# Spring 2016 MATH 528: Functional Analysis

## Branko Ćurgus

Wednesday, May 18, 2016

1. I will post summary of what was presented in class during last few weeks. In the theorems below we always assume that the vector spaces that we are dealing with are nontrivial; that is they contain a nonzero vector.
2. The following theorem provides several characterizations of continuity for linear operators.

Theorem. Let $(\mathcal E, \left\|\,\cdot\,\right\|_{\mathcal E})$ and $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ be a normed spaces. Let $T: \mathcal F \to \mathcal F$ be a linear operator. The following statements are equivalent.

1. $T$ is continuous.
2. $T$ is continuous at $0$.
3. There exists $c \in \mathbb R$ such that for all $x \in \mathcal E$ we have $\left\|Tx\right\|_{\mathcal F} \leq c \, \left\|x\right\|_{\mathcal E}$.
4. The set $\bigl\{ \left\|Tx\right\|_{\mathcal F} : x \in \mathcal E \ \text{and} \ \left\|x\right\|_{\mathcal E} \leq 1 \bigr\}$ is bounded.
5. The set $\bigl\{ \left\|Tx\right\|_{\mathcal F} : x \in \mathcal E \ \text{and} \ \left\|x\right\|_{\mathcal E} = 1 \bigr\}$ is bounded.

If $(\mathcal F, \left\langle\,\cdot,\,\cdot\right\rangle_{\mathcal F})$ is an inner product space then the following identities hold.

1. There exists $c \in \mathbb R$ such that forall $x \in \mathcal E$ and for all $y \in \mathcal F$ we have $\left| \left\langle Tx,y \right\rangle_{\mathcal F} \right| \leq c \, \left\|x\right\|_{\mathcal E} \left\|y\right\|_{\mathcal F}.$
2. The set $\bigl\{ \left|\left\langle Tx,y \right\rangle_{\mathcal F} \right| : x \in \mathcal E, \ y \in \mathcal F, \ \left\|x\right\|_{\mathcal E} \leq 1 \ \text{and} \ \left\|y\right\|_{\mathcal F} \leq 1 \bigr\}$ is bounded.
3. The set $\bigl\{ \left|\left\langle Tx,y \right\rangle_{\mathcal F} \right| : x \in \mathcal E, \ y \in \mathcal F, \ \left\|x\right\|_{\mathcal E} = 1 \ \text{and} \ \left\|y\right\|_{\mathcal F} = 1 \bigr\}$ is bounded.
4. The set $\bigl\{ \left\langle Tx,y \right\rangle_{\mathcal F} : x \in \mathcal E, \ y \in \mathcal F, \ \left\|x\right\|_{\mathcal E} = 1, \ \left\|y\right\|_{\mathcal F} = 1 \ \text{and} \ \left\langle Tx,y \right\rangle_{\mathcal F} \in \mathbb R \bigr\}$ is bounded.

3. The following definition introduces the concept of a bounded linear operator. Notice that this concepts is different from the natural generalization of the concept of a bounded function defined in calculus. Also notice that the theorem characterizing continuity of linear operators between normed spaces claims that for a linear operator between normed linear spaces boundedness (as defined below) is equivalent to continuity.

Definition. Let $(\mathcal E, \left\|\,\cdot\,\right\|_{\mathcal E})$ and $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ be a normed spaces. A linear operator $T: \mathcal F \to \mathcal F$ is said to be bounded if the set $\bigl\{ \left\|Tx\right\|_{\mathcal F} : x \in \mathcal E \ \text{and} \ \left\|x\right\|_{\mathcal E} \leq 1 \bigr\}$ is bounded. By ${\mathcal L}({\mathcal E},{\mathcal F})$ we denote the set of all bounded linear operators defined on $\mathcal E$ with values in $\mathcal F$. For $T \in {\mathcal L}({\mathcal E},{\mathcal F})$ we set $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} = \sup \bigl\{ \left\|Tx\right\|_{\mathcal F} : x \in \mathcal E \ \text{and} \ \left\|x\right\|_{\mathcal E} \leq 1 \bigr\}$

Corollary. Let $(\mathcal E, \left\|\,\cdot\,\right\|_{\mathcal E})$ and $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ be a normed spaces. Let $T: \mathcal F \to \mathcal F$ be a linear operator. Then $T$ is bounded if and only if it is continuous.

