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.
If $(\mathcal F, \left\langle\,\cdot,\,\cdot\right\rangle_{\mathcal F})$ is an inner product space then the following identities hold.
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.
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.
If $(\mathcal F, \left\langle\,\cdot,\,\cdot\right\rangle_{\mathcal F})$ is an inner product space then the following identities hold.
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 \[ \left\|TS\right\|_{{\mathcal L}({\mathcal E},{\mathcal G})} \leq \left\|T\right\|_{{\mathcal L}({\mathcal F},{\mathcal G})} \left\|S\right\|_{{\mathcal L}({\mathcal E},{\mathcal F})}. \]
Theorem. 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 $(TS)^{-1} = S^{-1}T^{-1}.$
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 $(a-b)^{-1} = a^{-1} (e - ba^{-1})^{-1}.$
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 \[ \left\| T \right\|_{{\mathcal L}(\mathcal H)} = \sup \bigl\{ |\langle Tv , v \rangle_{\mathcal H}| : v \in \mathcal H \ \text{and} \ \|v\|_{\mathcal H} = 1 \bigr\}. \]
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 $T \in {\mathcal L}({\mathcal F})$.
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.
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
\[
\forall\,\alpha, \beta \in \mathbb C \quad \forall\, v, w \in \mathcal H\qquad \Phi(\alpha v + \beta w) = \overline{\alpha} \Phi(v) + \overline{\beta} \Phi(w).
\]