[Math] Hilbert Adjoint Operator from Riesz Representation Theorem – $T^{*}y=\frac{\left\langle y,Tx\right\rangle }{\left\langle z_{0},z_{0}\right\rangle}z_0$

Question

Kreyszig's Functional analysis seems to introduce the hilbert adjoint operator by means of an explicit representation. I haven't seen this anywhere else and I would like to confirm this explicit solution for the Hilbert adjoint operator, that is $T^*:H_2\rightarrow H_1$ such that $\langle Tx,y\rangle=\langle x, T^*y\rangle, \forall x\in H_1, y\in H_2$.

The following discussion is pieced together from 2 theorems from Kreyszig's Introduction to Functional Analysis text.

First, statements of the 2 Theorems whose results piece together into the adjoint operator. I've left out some of their details that I didn't deem relevant.

3.8-1 Riesz's Theorem (Functionals on Hilber spaces): Every bounded linear functional $h$ on a Hilbert space $H$ can be
represented in terms of the inner product, namely,
$$f\left(x\right)=\left\langle x,z\right\rangle $$ where $z$ is
unique and given by
$$z=\frac{\overline{f\left(z_{0}\right)}}{\left\langle
z_{0},z_{0}\right\rangle
}z_{0},\hspace{1em}z_{0}\in\mathcal{N}\left(f\right)^{\perp} $$

($\mathcal{N(f)}$ is the null space of $f$. $z_0$ is chosen as such so that $\langle x, z \rangle = 0 $ for all $x : f(x)=0$)

3.8-4 Riesz representation: Let $H_{1} , H_{2} $ Hilbert spaces and $h:\ H_{1}\times H_{2}\rightarrow K $ a bounded sesquilinear form.
Then $h$ has a representation $$h\left(x,y\right)=\left\langle Sx,y\right\rangle $$

The proof from the text considers $\overline{h\left(x,y\right)} $, keeps $x$ fixed (so as if $\overline{h\left(x,y\right)}$ is a function of 1 variable, $y$ ), and then applies Riesz's Theorem (above) such that $$\overline{h\left(x,y\right)}=\left\langle y,z\right\rangle ,\hspace{1em}z=\frac{h\left(x,z_{0}\right)}{\left\langle z_{0},z_{0}\right\rangle }z_{0},\hspace{1em}z_{0}\in\mathcal{N}\left(h\left(x,y\right)\right)^{\perp} $$

and taking conjugates:$$ h\left(x,y\right)=\left\langle z,y\right\rangle $$

and now we let $z=Sx$

and there we have the unique representation.

Now an explicit form for $T^{\star}$: Working backwards from the previous proof (replacing $S$ with $T^{*}$ , $x$ with $y$ ) I have $$T^{*}y=z=\frac{h\left(y,z_{0}\right)}{\left\langle z_{0},z_{0}\right\rangle }z_0,\hspace{1em}z_{0}\in\mathcal{N}\left(h\left(y,x\right)\right)^{\perp} $$

with$$h\left(y,x\right)=\left\langle z,x\right\rangle =\left\langle T^{*}y,x\right\rangle $$

which, by definition of the Hilbert adjoint operator $$\cdots=\left\langle y,Tx\right\rangle $$

so does this give us an explicit form for the adjoing operator? i.e. $$T^{*}y=\frac{\left\langle y,Tx\right\rangle }{\left\langle z_{0},z_{0}\right\rangle }z_0,\hspace{1em}z_{0}\in\mathcal{N}\left(\left\langle y,Tx\right\rangle \right)^{\perp} $$

Is this a correct "explicit" form of the adjoint operator? the hard part seems to just find the $z_0$. If so, why can't I find this form of the adjoint operator anywhere? It seems difficult to "solve" for $T^*$ straight from the definition $\langle Tx,y\rangle=\langle x, T^*y\rangle$ (like as they did in here: Finding the adjoint of an operator, or as I've seen $V^* = A^H$ proved for matrices)

Thanks for your help.

Answer 1

In case anyone's still reading this, here's an elaboration of Martin Argerami's answer. In short, your formula isn't in any way a formula for actually computing anything, but rather an encapsulation of the proof of Riesz's theorem. Fleshing things out, if $f \in H^\ast$ is non-zero:

1. On entirely abstract grounds, $\mathbb{C} = \operatorname{Ran}(f) \cong H/\operatorname{Ker}(f) \cong \operatorname{Ker}(f)^\perp$, so that $\operatorname{Ker}(f)^\perp$ is $1$-dimensional, and hence $\operatorname{Ker}(f)^\perp = \mathbb{C} z_0$ for any non-zero $z_0 \in \operatorname{Ker}(f)^\perp$, so fix some (viz, any) such $z_0$ as a basis for $\operatorname{Ker}(f)^\perp$. Note that this is all that one knows about what $z_0$ is.
2. Writing $H = \operatorname{Ker}(f) \oplus \operatorname{Ker}(f)^\perp$, we see that $f$ is entirely determined by its restriction to $\operatorname{Ker}(f)^\perp$. However, $\operatorname{Ker}(f)^\perp$ is $1$-dimensional, so by finite-dimensional linear algebra, its dual $\left(\operatorname{Ker}(f)^\perp\right)^\ast$ is also $1$-dimensional, and thus spanned by any non-zero functional, e.g., $f_0 : x \mapsto \left\langle x,z_0 \right\rangle$. Thus, there exists some constant $\alpha \in \mathbb{C}$ such that $f|_{\operatorname{Ker}(f)^\perp} = \alpha f_0$. Plugging in our distinguished (but not explicitly known!) vector $z_0$, we find that $f(z_0) = \alpha f_0(z_0) = \alpha \left\langle z_0,z_0\right\rangle$, and hence that $$\alpha = \frac{f(z_0)}{\left\langle z_0,z_0 \right\rangle}.$$
3. Putting everything together, one has that $f(x) = \left\langle x, \overline{\alpha}z_0 \right\rangle$ for $x \in \operatorname{Ker}(f)^\perp$, and hence for all $x \in H$, yielding your formula.

So, to cut a long story short, the formula you gave for $z$ depends entirely on a vector $z_0$ whose existence, guaranteed on purely abstract grounds, is the only thing we know about it. As a result, unless by some miracle you should happen to know an explicit non-zero vector in $\operatorname{Ker}(f)^\perp$, in which case you can take that as your $z_0$, the formula for $z$ is of no real computational use to you at all. In particular, it cannot possibly help you get an explicit formula for the adjoint of an operator.