4. The following propositon gives several formulas for the norm of a bounded (continuous) linear operator.

Proposition. Let $(\mathcal E, \left\|\,\cdot\,\right\|_{\mathcal E})$ and $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ be a normed spaces. Let $T: \mathcal F \to \mathcal F$ be a bounded linear operator. The following identities hold.

1. $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} = \min\bigl\{ c \in \mathbb R : \forall\,x \in \mathcal E \ \ \left\|Tx\right\|_{\mathcal F} \leq c \, \left\|x\right\|_{\mathcal E} \bigr\}.$
2. $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} = \sup\bigl\{ \left\|Tx\right\|_{\mathcal F} : x \in \mathcal E \ \text{and} \ \left\|x\right\|_{\mathcal E} = 1 \bigr\}$.

If $(\mathcal F, \left\langle\,\cdot,\,\cdot\right\rangle_{\mathcal F})$ is an inner product space then the following identities hold.

1. $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} = \min\bigl\{ c \in \mathbb R : \forall\,x \in \mathcal E \ \forall\,y \in \mathcal F \ \ \left| \left\langle Tx,y \right\rangle_{\mathcal F} \right| \leq c \, \left\|x\right\|_{\mathcal E} \left\|y\right\|_{\mathcal F} \bigr\}.$
2. $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} = \sup\bigl\{ \left|\left\langle Tx,y \right\rangle_{\mathcal F} \right| : x \in \mathcal E, \ y \in \mathcal F, \ \left\|x\right\|_{\mathcal E} \leq 1 \ \text{and} \ \left\|y\right\|_{\mathcal F} \leq 1 \bigr\}$.
3. $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} = \sup\bigl\{ \left|\left\langle Tx,y \right\rangle_{\mathcal F} \right| : x \in \mathcal E, \ y \in \mathcal F, \ \left\|x\right\|_{\mathcal E} = 1 \ \text{and} \ \left\|y\right\|_{\mathcal F} = 1 \bigr\}$.
4. $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} = \sup\bigl\{ \left\langle Tx,y \right\rangle_{\mathcal F} : x \in \mathcal E, \ y \in \mathcal F, \ \left\|x\right\|_{\mathcal E} = 1, \ \left\|y\right\|_{\mathcal F} = 1 \ \text{and} \ \left\langle Tx,y \right\rangle_{\mathcal F} \in \mathbb R \bigr\}$.

Proof. i. Denote the set which appears in i. by $\Delta$. Clearly $\Delta$ consists of nonnegative numbers. Thus $\inf\Delta$ exists. From the definition of $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}$ it follows that for all $x \in \mathcal E$ we have $\left\| Tx \right\|_{\mathcal F} \leq \left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} \left\|x\right\|_{\mathcal E}.$ Therefore the real number $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}$ belongs to $\Delta$. Hence, $\inf\Delta \leq \left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}$. Let $c \in \Delta$ be arbitrary. Then, for arbitrary $x \in \mathcal E$ such that $\left\|x\right\|_{\mathcal E} \leq 1$ we have $\left\|Tx\right\|_{\mathcal F} \leq c \left\|x\right\|_{\mathcal E} \leq c.$ Taking the supremum over all $x \in \mathcal E$ such that $\left\|x\right\|_{\mathcal E} \leq 1$ we get $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} \leq c.$ Thus $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}$ is a lower bound of $\Delta$. Therefore $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} \leq \inf \Delta.$ Since we already proved that $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} \in \Delta$, the identity i. is proved.

