$$

$$

$$

Lecture 4: Bounded Operators

tags: 224a

We now begin to develop the spectral theory of bounded operators on Hilbert space. Besides giving a great deal of insight into the structure of these operators, it is an indispensible tool in solving "infinite dimensional linear equations" (such as differential equations), due to the absence of algorithms like Gaussian elimination.

Norms and Convergence

A linear transformation

$T:X\to Y$ for Banach spaces

$X,Y$ is called bounded if there is a constant

$C$ such that

$$\Vert Tx{\Vert }_Y\le C\Vert x{\Vert }_X\mathrm{\forall }x\in X,$$
and the least such constant is called its operator norm

$\Vert T\Vert$. We will denote the set of bounded operators

$X\to Y$ by

$\mathcal{L}(X,Y)$. It is shown on the homework that an operator

$T$ is continuous with respect to convergence in norm if and only if it is bounded.

The fact that

$\Vert T\Vert$ satisfies the triangle inequality follows from linearity and the triangle inequality for

$\Vert \cdot {\Vert }_Y$.

Here are some examples of bounded operators:

1. $X=Y={\ell }^2(\mathbb{Z})$ with $(Tx)n = x{n-1} (shift operator). Check that
   $\Vert Tx\Vert =\Vert x\Vert$ for all
   $x$.
2. $X=L^1(\mathbb{R}),Y=\mathbb{C}$ and
   $Tf={\int }_{\mathbb{R}}f(x)dx.$ The triangle inequality shows that
   $\Vert T\Vert \le 1$ and this is achieved for
   $f$ any nonnegative function.
3. $X=Y=L^2(0,1)$,
   $(Vf)(x):={\int }_0^{x}f(y)dy,$ the Volterra operator. A simple Cauchy-Schwartz argument reveals that the norm is bounded by
   $1/\sqrt{2}$. The true answer is
   $2/\pi ,$ though this is harder to show.
4. Any bounded linear functional
   ${\ell }_y:H\to \mathbb{C}$. By the Riesz theorem this is always equal to
   $(y,x)$ for some
   $y$, and it is easy to see that
   $\Vert {\ell }_y\Vert =\Vert y\Vert$.

We will often define operators in

$\mathcal{L}(X,Y)$ as absolutely convergent power series in other operators, so it is important to know that this space contains enough limit points. Let

$T_n\to T$ denote

$\Vert T_n-T\Vert \to 0$ as

$n\to \mathrm{\infty }$.

Theorem. If

$X$ is a normed linear space and

$Y$ is a Banach space,

$\mathcal{L}(X,Y)$ is complete with respect to the operator norm.
Proof. Suppose

$T_n$ is a sequence in

$\mathcal{L}(X,Y)$ satisfying

$\Vert T_n-T_m\Vert \to 0$ as

$n,m\to \mathrm{\infty }$. For a fixed vector

$x\in X$, the sequence

$T_{n}x\in Y$ satisfies

$\Vert T_{n}x-T_{m}x\Vert \le \Vert T_n-T_m\Vert \Vert x\Vert \to 0$ so it must have a limit point; define

$Tx$ to be equal to this limit point.

Linearity of

$T$ can be checked easily. We will now show

$\Vert T_n-T\Vert \to 0$. Fix

$ϵ>0$, and choose

$n$ such that

$\Vert T_n-T_m\Vert \le ϵ/2$ whenever

$m/gen$. Let

$\Vert x\Vert =1$ and observe that for every

$m\ge n$ we have

$$\Vert T_{n}x-Tx\Vert \le \Vert T_{n}x-T_{m}x\Vert +\Vert T_{m}x-Tx\Vert \le ϵ/2+ϵ/2,$$
by taking

$m$ sufficiently large and using

$\Vert T_{m}x-Tx\Vert \to 0$ (note that

$m$ can depend on

$x$). Since

$n$ depended only on

$ϵ$ we have shown that

$\underset{\Vert x\Vert =1}{sup}\Vert T_{n}x-T_x\Vert <ϵ$, as desired.

To see that

$T$ is bounded, observe that for every

$n$ one has

$\Vert T\Vert \le \Vert T_n-T\Vert +\Vert T_n\Vert$, and choose

$n$ such that

$\Vert T_n-T\Vert \le 1$ (say).

