Adjoint Operators and Reflexivity

Kreyszig Functional Analysis

Here are a collection of randomly assorted yet critically important results from Kreyszig 4.5, 4.6

Theorem 1. Let $T:X\to Y$ be a bounded linear operator between normed spaces. Its adjoint operator $T^{\times }:Y^{\prime }\to X^{\prime }$ is a bounded linear operator with norm $\Vert T^{\times }\Vert =\Vert T\Vert$.

Proof. Note that $T^{\times }$ must be linear as for any $g_1,g_2\in Y^{\prime }$ and $x\in X$,

$$(T^{\times }(\alpha g_1+\beta g_2))(x)=(\alpha g_1+\beta g_2)T(x)=\alpha g_{1}T(x)+\beta g_{2}T(x)=(\alpha T^{\times }{g}_1+\beta T^{\times }{g}_2)(x)$$

$T^{\times }$ is also bounded as for any $g\in Y^{\prime }$ and $x\in X$,

$$\Vert (T^{\times }g)(x)\Vert =\Vert gT(x)\Vert \le \Vert g\Vert \Vert T\Vert \Vert x\Vert \Rightarrow \Vert T^{\times }g\Vert \le \Vert g\Vert \Vert T\Vert \Rightarrow \Vert T^{\times }\Vert \le \Vert T\Vert$$

We want to show that $\Vert T^{\times }\Vert =\Vert T\Vert$, and so we want to find some $g\in Y^{\prime }$ and $x\in X$ that makes the inequalities $\Vert (T^{\times }g)(x)\Vert =\Vert gT(x)\Vert \le \Vert g\Vert \Vert Tx\Vert$ and $\Vert Tx\Vert \le \Vert T\Vert \Vert x\Vert$ airtight. Then, we obtain $\Vert g\Vert \Vert T\Vert \Vert x\Vert =(T^{\times }g)(x)\le \Vert T^{\times }\Vert \Vert g\Vert \Vert x\Vert \Rightarrow \Vert T\Vert \le \Vert T^{\times }\Vert$. Note the intuition is that if $g$ and $T$ scale in the same direction, then their composition should satisfy $\Vert T^{\times }g\Vert =\Vert gT\Vert =\Vert g\Vert \Vert T\Vert$ and thus $\Vert T\Vert \le \Vert T^{\times }\Vert$.

Note that for any nonzero $y\in Y$, by Hahn-Banach there exists some functional $g_y\in Y^{\prime }$ such that $\Vert g_y\Vert =1$ and $g_y(y)=\Vert y\Vert$. Thus, for any $x\in X$, we can take $g_{Tx}$ such that $\Vert T^{\times }\Vert \Vert x\Vert =\Vert T^{\times }\Vert \Vert g_{Tx}||\Vert x\Vert \ge \Vert (T^{\times }{g}_{Tx})(x)\Vert =\Vert g_{Tx}(Tx)\Vert =\Vert Tx\Vert$. Therefore, $\Vert T\Vert \le \Vert T^{\times }\Vert$.

Theorem 2. Let $H_1$ and $H_2$ be Hilbert spaces. Let $A_1:H_1^{\prime }\to H_1$ and $A_2:H_2^{\prime }\to H_2$ be the canonical bijections between $H_i^{\prime }$ and $H_i$. For any bounded linear operator $T:H_1\to H_2$, $T^{\ast }=A_1{T}^{\times }{A}_2^{-1}:H_2\to H_1$ is the Hilbert-adjoint operator of $T$.

Proof. For any $x\in H_1$ and $y\in H_2$,

$$⟨x,A_1{T}^{\times }{A}_2^{-1}y⟩=(T^{\times }{A}_2^{-1}y)(x)=A_2^{-1}y(Tx)=⟨Tx,y⟩\Rightarrow T^{\ast }=A_1{T}^{\times }{A}_2^{-1}$$

Theorem 3. Every Hilbert space $H$ is reflexive.

Proof. It suffices to show that for any $g\in H^{\prime \prime }$ there exists some $h\in H$ such that for any $f\in H^{\prime }$, $g(f)=f(h)$.