ii. Denote by $\alpha$ the supremum in ii. Then, for all $x \in \mathcal E$ we have $\left\| Tx \right\|_{\mathcal F} \leq \alpha \left\|x\right\|_{\mathcal E}$. Therefore for all $x \in \mathcal E$ such that $\left\|x\right\|_{\mathcal E} \leq 1$ we have $\left\| Tx \right\|_{\mathcal F} \leq \alpha$. That is, $\alpha$ is an upper bound of the set which appears in the definition of $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}$. Hence, $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} \leq \alpha$. Since the reverse inequality is a consequence of the set inclusion, ii. is proved.

vi. Notice first that the set that appears in vi. (which is a subset of $\mathbb R$) is symmetric with respect to $0$. Therefore its supremum, which we denote by $\beta$, is nonnegative. Let $x \in \mathcal E$, $y \in \mathcal F$, be such that $\left\|x\right\|_{\mathcal E} = 1$, $\left\|y\right\|_{\mathcal F} = 1$ and $\left\langle Tx,y \right\rangle_{\mathcal F} \in \mathbb R$. Then by the Cauchy-Bunyakovsky-Schwarz Inequality and the definition of $\left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}$ we have $\left\langle Tx,y \right\rangle_{\mathcal F} \leq \left\|Tx\right\|_{\mathcal F} \left\|y\right\|_{\mathcal F} \leq \left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}.$ This proves that $\beta \leq \left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}.$ Let $v \in \mathcal E$ such that $\left\| v \right\|_{\mathcal E} \leq 1$ be arbitrary. If $Tv = 0,$ then $\left\| Tv\right\|_{\mathcal F} \leq \beta.$ If $Tv \neq 0,$ then $\left\| v \right\|_{\mathcal E} \gt 0$ and $\left\| Tv \right\|_{\mathcal F} \gt 0.$ Now set $x = \frac{1}{\left\| v \right\|_{\mathcal E}} v \quad \text{and} \quad y = \frac{1}{\left\| T v \right\|_{\mathcal F}} T v,$ notice that $x \in \mathcal E$, $y \in \mathcal F$, $\left\| x \right\|_{\mathcal E} = 1$ and $\left\| y \right\|_{\mathcal F} = 1,$ and calculate $\left\langle Tx, y \right\rangle_{\mathcal F} = \frac{1}{\left\| v \right\|_{\mathcal E}} \frac{1}{\left\| T v \right\|_{\mathcal F}} \left\langle Tv, Tv \right\rangle_{\mathcal F} = \frac{\left\| T v \right\|_{\mathcal F}}{\left\| v \right\|_{\mathcal E}} \in \mathbb R.$ By the definition of $\beta$ we have that $\left\langle Tx, y \right\rangle_{\mathcal F} = \frac{\left\| T v \right\|_{\mathcal F}}{\left\| v \right\|_{\mathcal E}} \leq \beta.$ Thus, $\left\| T v \right\|_{\mathcal F} \leq \beta \left\| v \right\|_{\mathcal E}$ and hence, $\left\| T v \right\|_{\mathcal F} \leq \beta,$ since $\left\| v \right\|_{\mathcal E} \leq 1.$ As $v \in \mathcal E$ such that $\left\| v \right\|_{\mathcal E} \leq 1$ was arbitrary, the inequality $\left\| T v \right\|_{\mathcal F} \leq \beta$ proves that $\beta \geq \left\|T\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}.$ Together with already established reverse inequality, vi. is proved.

