### Advanced Analysis, Notes 4: Hilbert spaces (bounded operators, Riesz Theorem, adjoint)

Up to this point we studied Hilbert spaces as they sat there and did nothing. But the central subject in the study of Hilbert spaces is the theory of the operators that act on them. Paul Halmos, in his classic paper “Ten Problem in Hilbert Space“, wrote:

Nobody, except topologists, is interested in problems about Hilbert space; the people who work in Hilbert space are interested in problems about operators.

Of course, Halmos was exaggerating; topologists don’t really care much for Hilbert spaces for their own sake, and functional analysts have much more to say about the structure theory of Hilbert space then what we have learned. Nevertheless, this quote is very close to the truth. We proceed to study operators.

#### 1. Bounded operators

A linear transformation $T : H \rightarrow K$ mapping between two Hilbert spaces is said to be bounded if the operator norm $\|T\|$ of $T$, defined as

$\|T\| := \sup_{\|x\|_H=1} \|Tx\|_K$

satisfies $\|T\| < \infty$. An immediate observation is that for all $x \in H$ we have $\|T x\| \leq \|T \| \|x\|$.

A bounded linear transformation is usually referred to as a bounded operator, or simply as an operator. The set of bounded operators between two Hilbert spaces $H$ and $K$ is denoted $B(H,K)$ and $B(H,H)$ is usually abbreviated as $B(H)$.

In this course we will almost exclusively deal with bounded operators, so we will not bother using the adjective “bounded” over and over again. Although unbounded operators defined on linear subspaces of Hilbert spaces are of great importance and are studied a lot, unbounded operators defined on the entire Hilbert space $H$ are hardly ever of any interest.

Example: Every linear transformation on a finite dimensional space is bounded. Let $H = \mathbb{C}^n$ be a finite dimensional Hilbert space (with the standard inner product). Every linear transformation on $H$ is of the form $T_A$ given by $T_A (x) = Ax$, where $A \in M_n(\mathbb{C})$ is an $n \times n$ matrix. By elementary analysis, $T_A$ is continuous, and by compactness of the unit sphere $S = \{x \in \mathbb{C}^n : \|x\|=1\}$, $T_A$ is bounded.

Alternatively, one can find an explicit bound for $\|T_A\|$ by using the definition. Let $x \in S$. Then

$\|Ax\|^2 = \sum_{i=1}^n |\sum_{j=1}^n a_{ij} x_j|^2 .$

By Cauchy-Schwarz, the right hand side is less than

$\sum_{i=1}^n (\sum_{j=1}^n |a_{ij}|^2) \|x\|^2 \leq \|x\|^2 \sum_{i,j=1}^n |a_{ij}|^2 .$

Thus $\|T_A\| \leq \sqrt{\sum_{i,j=1}^n |a_{ij}|^2}$.

Exercise A: Prove that the above bound for $\|T_A\|$ may be attained in some cases, but that in general it is not sharp.

A simple and useful way of constructing bounded operators on Hilbert spaces is by applying the following exercise.

Exercise B: Let $D$ be a dense linear subspace of a Hilbert space $H$. Let $T$ be a linear transformation defined on $D$, and suppose that $\sup \|T x\|$, where $x$ runs over all $x \in D$ with $\|x\|=1$, exists and is equal to $M$. Then there is a unique bounded operator $\tilde{T}$ defined on $H$ that agrees with $T$ on $D$ and satisfies $\|T\| = M$. Moreover, prove that if $\|Tx\| = \|x\|$ for all $x \in D$, then  $\|\tilde{T}x\| = \|x\|$ for all $x \in H$.

Example: Let $H = L^2[0,1]$, and let $g \in C[0,1]$. We define the multiplication operator $M_g : L^2[0,1] \rightarrow L^2[0,1]$ by

$M_g f = gf .$

The operator $M_g$ is clearly defined and bounded on $C[0,1]$, so one may use Exercise B to extend $M_g$ to a bounded operator on all of $L^2[0,1]$ (if one knows measure theory, then this extension procedure is not needed). It is not difficult to compute $\|M_g\| = \|g\|_\infty$.

Exercise C: Prove that there exists an unbounded operator on $\ell^2$.

Proposition 1: For a linear transformation $T : H \rightarrow K$ mapping between two inner product spaces, the following are equivalent:

1. $T$ is bounded.
2. $T$ is continuous.
3. $T$ is continuous at some $x_0 \in H$

Exercise D: Prove Proposition 1. (Of course, Proposition 1 holds in any normed space, and has nothing in particular to do with inner products).

