Hilbert-Adjoint Operators

Kreyszig Functional Analysis

We first prove the following foundational theorem:

Riesz’s Theorem (Hilbert Dual Space) For any Hilbert space $H$, every bounded linear functional $f$ on $H$ (i.e. in the dual space $H^{\prime }$) can be represented in terms of

$$f(x)=⟨x,z⟩$$

where $z$ is some fixed element of $H$ that is uniquely determined by $f$ and has norm $\Vert z\Vert =\Vert f\Vert$. This defines a bijection between $H$ and its dual space $H^{\prime }$.

Proof We first think about what properties such as $z$ would satisfy if it exists. Clearly, $z$ would have to be orthogonal to the null space $\mathcal{N}(f)$. Thus, a natural starting point is taking some nonzero $z_0$ such that $z_0\perp \mathcal{N}(f)$. If such a $z_0$ doesn’t exist, then $\mathcal{N}(f{)}^{\perp }=\{0\}$ and $\mathcal{N}(f)=(\mathcal{N}(f{)}^{\perp }{)}^{\perp }$ must be the entire space, meaning $f$ is simply the zero map. In this case we can trivially see $z=0$ is the desired unique element of $H$.

Note that for any $x\in H$, $xf(z_0)-z_{0}f(x)\in \mathcal{N}(f)$. Thus, $z_0\perp (xf(z_0)-z_{0}f(x))$, and

$$0=⟨xf(z_0)-z_{0}f(x),z_0⟩=f(z_0)⟨x,z_0⟩-f(x)⟨z_0,z_0⟩\Rightarrow f(x)=⟨x,z_0⟩f(z_0)/⟨z_0,z_0⟩$$

Therefore, if we take $z=z_0\overline{f(z_0)}/⟨z_0,z_0⟩$, then for any $x\in H$

$$f(x)=⟨x,z⟩$$

Assume such a $z$ is not unique. Then, there must be two distinct $z,z^{\prime }\in H$ such that for any $x\in H$, $⟨x,z⟩=f(x)=⟨x,z^{\prime }⟩$. Thus, $⟨x,z-z^{\prime }⟩=0$ for all $x\in H$. Plugging in $x=z-z^{\prime }$ gives $\Vert z-z^{\prime }\Vert =0$, which by fundamental norm properties forces $z=z^{\prime }$. Thus, $z$ is unique.

Also, $\Vert f(z)\Vert =⟨z,z⟩=\Vert z{\Vert }^2\Rightarrow \Vert f(z)\Vert /\Vert z\Vert =\Vert z\Vert$, $\Vert f\Vert \ge \Vert z\Vert$. However, $\Vert f(x)\Vert =\Vert ⟨x,z⟩\Vert \le \Vert x\Vert \Vert z\Vert \Rightarrow \Vert f\Vert \le \Vert z\Vert$. Therefore, $\Vert f\Vert =\Vert z\Vert$.

Note this defined bijection between $H$ and $H^{\prime }$ seems like a norm space isomorphism, as it is norm-preserving. However, the inner product is sesquilinear and thus the mapping isn’t linear.

Riesz’s Theorem (Bounded Sesquilinear Forms) Let $H_1$, $H_2$ be Hilbert spaces and $h:H_1\times H_2\to K$ be a bounded sesquilinear form ($K$ is the scalar field of $H_1$ and $H_2$). Then there is a unique map $S:H_1\to H_2$ determined by $h$ such that

$$h(x,y)=⟨Sx,y⟩$$

This $S$ is a linear bounded operator with norm $\Vert S\Vert =\Vert h\Vert$. This defines a normed space isomorphism between the bounded sesquilinear forms $H_1\times H_2\to K$ and the space of bounded linear operators from $H_1$ to $H_2$, $B(H_1,H_2)$.

Proof Imagine fixing $x$ and treat $h$ as a single-variable function on $y$. Note that $\overline{h(x,y)}$ is linear in terms of $y$, and thus is a bounded linear functional on $H_2$. By Riesz’s for the dual of Hilbert spaces, there must exist some $Sx\in H_2$ such that

$$\overline{h(x,y)}=⟨y,Sx⟩\Rightarrow h(x,y)=⟨Sx,y⟩$$

For now, all we know is that $S$ is some map from $H_1$ to $H_2$. We must show $S$ is a bounded linear operator. Linearity follows from

$$⟨S(\alpha x_1+\beta x_2),y⟩=h(\alpha x_1+\beta x_2,y)=\alpha h(x_1,y)+\beta h(x_2,y)=⟨\alpha S{x}_1+\beta S{x}_2,y⟩$$

As this holds for all $y\in H_2$, this forces $S(\alpha x_1+\beta x_2)=\alpha S{x}_1+\beta S{x}_2$, proving linearity. As for boundedness, note that if we plug in $y=Sx$, then

$$\Vert h\Vert \Vert x\Vert \Vert Sx\Vert \ge h(x,Sx)=⟨Sx,Sx⟩=\Vert Sx{\Vert }^2\Rightarrow \Vert Sx\Vert \le \Vert h\Vert \Vert x\Vert$$

Thus, $S$ is bounded with $\Vert S\Vert \le \Vert h\Vert$. However, $\Vert h(x,y)\Vert =⟨Sx,y⟩\le \Vert Sx\Vert \Vert y\Vert \le \Vert S\Vert \Vert x\Vert \Vert y\Vert$ implies $\Vert h\Vert \le \Vert S\Vert$, and thus $\Vert h\Vert =\Vert S\Vert$. Uniqueness is trivial.

Hilbert-Adjoint Operator For any bounded linear operator $T:H_1\to H_2$ where $H_1$ and $H_2$ are Hilbert spaces, there exists a unique map $T^{\ast }:H_2\to H_1$ (the “Hilbert-adjoint operator”) such that for any $x\in H_1$ and $y\in H_2$,

$$⟨Tx,y⟩=⟨x,T^{\ast }y⟩$$

This $T^{\ast }$ is a bounded linear operator with norm $\Vert T^{\ast }\Vert =\Vert T\Vert$.

Proof Note that $h(y,x)=⟨y,Tx⟩$ is a sesquilinear form. $h$ must be bounded as it satisfies $\Vert h(y,x)\Vert \le \Vert y\Vert \Vert Tx\Vert \le \Vert y\Vert \Vert T\Vert \Vert x\Vert \Rightarrow \Vert h\Vert \le \Vert T\Vert$. Also, plugging in $y=Tx$ gives $\Vert h\Vert \Vert Tx\Vert \Vert x\Vert \ge \Vert h(Tx,x)\Vert =\Vert Tx{\Vert }^2\Rightarrow \Vert Tx\Vert \le \Vert h\Vert \Vert x\Vert \Rightarrow \Vert T\Vert \le \Vert h\Vert$. Thus, $\Vert T\Vert =\Vert h\Vert$.

By Riesz’s theorem there must exist some linear bounded operator $T^{\ast }:H_2\to H_1$ such that $⟨y,Tx⟩=h(y,x)=⟨T^{\ast }y,x⟩$. This $T^{\ast }$ also satisfies $\Vert T^{\ast }\Vert =\Vert h\Vert =\Vert T\Vert$. Uniqueness is trivial; we are done.

Note that for the above two proofs, uniqueness can be proven quite easily at the end. The hard part is showing the existence of some bounded linear operator that satisfies the given conditions. Thus, there is no reason to be concerned with uniqueness while going through the proof.