Proposition. Let $(\mathcal E, \left\|\,\cdot\,\right\|_{\mathcal E})$, $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ and $(\mathcal G, \left\|\,\cdot\,\right\|_{\mathcal G})$ be normed spaces. If $S \in {\mathcal L}({\mathcal E},{\mathcal F})$ and $T \in {\mathcal L}({\mathcal F},{\mathcal G}),$ then $TS \in {\mathcal L}({\mathcal E},{\mathcal G})$ and $\left\|TS\right\|_{{\mathcal L}({\mathcal E},{\mathcal G})} \leq \left\|T\right\|_{{\mathcal L}({\mathcal F},{\mathcal G})} \left\|S\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}.$

5. Since we are desperate for completeness, the following theorem is useful.

Theorem. Let $(\mathcal E, \left\|\,\cdot\,\right\|_{\mathcal E})$ and $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ be normed spaces.

1. The space ${\mathcal L}({\mathcal E},{\mathcal F})$ of all bounded linear operators from $\mathcal E$ to $\mathcal F$ is a vector space and the function $\left\|\,\cdot\,\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} : {\mathcal L}({\mathcal E},{\mathcal F}) \to \mathbb R$ is a norm on that vector space.
2. Let $T, T_n \in {\mathcal L}({\mathcal E},{\mathcal F})$ for all $n\in \mathbb N$. If $T = \lim_{n\to\infty}T_n$ in the normed space $\bigl({\mathcal L}({\mathcal E},{\mathcal F}), \left\|\,\cdot\,\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}\bigr),$ then for every $x \in \mathcal E$ we have that $Tx = \lim_{n\to\infty}T_nx$ in the normed space $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$.
3. If $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ is a Banach space, then the normed space $\Bigl( {\mathcal L}({\mathcal E},{\mathcal F}), \left\|\,\cdot\,\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})} \Bigr)$ is a Banach space.

6. Next we defined the concept of an invertible operator and the operator inverse. Let $(\mathcal E, \left\|\,\cdot\,\right\|_{\mathcal E})$ and $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ be normed spaces.

Definition. An operator $T \in {\mathcal L}({\mathcal E},{\mathcal F})$ is said to be invertible if there exists an operator $S \in {\mathcal L}({\mathcal F},{\mathcal E})$ such that $\tag{*} \forall\, x \in \mathcal E \quad STx = x \quad \text{and}\quad \forall\, y \in \mathcal F \quad TSy = y.$

Proposition. If $T \in {\mathcal L}({\mathcal E},{\mathcal F})$ is invertible, then (*) detrmines $S \in {\mathcal L}({\mathcal F},{\mathcal E})$ uniquely.

Definition. For an invertible $T \in {\mathcal L}({\mathcal E},{\mathcal F})$ the operator $S \in {\mathcal L}({\mathcal F},{\mathcal E})$ which satisfies (*) is called the inverse of $T$ and it is denoted by $T^{-1}$.

Proposition. Let $(\mathcal E, \left\|\,\cdot\,\right\|_{\mathcal E})$, $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ and $(\mathcal G, \left\|\,\cdot\,\right\|_{\mathcal G})$ be normed spaces. If $S \in {\mathcal L}({\mathcal E},{\mathcal F})$ and $T \in {\mathcal L}({\mathcal F},{\mathcal G})$ are invertible, then $TS \in {\mathcal L}({\mathcal E},{\mathcal G})$ is invertible and $(TS)^{-1} = S^{-1}T^{-1}.$

7. Let $(\mathcal E, \left\|\,\cdot\,\right\|_{\mathcal E})$ be a normed space. The space ${\mathcal L}({\mathcal E},{\mathcal E})$ is denoted by ${\mathcal L}({\mathcal E}).$ In addition to being a vector space, the space ${\mathcal L}({\mathcal E})$ has an additional operation: we can compose operators in ${\mathcal L}({\mathcal E}).$ Such a structure is called an algebra.