$◻$

Besides convergence in norm, there are two other useful notions of convergence for sequences of operators, called strong and weak convergence; for definitions and examples, see Sec VI.1 of Reed and Simon. Roughly speaking, weak convergence corresponds to "entrywise" convergence of matrices, and strong convergence corresponds to "columnwise" convergence. For the purposes of this course, the word "topology" just means notion of convergence. As seen from the examples, these notions of convergence are very different, and it is important to be mindful of which one is meant. We remark that

$\mathcal{L}(X,Y)$ is also closed with respect to weak convergence, and this is proven in Reed and Simon Theorem VI.1.

Adjoints and Positivity

We will use

$\mathcal{L}(H)$ to denote bounded operators from a Hilbert space to itself.

Theorem. If

$A\in \mathcal{L}(X,Y)$ for complex separable Hilbert spaces

$X,Y$, there is a unique operator

$A^{\ast }\in \mathcal{L}(Y,X)$ called its adjoint which satisfies

$$(y,Ax{)}_Y=(A^{\ast }y,x{)}_X(\ast )$$
for all

$x\in X,y\in Y$.
Proof. Given

$y\in Y$, consider the linear functional

${\ell }_{A^{\ast }y}(x):=(y,Ax)$ on

$X$. This is bounded by Cauchy-Schwartz, so by the Riesz theorem there is a vector

$z$ such that

${\ell }_{A^{\ast }y}(x)=(z,x)$ for all

$x\in X$; define

$A^{\ast }y=z$ for this

$z$. By construction, the property

$(\ast )$ is satisfied for all

$x\in X$. Linearity is then implied by the properties of the inner product, as

$$(A^{\ast }(ay+b{y}^{\prime }),x)=(ay+b{y}^{\prime },Ax)=\overline{a}(y,Ax)+\overline{b}(y^{\prime },Ax)$$

$$=\overline{a}(A^{\ast }y,x)+\overline{b}(A^{\ast }{y}^{\prime },x)=(a{A}^{\ast }y+b{A}^{\ast }{y}^{\prime },x)$$
for all

$x\in X$, so by uniqueness in the Riesz theorem we must have

$A^{\ast }(ay+b{y}^{\prime })=a{A}^{\ast }y+b{A}^{\ast }{y}^{\prime }$.

Boundedness follows by observing that

$$\Vert A^{\ast }y{\Vert }^2=|(A^{\ast }y,A^{\ast }y)|=\Vert (y,A{A}^{\ast }y)|\le \Vert y\Vert \Vert A\Vert \Vert A^{\ast }y\Vert$$
for all

$y$, whence

$\Vert A^{\ast }\Vert \le \Vert A\Vert$.

Uniqueness of

$A^{\ast }$ follows because if

$(Cy,x)=(Dy,x)$ for all

$x,y$ then we must have

$Cy=Dy$ for all

$y$ or

$C=D$.

$◻$

The properties of the adjoint operation with respect to other operations like addition, multiplication, and norm are listed in Reed and Simon Theorem VI.3.

An operator

$A\in \mathcal{L}(H)$ is called self-adjoint if

$A=A^{\ast }$. Self-adjointness is an important property both from a mathematical (becuase it allows a much better understanding of the spectrum) and physical (because such operators correspond to observables in quantum mechanics) point of view. A homework problem asks you to prove that

$$\Vert A\Vert =\underset{\Vert x\Vert =1}{sup}(x,Ax)$$
whenever

$A=A^{\ast }$.

An operator

$A\in \mathcal{L}(H)$ is called positive, denoted

$A\ge 0$, if

$(x,Ax)\ge 0$ for all

$x\in H$. On the homework it is shown in a complex Hilbert space, a positive operator is always self-adjoint.

We conclude the lecture by recovering a familiar fact from finite dimensional linear algebra: a positive operator has a square root. The proof constructs this square root using an absolutely convergent power series. This is a common technique in the infinite dimensional setting; at a high level, the techniques of analytic function theory are the ones that generalize best beyond finite dimensions.

Theorem. If

$A\ge 0$ then there is a unique

$B\in \mathcal{L}(H)$ with

$B^2=A$,

$B\ge 0$, and

$AB=BA$.
See Reed and Simon Theorem VI.9 for a proof.