Adjoints and Riesz Representation Theorem

Riesz Representation Theorem

Linear Functional

A linear functional $\ell$ is a linear transformation from a Hilbert space to the underlying field of scalars.

Riesz Representation Theorem

Let $\ell$ be a continuous linear functional on a Hilbert space \newcommand{\H}{\mathcal{H}}\H. Then, there exists a unique g\in\H, such that $\ell (f)=⟨f,g⟩$ for all f\in \H. Moreover, $\Vert \ell \Vert =\Vert g\Vert$.

Proof Consider the kernel of $\ell$, say $\S :=\ker \ell =\{f\in \stackrel{˝}{:}\ell (f)=0\}$, which is a closed subspace of \H, so $\stackrel{˝}{=}\S \oplus {\S }^{\perp }$. If ${\S }^{\perp }=\varnothing$, then $\ell =0$. Otherwise, pick some $h\in {\S }^{\perp }$ with $\Vert h\Vert =1$, and let $g:=\overline{\ell (h)}h$. Then if we let $u:=\ell (f)h-\ell (h)f$, observe that $u\in \S$ because $\ell (u)=\ell (f)\ell (h)-\ell (h)\ell (h)=0$, therefore, $0=⟨u,h⟩=\ell (f)⟨h,h⟩-⟨f,\overline{\ell (h)}h⟩=\ell (f)⟨h,h⟩-⟨f,g⟩,$which shows that $\ell (f)=⟨f,g⟩$. Moreover, if $g^{\prime }\in {\S }^{\perp }$ is another element such that $\ell (f)=⟨f,g^{\prime }⟩$, then we have $\ell (f)=⟨f,g⟩=⟨f,g^{\prime }⟩,$it follows that $⟨f,g-g^{\prime }⟩=0$ for all f\in\H, which implies that $g=g^{\prime }$. $\square$

Adjoint

Proposition

Let T\colon \H_{1}\to\H_{2} be a bounded linear map on separable Hilbert spaces. Then there exists a unique adjoint bounded linear map T^{*}\colon \H_{2}\to\H_{1} such that $⟨Tf,g⟩=⟨f,T^{\ast }g⟩$ for all f\in\H_{1} and g\in\H_{2}. This map $T^{\ast }$ is called the adjoint of $T$.

Proof Given g\in \H_{2}, the map $x\mapsto ⟨Tx,g⟩$ is a linear functional on \H_{1}. So by the Riesz representation theorem, there exists a unique w\in \H_{2}, such that $⟨Tx,g⟩=⟨x,w⟩,$ for all x\in \H_{1}. We define $T^{\ast }(g):=w$. This is a linear map from \H_{2} to \H_{1}, because the inner product is linear. It is also bounded because $\Vert T\Vert =\Vert T^{\ast }\Vert$ (see below). $\square$

Proposition

For any bounded linear operator T\colon \H_{1}\to \H_{2}, there holds $\Vert T\Vert =\Vert T^{\ast }\Vert$, and $\Vert T{\Vert }^2=\Vert T^{\ast }{\Vert }^2=\Vert T\circ T^{\ast }\Vert =\Vert T^{\ast }\circ T\Vert .$

Proof We shall first show that $\Vert T\Vert =\Vert T^{\ast }\Vert$. In fact, by the lemma $\Vert T\Vert ={\sup }_{\Vert f\Vert =1,\Vert g\Vert =1}|⟨Tf,g⟩|={\sup }_{\Vert f\Vert =1,\Vert g\Vert =1}|⟨f,T^{\ast }g⟩|={\sup }_{\Vert f\Vert =1,\Vert g\Vert =1}|⟨T^{\ast }g,f⟩|=\Vert T^{\ast }\Vert .$Now, for any $\Vert f\Vert =1$ and $\Vert g\Vert =1$, we have $|⟨T(T^{\ast }f),g⟩|=|⟨T^{\ast }f,T^{\ast }g⟩|\le \Vert T^{\ast }f\Vert \Vert T^{\ast }g\Vert \le \Vert T^{\ast }{\Vert }^2,$thus $\Vert T\circ T^{\ast }\Vert =\sup |⟨T(T^{\ast }f),g⟩|\le \Vert T^{\ast }{\Vert }^2$. For the other direction, note that $\Vert T\circ T^{\ast }\Vert \ge ⟨T(T^{\ast }f),f⟩=⟨T^{\ast }f,T^{\ast }f⟩=\Vert T^{\ast }f{\Vert }^2,$so $\Vert T\circ T^{\ast }\Vert \ge \sup \Vert T^{\ast }f{\Vert }^2=\Vert T^{\ast }{\Vert }^2,$ hence $\Vert T\circ T^{\ast }\Vert =\Vert T^{\ast }{\Vert }^2$. The same argument applies to $\Vert T^{\ast }\circ T\Vert$. $\square$

Infinite Diagonal Matrix

Diagonalised Linear Operator

Suppose \H is a separable Hilbert space with orthonormal basis $\{e_i{\}}_{i=1}^{\infty }$. A linear operator T\colon \H\to\H is diagonalised if there exists a sequence of scalars $\{{\lambda }_i{\}}_{i=1}^{\infty }\subset \text{C}$ such that $T(e_i)={\lambda }_i{e}_i,\text{for all }i.$

Proposition

Suppose T\colon \H\to\H is a diagonalised linear operator on a separable Hilbert space \H with orthonormal basis $\{e_i{\}}_{i=1}^{\infty }$. Then the following holds:

• $\Vert T\Vert ={\sup }_k|{\lambda }_k|$,
• $T^{\ast }$ is also diagonalised with ${\bar{\lambda }}_k$ as the diagonal entries. Hence, $T=T^{\ast }$ if and only if ${\lambda }_k\in \mathbb{R}$ for all $k$.
• $T$ is unitary if and only if $|{\lambda }_k|=1$ for all $k$.
• $T$ is an orthogonal projection if and only if ${\lambda }_k\in \{0,1\}$ for all $k$.