Definition. A vector space $\mathcal A$ over $\mathbb C$ is said to be an algebra over $\mathbb C$ if there is a binary operation $\circ:\mathcal A \times \mathcal A \to \mathcal A$ such that $\bigl(\mathcal A, +, \circ \bigr)$ is a ring and $\forall\, \alpha \in \mathbb C \ \ \forall\, x, y \in \mathcal A \quad (\alpha x)\circ y = x \circ (\alpha y) = \alpha (x\circ y).$ An algebra $\mathcal A$ is said to be unital if there exists $e \in \mathcal A$ such that for all $a \in \mathcal A$ we have $e\circ a = a \circ e = a$. In a unital algebra $\mathcal A$, an element $a \in \mathcal A$ is said to be invertible if there exists $b \in \mathcal A$ such that $b\circ a = a\circ b = e$. For invertible $a \in \mathcal A$, $b \in \mathcal A$ such that $b\circ a = a\circ b = e$ is called the inverse of $a$ and it is denoted by $a^{-1}.$

In an algebra it is common to write $xy$ insted of $x\circ y$.

Definition. A normed vector space $\bigl(\mathcal A,\left\|\,\cdot\,\right\|\bigr)$ which is an algebra and $\forall\, x, y \in \mathcal A \quad \left\| x\circ y \right\| \leq \left\|x\right\| \, \left\|y\right\|$ is said to be normed algebra. A unital normed algebra is a normed algebra which is a unital algebra and $\left\|e\right\| = 1.$

Theorem. Let $(\mathcal A, \left\|\,\cdot\,\right\|)$ be a normed algebra. The binary operation $\circ:\mathcal A \times \mathcal A \to \mathcal A$ is a continuous function with the product topology on $\mathcal A \times \mathcal A$ and the normed topology on $\mathcal A.$

Definition. A normed algebra which is a complete normed vector space is said to be Banach algebra.

Corollary. If $(\mathcal E, \left\|\,\cdot\,\right\|_{\mathcal E})$ is a Banach space, then $\bigl({\mathcal L}({\mathcal E}), \left\|\,\cdot\,\right\|_{{\mathcal L}({\mathcal E})}\bigr)$ is a unital Banach algebra.

Theorem. Let $(\mathcal A, \left\|\,\cdot\,\right\|)$ be a unital Banach algebra. If $a \in \mathcal A$ is invertible, then $a - b$ is invertible whenever $\left\|b\right\| \lt 1/\left\|a^{-1}\right\|$. In this case we have $(a-b)^{-1} = a^{-1} \sum_{k=0}^\infty \bigl(b a^{-1}\bigr)^k \quad \text{and} \quad \left\| (a-b)^{-1} \right\| \leq \frac{\left\|a^{-1}\right\|}{1-\left\|a^{-1}\right\|\left\|b\right\|}.$ In particular, the set of all invertible elements of $\mathcal A$ is open in $(\mathcal A, \left\|\,\cdot\,\right\|)$.

Proof. We first prove the special case of the theorem for $a = e$. Since $e^{-1} = e$, let $b \in \mathcal A$ be such that $\left\|b\right\| \lt 1.$ Consider the following sequence $\forall\, n\in\mathbb N \quad c_n = \sum_{k=0}^n b^k.$ Let $m,n\in\mathbb N$ be such that $m \lt n$. Then $\left\|c_n - c_m\right\| = \left\|\sum_{k=m+1}^n b^k\right\| \leq \sum_{k=m+1}^n \left\|b\right\|^k = \frac{\left\|b\right\|^{m+1}- \left\|b\right\|^{n+1}}{1-\left\|b\right\|} \leq \frac{\left\|b\right\|^{m+1}}{1-\left\|b\right\|}.$ Since $\left\|b\right\| \lt 1$, the sequence $n\mapsto c_n$ is a Cauchy sequence in $(\mathcal A, \left\|\,\cdot\,\right\|)$. Since $(\mathcal A, \left\|\,\cdot\,\right\|)$ is a Banach space this sequence converges. Set $c= \lim_{n\to\infty} c_n$. Now for arbitrary $n\in\mathbb N$ calculate $(e-b) c_n = c_n (e-b) = e - b^{n+1}.$ Since the binary operation $\circ$ is a continuous function, we have $\lim_{n\to\infty}\bigl((e-b) c_n \bigr) = (e-b)c,$ $\lim_{n\to\infty}\bigl( c_n (e-b) \bigr) = c (e-b).$ Also, the assumption $\left\|b\right\| \lt 1$ implies that $\lim_{n\to\infty}\bigl( e - b^{n+1} \bigr) = e.$ The last three equalities and $(e-b) c_n = c_n (e-b) = e - b^{n+1}$ yield that $(e-b)c = c (e-b) = e.$ This proves the invertibility of $e-b$. If $a$ is invertible and if we assume $\left\|b\right\| \lt 1/\left\|a^{-1}\right\|,$ then we have that $\left\| a^{-1}b\right\| \lt 1$. Thus $e - b a^{-1}$ is invertible by the special case which we just proved. Since $a - b = (e - b a^{-1}) a,$ the element $a-b$ is invertible as a product of invertible elements and $(a-b)^{-1} = a^{-1} (e - ba^{-1})^{-1}.$