Definition 2: For every $T \in B(H,K)$, we define the kernel of $T$ to be the space $N(T) = \{x \in H : Tx = 0\}$. We define the range of $T$ to be the space $R(T) = \{Tx : x \in H\}$

Proposition 3: Let $S, T \in B(H,K)$. Then $S = T$ if and only if $(Tx,y) = (Sx,y)$ for all $x \in H$ and $y \in K$. In case $H = K$, then $T = S$ if and only if $(Sx,x) = (Tx,x)$ for all $x \in H$

Exercise E: Prove the proposition, and show that the second part fails for real Hilbert spaces.

#### 2. Linear functionals and the Riesz representation theorem

Definition 4: A linear operator $F : H \rightarrow \mathbb{C}$ is said to be a linear functional.

Example: Let $H$ be a Hilbert space, and let $y \in H$. We define a linear functional $F_y$ on $H$ as follows:

$F_y(x) = (x,y) \,\, ,\,\, x \in H .$

By the linearity of the inner product in the first component, $F_y$ is a linear functional. For every $x \in H$, Cauchy-Schwarz implies that $|F_y(x)| = |(x,y)| \leq \|x\| \|y\|$, thus $\|F_y\| \leq \|y\|$. When $y \neq 0$, we apply $F_y$ to $u = y/\|y\|$ and obtain $F_y(u) = \|y\|$, thus $\|F_y\| = \|y\|$. In fact, this example covers all the possibilities.

Theorem 5 (Riesz representation theorem): If $F$ is a bounded linear functional on a Hilbert space $H$, then there exists a unique $y \in H$ with $\|y \| = \|F\|$, such that $F(x) = (x,y)$ for all $x \in H$

Proof: If $F = 0$ then we choose $y = 0$ and the proof is complete, so assume that $F \neq 0$. Then $N(F)$ is a closed subspace which is not $H$. We have $N(F) \oplus N(F)^\perp$. Since the restriction of $F$ to $N(F)^\perp$ is a linear functional with trivial kernel, the dimension of $N(F)^\perp$ is equal to $1$. Choose $u \in N(F)^\perp$ that is a unit vector (i.e., $\|u\| = 1$). We claim that $y = \overline{F(u)} u$ does the trick. Indeed, let $x \in H$. Let $x = z + w$ be the decomposition of $x$ with respect to $H = N(F)^\perp \oplus N(F)$. Then by Theorem 15 in Notes 2, $z = (x,u)u$. Then we have

$F(x) = F(z) + F(w) = F((x,u)u) = (x,u) F(u) = (x,y).$

The equality $\|y\| = \|F\|$ has been worked out in the above example. Finally, $y$ is unique, because $y'$ is another candidate for this role, then $(x, y - y') = F(x) - F(x) = 0$ for all $x \in H$, whence $y=y'$ (because, in particular, $(y-y',y-y') = 0$).

Exercise F: Where did we use the fact that $H$ is a Hilbert space? Where did we use the fact that $F$ is bounded? Construct examples showing that neither of these assumptions can be dropped.

Exercise G: Prove that a linear functional on a Hilbert space is bounded if and only if its kernel is closed.

#### 3. The adjoint of a bounded operator

Proposition 6 (existence of the adjoint): For every $T \in B(H)$ there exists a unique $S \in B(H)$ such that $(Tx,y) = (x, S y)$ for all $x, y \in H$

Proof: Fix $y \in H$, and consider the functional $F(x) = (Tx,y)$. It is clear that $F$ is linear, and the estimate

$|F(x)| \leq \|T x \| \|y\| \leq \|T\| \|y\| \|x\|$

shows that $F$ is bounded and $\|F\| \leq \|T\| \|y\|$. By Theorem 5, there exists a unique element in $H$, call it $Sy$, such that $(Tx,y) = F(x) = (x, Sy)$ for all $x$. What remains to be proved is that the mapping $y \mapsto S y$ is a bounded linear map. Linearity is easy. For example, if $y_1, y_2, \in H$, then

$(x, S(y_1 + y_2)) = (Tx, y_1 + y_2) = (Tx, y_1) + (Tx, y_2) = (x, Sy_1 + Sy_2) ,$

so $S(y_1 + y_2) = Sy_1 + Sy_2$. As for boundedness, Theorem 5 also tells us that $\|Sy \| = \|F\| \leq \|T\| \|y\|$, thus $S$ is bounded and $\|S\| \leq \|T\|$. The uniqueness of $S$ also follows, from the uniqueness of $Sy$ for each $y$.

