Confusion about equivalent definitions of adjoint linear operators

Question

The definition of an adjoint I have is that if $(V,|| \ ||_V)$ and $(W,|| \ ||)_W)$ are normed vector spaces, and $T:V\to W$ is a bounded linear operator, then the adjoint of $T$ is $T^*:W^*\to V^*$ given such that $(T^*\varphi)(v)=(\varphi\circ T)(v)$, where $V^*$ and $W^*$ are the duals of $V$ and $W$ respectively, and $\varphi$ is a linear functional.

From this, I am told we are able to show that for Hilbert spaces, $H_1$ and $H_2$, the adjoint of the linear operator $T:H_1\to H_2$, is such that $\langle Tx,y\rangle_2=\langle x,T^*y\rangle_1$ if $x\in H_1$ and $y\in H_2$.

I cannot seem to link the two definitions at all, and I feel as if I am missing something. Any help would be greatly appreciated.

Answer 1

In fact, these are two different - though closely related - things. I prefer to call the thing "dual" or "transpose" for normed spaces to avoid that confusion. So let's call it the transpose for normed spaces and denote it by ${}^tT$ rather than $T^{\ast}$, and reserve the name adjoint and the notation $T^{\ast}$ for the map characterised by $\langle Tx,y\rangle_2 = \langle x, T^{\ast} y\rangle_1$ for $x\in H_1,\, y\in H_2$ for now.

Then for the linear operator $T \colon H_1 \to H_2$ we have

1. the transpose ${}^tT \colon H_2^{\ast} \to H_1^{\ast}$, and
2. the adjoint $T^{\ast} \colon H_2 \to H_1$.

These are maps between different spaces, and so different maps.

However, for Hilbert spaces, we have an identification between the space and its dual by the Riesz map,

$$R_H(x) = \langle\,\cdot\,,x\rangle_H = y \mapsto \langle y,x\rangle_H.$$

(Assuming the mathematical convention that the inner product is linear in the first and antilinear in the second argument. Make it $R_H(x) = \langle x,\,\cdot\,\rangle_H$ for the physicists' convention.)

The Riesz map is linear in the case of real Hilbert spaces, and antilinear for complex Hilbert spaces, and the adjoint and the transpose of $T$ are connected via the Riesz maps of $H_1$ and $H_2$, namely we have

\begin{align} \langle Tx, y\rangle_2 &= R_{H_2}(y)(Tx) = \Bigl({}^t T\bigl(R_{H_2}(y)\bigr)\Bigr)(x),\\ \langle x, T^{\ast} y\rangle_1 &= R_{H_1}\bigl(T^{\ast}y\bigr)(x), \end{align}

for all $x\in H_1, \, y\in H_2$, so

$${}^tT\bigl(R_{H_2}(y)\bigr) = R_{H_1}\bigl(T^{\ast}y\bigr)$$

for all $y\in H_2$, and finally

$$T^{\ast} = R_{H_1}^{-1}\circ {}^tT \circ R_{H_2}.$$

Note that the map $T \mapsto {}^tT$ is always linear, while the map $T \mapsto T^{\ast}$ is antilinear for complex Hilbert spaces.