Let $A:H^{\prime }\to H$ be the canonical bijection between $H$ and its dual space. $H^{\prime }$ is an inner product space, with inner product given by

$$⟨f_1,f_2{⟩}_1=⟨A{f}_2,A{f}_1⟩$$

As $H^{\prime }$ is complete, this means $H^{\prime }$ is also a Hilbert space. Therefore, for any $g\in H^{\prime \prime }$ there exists some $g_1\in H^{\prime }$ such that $g(f)=⟨f,g_1{⟩}_1=⟨A{g}_1,Af⟩=f(A{g}_1)$. We are done.

Theorem 4. If the dual space $X^{\prime }$ of a normed space $X$ is separable, then $X$ is separable.

Proof. A generally useful theorem when discussing separability in normed spaces is the following

Lemma: A normed space $N$ is separable iff its $1$-sphere $S(N)=\{n\in N|\Vert n{\Vert }_N=1\}$ is separable If $N$ is separable, it contains an at most countable dense subset $D=\{d_i\}$. Let $D^{\prime }=\{d_i/\Vert d_i\Vert \}$. Then, $D^{\prime }$ is an at most countable dense subset of $S(N)$, and we are done. If $S(N)$ is separable, there is an at most countable dense subset of $S(N)$, which we call $D^{\prime }=\{d_i^{\prime }\}$. Let $D=\{d|d=r{d}^{\prime }\text{ for }r\in \mathbb{Q}\}$. Then $D$ is at most countable dense subset of $N$, and we are done.

Thus, we are not changing the problem in any way by restating it as: If the set of norm-$1$ linear functionals on $X$ is separable, then $X$ is separable. Let $(f_n)$ be a countable dense subset of the norm-$1$ linear functionals on $X$.

It is sufficient to find some countable subset of $X$ such that the closure of its span is the entire set $X$, as then the set of finite rational-coefficient sums of those elements is a countable dense subset of $X$. Our intuition is as follows: for any element $x\in X$, by Hahn-Banach there is some corresponding norm-$1$ functional $f$ such that $f$ preserves the norm of $x$ $f(x)=\Vert x\Vert$ and roughly destroys all other directions. Furthermore, as $(f_n)$ is dense in the norm-$1$ sphere, there must be some $f_n$ that roughly only preserves the $x$ direction as well. Therefore, if for every $f_n$ we choose some element $x_n$ such that $\Vert x_n\Vert =1$ and $\Vert f_n(x_n)\Vert >0.5$, then $Y=\overline{Span\{x_n\}}$ should comfortably cover the entire set $X$, and we are done.

Assume, FTSOC, that $Y\ne X$, and there is some $z\notin Y$. As $Y$ is a closed subspace of $X$ (the span of a set and the closure of the span are always vector subspaces), we consider the set $Z=Span\{Y,z\}$. Let $\delta ={\inf }_{y\in Y}d(y,z)$; any element $k\in Z$ has a unique representation $k=y_k+\alpha z$ where $\alpha$ is a scalar and $y_k\in Y$, and we define the linear functional $\tilde{f}$ as $\tilde{f}(k)=\alpha \delta$. Clearly, $\Vert \tilde{f}\Vert =1$ and $\tilde{f}(y)=0$ for any $y\in Y$, and by Hahn-Banach we can extend $\tilde{f}$ to make some norm-$1$ functional $g$ on $X$ such that $g(y)=0$ for all $y\in Y$. For any $f_n$,

$$0.5=|f_n(x_n)|=|f_n(x_n)-g(x_n)|=|(f_n-g)(x_n)|\le \Vert x_n\Vert \Vert f_n-g\Vert =\Vert f_n-g\Vert$$

This contradicts the denseness of $f_n$ in the $1$-sphere, and we are done.

Courtesy of ChatGPT, here is a nice line that captures the intuition of this proof: “For any direction in $X$, Hahn–Banach provides a functional that strongly detects that direction. Since the functionals $f_n$​ approximate all norm-$1$ functionals, we can choose vectors $x_n$ that each $f_n$​ detects strongly. Any functional annihilating the span of the $x_n$​ would therefore remain at least some fixed distance away from every $f_n$​, contradicting density.”