8. Next we introduced the concept of the operator adjoint. Here we consider two Hilbert spaces $(\mathcal H, \langle\cdot\,,\cdot\rangle_{\mathcal H})$ and $(\mathcal K, \langle\cdot\,,\cdot\rangle_{\mathcal K})$. Denote by $\Phi_{\mathcal H}$ and $\Phi_{\mathcal K}$ the corresponding conjugate-linear bijections introduced in the Theorem posted on May 9.

Theorem. Let $(\mathcal H, \langle\cdot\,,\cdot\rangle_{\mathcal H})$ and $(\mathcal K, \langle\cdot\,,\cdot\rangle_{\mathcal K})$ be Hilbert spaces. Let $T \in {\mathcal L}(\mathcal H,\mathcal K)$. Then there exists a unique operator $T' \in {\mathcal L}(\mathcal K,\mathcal H)$ such that $\forall\, u \in \mathcal H \ \ \forall\, x \in \mathcal K \qquad \left\langle Tu , x \right\rangle_{\mathcal K} = \left\langle u , T' x \right\rangle_{\mathcal H}.$ The operator $T'$ is given by $\forall\, x \in \mathcal K \qquad T'x = \Phi_{\mathcal H}^{-1} \bigl(\Phi_{\mathcal K}(x)\circ T\bigr)$ and $\left\| T' \right\|_{{\mathcal L}(\mathcal K,\mathcal H)} = \left\| T \right\|_{{\mathcal L}(\mathcal H,\mathcal K)}.$

Definition. Let $(\mathcal H, \langle\cdot\,,\cdot\rangle_{\mathcal H})$ and $(\mathcal K, \langle\cdot\,,\cdot\rangle_{\mathcal K})$ be Hilbert spaces. Let $T \in {\mathcal L}(\mathcal H,\mathcal K)$. The adjoint of $T$, denoted by $T^*$, is the unique operator in ${\mathcal L}(\mathcal K,\mathcal H)$ which satisfies $\forall\, u \in \mathcal H \ \ \forall\, x \in \mathcal K \qquad \left\langle Tu , x \right\rangle_{\mathcal K} = \left\langle u , T^* x \right\rangle_{\mathcal H}.$

Theorem. Let $(\mathcal G, \langle\cdot\,,\cdot\rangle_{\mathcal G})$, $(\mathcal H, \langle\cdot\,,\cdot\rangle_{\mathcal H})$ and $(\mathcal K, \langle\cdot\,,\cdot\rangle_{\mathcal K})$ be Hilbert spaces. Let $S \in {\mathcal L}(\mathcal G,\mathcal H)$ and $T, V \in {\mathcal L}(\mathcal H,\mathcal K)$. For $\alpha, \beta \in \mathbb C$ we have $(\alpha T + \beta V)^* = \overline{\alpha} T^* + \overline{\beta} V^*$ and $(TS)^* = S^* T^*.$ If $T$ is invertible, then $T^*$ is also invertible and $(T^*)^{-1} = \bigl(T^{-1}\bigr)^*.$

