces are over $\mathbb{C}.$ It's easy to prove that $||T|| = ||c||,$ but I'm getting 2 different answers for $T^*.$

According to the fact $\langle Tx, y \rangle = \langle x, T^* y \rangle,$ we get $T^*$ is the same but multiplies $x$ by $\overline{c}$ instead.

However, according to the isomorphism $F : \ell^2 \to (\ell^2)^*$ given by $F(x) = f_x$ where $f_x$ is defined by $f_x(y) = \sum\limits_n x_n y_n$ and the fact $T^* : (\ell^2)^* \to (\ell^2)^*$ is defined by $f \to f \circ T,$ we get $T^* = T.$

I'm confident about the details of the 1st approach, so here's the 2nd approach:

To compute $T^*(x),$ we apply the isomorphism and then undo it, i.e. $T^*(x) "=" F^{-1}(T^*(F(x))).$ The RHS becomes $F^{-1}(f_x \circ T).$ Since $(f_x \circ T)(y) = f_x(\{c_n y_n\}) = \sum\limits_n x_n c_n y_n = \sum\limits_n (c_n x_n) y_n = f_{T(x)}(y),$ we have $f_x \circ T = f_{T(x)}$ and the RHS simplifies to $F^{-1}(f_{T(x)}) = T(x),$ proving $T^* = T.$

Where's the mistake?

Answer 1

You are making no mistake.

Adjoint in the sense of linear operators on a Hilbert space is not the same as adjoint in the sense of Banach spaces.

What happens is that when you consider $\ell^2$ as its own dual, there is more than one way to implement the equivalence:

• one way, the "Hilbert space way" uses the duality $x\longmapsto g_x$, where $$ g_x(y)=\sum_ny_n\overline{x_n}, $$ and this gives you the Hilbert space adjoint $T^*$ (conjugate transpose if $T$ is a matrix);

• the other way, the "Banach space way" uses the duality $x\longmapsto f_x$, where $$ f_x(y)=\sum_ny_n{x_n}, $$ and this gives you the Banach space adjoint $T^*$ (transpose if $T$ is a matrix).

The latter one is simpler, but the former is way more convenient to deal with positivity of operators, among other things.