Definition 7: For every $T \in B(H)$, we denote by $T^*$ the unique operator for which $(Tx, y) = (x, T^* y)$ for all $x,y \in H$. The operator $T^*$ is said to be the adjoint of $T$

Example: Let $H = \mathbb{C}^n$, let $A \in M_n(\mathbb{C})$ and let $T_A \in B(H)$ be multiplication by $A$. If $A = (a_{ij})$, then denote by $A^*$ the matrix with $ij$th entry $\overline{a_{ji}}$. Then $T_A^* = T_{A^*}$. This should be well known from a course in linear algebra, but is also immediate:

$(T_A x,y) = y^* A x = (A^* y)^* x = (x,T_{A^*}y) .$

Example: Let $H = L^2[0,1]$, and let $g \in C[0,1]$. Then $M_g^* = M_{\overline{g}}$.

Proposition 8: For $S,T \in B(H)$ and $a,b \in \mathbb{C}$ the following hold:

1. $(T^*)^* = T$
2. $\|T\| = \|T^*\|$
3. $(aS+bT)^* = \overline{a} S^* + \overline{b} T^*$
4. $(ST)^* = T^* S^*$
5. $\|T\|^2 = \|T^* T\|$
6. If $T$ is bijective and $T^{-1}$ is bounded then $T^*$ is invertible and $(T^*)^{-1} = (T^{-1})^*$

Remark: We will later learn that when $T \in B(H)$ is invertible, then $T^{-1}$ is automatically bounded.

Exercise H: Prove Proposition 8.

Proposition 9: Let $T \in B(H)$. We have the following:

1. $R(T)^\perp = N(T^*)$
2. $R(T^*)^\perp = N(T)$
3. $N(T)^\perp = \overline{R(T^*)}$
4. $N(T^*)^\perp = \overline{R(T)}$

Proof: 1 and 2 are equivalent, and so are 3 and 4. To prove 1, let $x \in H$. Then $T^*x = 0$ if and only if for all $y$$0 = (T^* x, y) = (x, Ty)$, and this happens if and  only if $x \perp R(T)$

To prove 4, apply $\perp$ to both sides of 1 to obtain $N(T^*)^\perp = R(T)^{\perp \perp}$. Thus the result follows from the following lemma.

Lemma 10: For every linear subspace $N$ in a Hilbert space, $N^{\perp\perp} = \overline{N}$

Proof: It is easy to see that $N^\perp = \overline{N}^\perp$, thus (as we know already from Lecture 2 that for a closed subspace $M$, it holds that $M = M^{\perp \perp}$) $N^{\perp\perp} = \overline{N}^{\perp\perp} = \overline{N}$.

#### 4. Special classes of operators

Definition 11: An operator $T \in B(H)$ is said to be

1. normal if $T^*T = TT^*$
2. self adjoint if $T = T^*$
3. isometric (or an isometry) if $\|Tx\| = \|x\|$ for all $x \in H$
4. unitary if it is a bijection satisfying $(T x, T y) = (x,y)$ for all $x,y \in H$
5. positive if $(Tx,x) \geq 0$ for all $x \in H$

Example:

1. Let $I$ denote that identity operator on $H$, that is $Ix = x$ for all $x \in H$. Then $I$ belongs to all of the above categories.
2. Let $M$ be a closed subspace. Then $P_M$ is positive and self adjoint. Indeed, if $x \in H$, write $x = y + z$ where $y \in M, z \in M^\perp$. Then $(P_Mx,x) = (y, y+z) = (y,y) \geq 0$. By Proposition 12 below, $P_M$ is self adjoint.
3. Let $S$ be the right shift on $\ell^2$: that is, $S(x_1, x_2, x_3, \ldots, ) = (0, x_1, x_2, \ldots)$. Then $S$ is an isometry. A computations shows that $S^*$ is given by $S^*(x_1, x_2, x_3, \ldots) = (x_2, x_3, \ldots)$. Further computations show that $S^* S = I$, and $S S^* = P_M$, where $M = \{(1, , 0, 0, \ldots)\}^\perp$. So $S$ is not normal.

Proposition 12: $T \in B(H)$ is self adjoint if and only if $(T x, x) \in \mathbb{R}$

Proposition 13: For $T \in B(H)$, the following are equivalent:

1. $T$ is an isometry.
2. $(T x, T y) = (x, y)$ for all $x,y \in H$.
3. $T^* T = I$

Proposition 14: For $T \in B(H)$, the following are equivalent:

1. $T$ is unitary.
2. $T^* T = T T^* = I$

Exercise I: Prove the above propositions.