Added the last section to functional analysis.
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- 'Legendre polynomials': mathematics/functional-analysis/inner-product-spaces/polynomials/legendre-polynomials.md
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- 'Hermite polynomials': mathematics/functional-analysis/inner-product-spaces/polynomials/hermite-polynomials.md
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- 'Laguerre polynomials': mathematics/functional-analysis/inner-product-spaces/polynomials/laguerre-polynomials.md
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- 'Representations of functionals': mathematics/functional-analysis/inner-product-spaces/representations-of-functionals.md
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- 'Operator classes': mathematics/functional-analysis/inner-product-spaces/operator-classes.md
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- 'Topology':
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- 'Fiber bundles': mathematics/topology/fiber-bundles.md
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- 'Calculus':
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# Operator classes
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## Hilbert-adjoint operator
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> *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 $y \in Y$
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>
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> $$
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> \langle Tx, y \rangle_Y = \langle x, T^* y \rangle.
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> $$
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We should first prove that for a given $T$ such a $T^*$ exists.
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> *Proposition 1*: the Hilbert-adjoint operator $T^*$ of $T$ exists is unique and is a bounded linear operator with norm
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>
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> $$
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> \|T^*\| = \|T\|.
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> $$
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??? note "*Proof*:"
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Will be added later.
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The Hilbert-adjoint operator has the following properties.
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> *Proposition 2*: let $T,S: X \to Y$ be bounded linear operators, then
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>
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> 1. $\forall x \in X, y \in Y: \langle T^* y, x \rangle_X = \langle y, Tx \rangle_Y$,
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> 2. $(S + T)^* = S^* + T^*$,
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> 3. $\forall \alpha \in F: (\alpha T)^* = \overline \alpha T^*$,
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> 4. $(T^*)^* = T$,
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> 5. $\|T^* T\| = \|T T^*\| = \|T\|^2$,
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> 6. $T^*T = 0 \iff T = 0$,
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> 7. $(ST)^* = T^* S^*, \text{ when } X = Y$.
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??? note "*Proof*:"
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Will be added later.
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## Self-adjoint operator
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> *Definition 2*: a bounded linear operator $T: X \to X$ on a Hilbert space $X$ is **self-adjoint** if
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>
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> $$
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> T^* = T.
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> $$
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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).
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Proposition 3, 4 and 5 pose some interesting results of self-adjoint operators.
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> *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
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>
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> $$
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> T \text{ is self-adjoint} \iff \forall x \in X: \langle Tx, x \rangle \in \mathbb{R}.
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> $$
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??? note "*Proof*:"
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Will be added later.
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> *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
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>
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> $$
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> ST = TS.
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> $$
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??? note "*Proof*:"
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Will be added later.
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Commuting operators therefore imply self-adjointness.
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> *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 $X$.
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??? note "*Proof*:"
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Will be added later.
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## Unitary operator
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> *Definition 3*: a bounded linear operator $T: X \to X$ on a Hilbert space $X$ is **unitary** if $T$ is bijective and $T^* = T^{-1}$.
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A bounded unitary linear operator has the following properties.
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> *Proposition 6*: let $U, V: X \to X$ be bounded unitary linear operators on a Hilbert space $X$, then
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>
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> 1. $U$ is isometric,
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> 2. $\|U\| = 1 \text{ if } X \neq \{0\}$,
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> 3. $UV$ is unitary,
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> 4. $U$ is normal, that is $U U^* = U^* U$,
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> 5. $T \in \mathscr{B}(X,X)$ is unitary $\iff$ $T$ is isometric and surjective.
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??? note "*Proof*:"
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Will be added later.
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# Representations of functionals
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> *Lemma 1*: let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space, if
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>
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> $$
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> \forall z \in X: \langle x, z \rangle = \langle y, z \rangle \implies x = y,
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> $$
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>
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> and if
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>
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> $$
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> \forall z \in X: \langle x, z \rangle = 0 \implies x = 0.
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> $$
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??? note "*Proof*:"
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Will be added later.
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Lemma 1 will be used in the following theorem.
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> *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
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>
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> $$
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> f(x) = \langle x, z \rangle,
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> $$
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>
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> for all $x \in x$, with $z$ uniquely dependent on $f$ and $\|z\| = \|f\|$.
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??? note "*Proof*:"
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Will be added later.
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## Sequilinear form
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> *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
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>
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> 1. $\forall x_{1,2} \in X, y \in Y: h(x_1 + x_2, y) = h(x_1, y) + h(x_2, y)$.
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> 2. $\forall x \in X, y_{1,2} \in Y: h(x, y_1 + y_2) = h(x_1, y_1) + h(x_2, y_2)$.
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> 3. $\forall x \in X, y \in Y, \alpha \in F: h(\alpha x, y) = \alpha h(x,y)$.
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> 4. $\forall x \in X, y \in Y, \beta \in F: h(x, \beta y) = \overline \beta h(x,y)$.
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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 $F$.
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> *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
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>
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> $$
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> \exists c \in F: |h(x,y)| \leq c \|x\| \|y\|,
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> $$
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>
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> for all $(x,y) \in X \times Y$ and the norm of $h$ is given by
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>
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> $$
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> \|h\| = \sup_{\substack{x \in X \backslash \{0\} \\ y \in Y \backslash \{0\}}} \frac{|h(x,y)|}{\|x\| \|y\|} = \sup_{\|x\|=\|y\|=1} |h(x,y)|.
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> $$
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For example, the inner product is sesquilinear and bounded.
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> *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
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>
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> $$
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> h(x,y) = \langle Tx, y \rangle_Y = \langle x, Sy \rangle_X,
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> $$
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>
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> for all $(x,y) \in X \times Y$, with $T$ and $S$ uniquely determined by $h$ with norms $\|T\| = \|S\| = \|h\|$.
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??? note "*Proof*:"
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Will be added later.
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