Definition. Let $(\mathcal H, \langle\cdot\,,\cdot\rangle_{\mathcal H})$ be a Hilbert space. An operator $T \in {\mathcal L}(\mathcal H)$ is said to be self-adjoint if $T = T^*.$

Theorem. Let $(\mathcal H, \langle\cdot\,,\cdot\rangle_{\mathcal H})$ be a Hilbert space and let $T \in {\mathcal L}(\mathcal H)$ be a self-adjoint operator. Then $\left\| T \right\|_{{\mathcal L}(\mathcal H)} = \sup \bigl\{ |\langle Tv , v \rangle_{\mathcal H}| : v \in \mathcal H \ \text{and} \ \|v\|_{\mathcal H} = 1 \bigr\}.$

9. In this item we define the spectrum and the resolvent set of a bounded operator on a Banach space. We also prove several properties of the resolvent operator function.

Definition. Let $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ be a Banach space. For $T \in {\mathcal L}({\mathcal F})$ we define the resolvent set, denoted by $\rho(T)$, to be the set of all $\lambda \in \mathbb C$ such that the operator $\lambda I - T$ is invertible. The complement of the resolvent set in $\mathbb C$ is called the spectrum of $T$, denoted by $\sigma(T).$ We define the point spectrum of $T$ to be $\sigma_p(T) = \bigl\{ \lambda \in \mathbb C : \operatorname{ker} ( \lambda I - T) \neq \{0\} \bigr\},$ the continuous spectrum of $T$ to be $\sigma_c(T) = \bigl\{ \lambda \in \mathbb C : \operatorname{ker} ( \lambda I - T) = \{0\}, \ \ \operatorname{ran} ( \lambda I - T) \neq \mathcal F \ \ \text{and} \ \ \overline{\operatorname{ran} ( \lambda I - T)} = \mathcal F \bigr\}$ and the residual spectrum of $T$ to be $\sigma_r(T) = \bigl\{ \lambda \in \mathbb C : \operatorname{ker} ( \lambda I - T) = \{0\} \ \ \text{and} \ \ \overline{\operatorname{ran} ( \lambda I - T)} \neq \mathcal F \bigr\}.$ The complex numbers $\lambda \in \sigma_p(T)$ are called eigenvalues of $T$.

Remark. In the above definition $\lambda \in \sigma_p(T)$ if and only if there exists $x \in \mathcal F \setminus\{0\}$ such that $Tx = \lambda x.$ Thus, the above definition of an eigenvalue is an extension of the familiar definition from linear algebra.

The sets $\sigma_p(T)$, $\sigma_c(T)$ and $\sigma_r(T)$ are disjoint and their union is the spectrum of $T$.

Definition. Let $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ be a Banach space and let $T \in {\mathcal L}({\mathcal F})$. The function $\forall\, \lambda \in \rho(T) \qquad R_\lambda(T) = (\lambda I - T)^{-1}$ defined on the resolvent set of $T$ is called the resolvent of $T$.

Theorem. Let $(\mathcal F, \left\|\,\cdot\,\right\|_{\mathcal F})$ be a Banach space and let $T \in {\mathcal L}({\mathcal F})$.

1. If $\lambda \in \mathbb C$ and $|\lambda| \gt \left\|T\right\|_{{\mathcal L}({\mathcal F})}$, then $\lambda \in \rho(T)$; $R_\lambda(T) = \sum_{k=0}^\infty \frac{1}{\lambda^{k+1}} T^k \quad \text{and} \quad \left\|R_\lambda(T)\right\|_{{\mathcal L}({\mathcal F})} \leq \frac{1}{|\lambda| - \left\|T\right\|_{{\mathcal L}({\mathcal F})}}.$
2. If $\lambda_0 \in \rho(T)$ and $\lambda \in \mathbb C$ is such that $|\lambda - \lambda_0| \lt 1/\left\|R_{\lambda_0}(T)\right\|_{{\mathcal L}({\mathcal F})}$, then $\lambda \in \rho(T)$ and $R_\lambda(T) = \sum_{k=0}^\infty (-1)^k (\lambda - \lambda_0)^k \bigl(R_{\lambda_0}(T)\bigr)^{k+1}.$

