From 3a4f13d677d49723dcd9ca993c1c700f480ec747 Mon Sep 17 00:00:00 2001 From: Luc Date: Fri, 9 Aug 2024 13:27:38 +0200 Subject: [PATCH] Added the last section to functional analysis. --- config/en/mkdocs.yaml | 2 + .../inner-product-spaces/operator-classes.md | 95 +++++++++++++++++++ .../representations-of-functionals.md | 68 +++++++++++++ 3 files changed, 165 insertions(+) create mode 100644 docs/en/mathematics/functional-analysis/inner-product-spaces/operator-classes.md create mode 100644 docs/en/mathematics/functional-analysis/inner-product-spaces/representations-of-functionals.md diff --git a/config/en/mkdocs.yaml b/config/en/mkdocs.yaml index 2959957..2772fb9 100755 --- a/config/en/mkdocs.yaml +++ b/config/en/mkdocs.yaml @@ -122,6 +122,8 @@ nav: - 'Legendre polynomials': mathematics/functional-analysis/inner-product-spaces/polynomials/legendre-polynomials.md - 'Hermite polynomials': mathematics/functional-analysis/inner-product-spaces/polynomials/hermite-polynomials.md - 'Laguerre polynomials': mathematics/functional-analysis/inner-product-spaces/polynomials/laguerre-polynomials.md + - 'Representations of functionals': mathematics/functional-analysis/inner-product-spaces/representations-of-functionals.md + - 'Operator classes': mathematics/functional-analysis/inner-product-spaces/operator-classes.md - 'Topology': - 'Fiber bundles': mathematics/topology/fiber-bundles.md - 'Calculus': diff --git a/docs/en/mathematics/functional-analysis/inner-product-spaces/operator-classes.md b/docs/en/mathematics/functional-analysis/inner-product-spaces/operator-classes.md new file mode 100644 index 0000000..ba2f456 --- /dev/null +++ b/docs/en/mathematics/functional-analysis/inner-product-spaces/operator-classes.md @@ -0,0 +1,95 @@ +# Operator classes + +## Hilbert-adjoint operator + +> *Definition 1*: let $(X, \langle \cdot, \cdot \rangle_X)$ and $(Y, \langle \cdot, \cdot \rangle_Y)$ be Hilbert spaces over the field $F$ and let $T: X \to Y$ be a bounded linear operator. The **Hilbert-adjoint operator** $T^*$ of $T$ is the operator $T^*: Y \to X$ such that for all $x \in X$ amd $y \in Y$ +> +> $$ +> \langle Tx, y \rangle_Y = \langle x, T^* y \rangle. +> $$ + +We should first prove that for a given $T$ such a $T^*$ exists. + +> *Proposition 1*: the Hilbert-adjoint operator $T^*$ of $T$ exists is unique and is a bounded linear operator with norm +> +> $$ +> \|T^*\| = \|T\|. +> $$ + +??? note "*Proof*:" + + Will be added later. + +The Hilbert-adjoint operator has the following properties. + +> *Proposition 2*: let $T,S: X \to Y$ be bounded linear operators, then +> +> 1. $\forall x \in X, y \in Y: \langle T^* y, x \rangle_X = \langle y, Tx \rangle_Y$, +> 2. $(S + T)^* = S^* + T^*$, +> 3. $\forall \alpha \in F: (\alpha T)^* = \overline \alpha T^*$, +> 4. $(T^*)^* = T$, +> 5. $\|T^* T\| = \|T T^*\| = \|T\|^2$, +> 6. $T^*T = 0 \iff T = 0$, +> 7. $(ST)^* = T^* S^*, \text{ when } X = Y$. + +??? note "*Proof*:" + + Will be added later. + +## Self-adjoint operator + +> *Definition 2*: a bounded linear operator $T: X \to X$ on a Hilbert space $X$ is **self-adjoint** if +> +> $$ +> T^* = T. +> $$ + +If a basis for $\mathbb{C}^n$ $(n \in \mathbb{N})$ is given and a linear operator on $\mathbb{C}^n$ is represented by a matrix, then its Hilbert-adjoint operator is represented by the complex conjugate transpose of that matrix (the Hermitian). + +Proposition 3, 4 and 5 pose some interesting results of self-adjoint operators. + +> *Proposition 3*: let $T: X \to X$ be a bounded linear operator on a Hilbert space $(X, \langle \cdot, \cdot \rangle_X)$ over the field $\mathbb{C}$, then +> +> $$ +> T \text{ is self-adjoint} \iff \forall x \in X: \langle Tx, x \rangle \in \mathbb{R}. +> $$ + +??? note "*Proof*:" + + Will be added later. + +> *Proposition 4*: the product of two bounded self-adjoint linear operators $T$ and $S$ on a Hilbert space is self-adjoint if and only if +> +> $$ +> ST = TS. +> $$ + +??? note "*Proof*:" + + Will be added later. + +Commuting operators therefore imply self-adjointness. + +> *Proposition 5*: let $(T_n)_{n \in \mathbb{N}}$ be a sequence of bounded self-adjoint operators $T_n: X \to X$ on a Hilbert space $X$. If $T_n \to T$ as $n \to \infty$, then $T$ is a bounded self-adjoint linear operator on $X$. + +??? note "*Proof*:" + + Will be added later. + +## Unitary operator + +> *Definition 3*: a bounded linear operator $T: X \to X$ on a Hilbert space $X$ is **unitary** if $T$ is bijective and $T^* = T^{-1}$. + +A bounded unitary linear operator has the following properties. + +> *Proposition 6*: let $U, V: X \to X$ be bounded unitary linear operators on a Hilbert space $X$, then +> +> 1. $U$ is isometric, +> 2. $\|U\| = 1 \text{ if } X \neq \{0\}$, +> 3. $UV$ is unitary, +> 4. $U$ is normal, that is $U U^* = U^* U$, +> 5. $T \in \mathscr{B}(X,X)$ is unitary $\iff$ $T$ is isometric and surjective. + +??? note "*Proof*:" + + Will be added later. \ No newline at end of file diff --git a/docs/en/mathematics/functional-analysis/inner-product-spaces/representations-of-functionals.md b/docs/en/mathematics/functional-analysis/inner-product-spaces/representations-of-functionals.md new file mode 100644 index 0000000..be8c7e5 --- /dev/null +++ b/docs/en/mathematics/functional-analysis/inner-product-spaces/representations-of-functionals.md @@ -0,0 +1,68 @@ +# Representations of functionals + +> *Lemma 1*: let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space, if +> +> $$ +> \forall z \in X: \langle x, z \rangle = \langle y, z \rangle \implies x = y, +> $$ +> +> and if +> +> $$ +> \forall z \in X: \langle x, z \rangle = 0 \implies x = 0. +> $$ + +??? note "*Proof*:" + + Will be added later. + +Lemma 1 will be used in the following theorem. + +> *Theorem 1*: for every bounded linear functional $f$ on a Hilbert space $(X, \langle \cdot, \cdot \rangle)$, there exists a $z \in X$ such that +> +> $$ +> f(x) = \langle x, z \rangle, +> $$ +> +> for all $x \in x$, with $z$ uniquely dependent on $f$ and $\|z\| = \|f\|$. + +??? note "*Proof*:" + + Will be added later. + +## Sequilinear form + +> *Definition 1*: let $X$ and $Y$ be vector spaces over the field $F$. A **sesquilinear** form $h$ on $X \times Y$ is an operator $h: X \times Y \to F$ satisfying the following conditions +> +> 1. $\forall x_{1,2} \in X, y \in Y: h(x_1 + x_2, y) = h(x_1, y) + h(x_2, y)$. +> 2. $\forall x \in X, y_{1,2} \in Y: h(x, y_1 + y_2) = h(x_1, y_1) + h(x_2, y_2)$. +> 3. $\forall x \in X, y \in Y, \alpha \in F: h(\alpha x, y) = \alpha h(x,y)$. +> 4. $\forall x \in X, y \in Y, \beta \in F: h(x, \beta y) = \overline \beta h(x,y)$. + +Hence, $h$ is linear in the first argument and conjugate linear in the second argument. Bilinearity of $h$ is only true for a real field $F$. + +> *Definition 2*: let $X$ and $Y$ be normed spaces over the field $F$ and let $h: X \times Y \to F$ be a sesquilinear form, then $h$ is a **bounded sesquilinear form** if +> +> $$ +> \exists c \in F: |h(x,y)| \leq c \|x\| \|y\|, +> $$ +> +> for all $(x,y) \in X \times Y$ and the norm of $h$ is given by +> +> $$ +> \|h\| = \sup_{\substack{x \in X \backslash \{0\} \\ y \in Y \backslash \{0\}}} \frac{|h(x,y)|}{\|x\| \|y\|} = \sup_{\|x\|=\|y\|=1} |h(x,y)|. +> $$ + +For example, the inner product is sesquilinear and bounded. + +> *Theorem 2*: let $(X, \langle \cdot, \cdot \rangle_X)$ and $(Y, \langle \cdot, \cdot \rangle_Y)$ be Hilbert spaces over the field $F$ and let $h: X \times Y \to F$ be a bounded sesquilinear form. Then there exists a bounded linear operators $T: X \to Y$ and $S: Y \to X$, such that +> +> $$ +> h(x,y) = \langle Tx, y \rangle_Y = \langle x, Sy \rangle_X, +> $$ +> +> for all $(x,y) \in X \times Y$, with $T$ and $S$ uniquely determined by $h$ with norms $\|T\| = \|S\| = \|h\|$. + +??? note "*Proof*:" + + Will be added later. \ No newline at end of file