Theorem. Let $(\mathcal H, \langle\cdot\,,\cdot\rangle_{\mathcal H})$ be a Hilbert space and let $T \in {\mathcal L}(\mathcal H)$. Then $\sigma(T) \neq \emptyset.$

Proof.

Monday, May 9, 2016

• We finished Chapter 6 today.
• The most important theorem in this chapter is the Riesz-Frechet theorem.

Theorem. Let $(\mathcal H, \langle\cdot\,,\cdot\rangle)$ be a Hilbert space. The mapping $\Phi$ defined on $\mathcal H$ by $\tag{*} \forall\, v\in\mathcal H \quad \forall\, u \in\mathcal H \quad \bigl(\Phi(v)\bigr)(u) = \langle u, v \rangle$ is a bijection between $(\mathcal H, \langle\cdot\,,\,\cdot\rangle)$ and its dual space $\mathcal H^*$. The bijection $\Phi: \mathcal H \to \mathcal H^*$ is conjugate-linear and $\tag{**} \forall\, v\in\mathcal H \quad \bigl\| (\Phi(v) \bigr\|_{\mathcal H^*} = \|v\|_{\mathcal H}.$

Proof. The first step is to prove that for every $v\in \mathcal H$ the mapping $\Phi(v)$ defined by (*) is a continuous linear functional on $\mathcal H$. To prove continuity one uses the Cauchy-Bunakovky-Schartz inequality. At this point one can also prove that (**) holds.

The second step is to prove that $\Phi: \mathcal H \to \mathcal H^*$ is an injection.

The third step is to prove that $\Phi: \mathcal H \to \mathcal H^*$ is a surjection. (This is the trickiest part of the proof.)

The fourth step is to prove that $\Phi: \mathcal H \to \mathcal H^*$ is conjugate linear. That is $\forall\,\alpha, \beta \in \mathbb C \quad \forall\, v, w \in \mathcal H\qquad \Phi(\alpha v + \beta w) = \overline{\alpha} \Phi(v) + \overline{\beta} \Phi(w).$

Monday, April 18, 2016

• We finished Chapter 3 today. The relevant Exercise in Chapter 3 are 3.3. The relevant problems in Chapter 3 are the problems 3.1 through 3.4, 3.6, 3.7, 3.8, 3.9, 3.10. In addition to that the book calls the "closest point property" we did a characterization of the best approximation property for subspaces using orthogonality. A proof of that characterization is on my Cauchy-Bunyakovsky-Schwarz inequality page.
• In addition to what is covered in Chapter 2, I proved that each normed space has a completion and I proved the characterization of completeness of a normed space in terms of the absolutely convergent series. This material is on my normed spaces page. The relevant exercises in Chapter 2 are 2.4, 2.8, 2.11 The relevant problems in Chapter 2 are the problems 2.1 through 2.4, 2.7, 2.9, 2.11, 2.12, 2.13,

• Thursday, April 14, 2016

• I wrote a page about normed spaces. On this page I prove that every normed space has a completion. I also prove that a normed space is complete if and only if every absolutely convergent series is convergent.

• Monday, April 4, 2016

• We finished Chapter 1. The relevant Exercises in Chapter 1 are 1.3, 1.4, 1.10, 1.12, 1.15. The relevant problems in Chapter 1 are the problems 1.1 through 1.10.
• I rewrote the page about Cauchy-Bunyakovsky-Schwarz inequality. I aligned the terminology that I use on this page with the terminology used in the book.

• Tuesday, March 29, 2016