Notes on Inner Product & Hilbert Space
An inner product is a positive definite bilinear form (双线性形式) on a vector space: for a vector space $X$ with underlying scalar field $\mathbb{F}$, a map $\langle \cdot, \cdot \rangle: X \times X \to \mathbb{F}$ that satisfies
An inner product space is a set with an inner product, $(X, \langle \cdot, \cdot \rangle)$. Inner product specifies the geometry of a vector space. Inner product space has norm $|x| = \sqrt{\langle x, x \rangle}$. Inner product space may have a finite or infinite number of dimensions.
A Hilbert Space is a complete inner product space. Hilbert spaces typically arise as infinite-dimensional function spaces, e.g. the $l^2$ space of square summable infinite sequences, $L^2$ spaces of square-integrable functions, $H^s$ Sobolev spaces of twice square-integrably weakly-differentiable functions, and Hardy spaces $H^2(U)$ and Bergman spaces $L^2_a(G)$ of holomorphic functions.
A Euclidean space $(\mathbb{R}^n, (\cdot,\cdot))$ is a finite-dimensional real vector space $\mathbb{R}^n$, $n \in \mathbb{N}$, with an inner product $(x,y) = \sum_{i=1}^n x_i y_i$.
Two elements of an inner product space are orthogonal if their inner product is zero: $\langle x, y \rangle = 0$.
Orthogonal complement
Orthogonal projection
Orthogonal set; Orthogonal set; Maximal/complete orthogonal set; Orthonormal basis
Gram-Schmidt orthogonalization process
For a vector space, an approximation of a point on a (closed) subspace is a point on the subspace that is closest to the point.
Theorem (Approximation in Banach and Hilbert spaces): For a Banach space, an approximation may not exist. For a Hilbert space, the approximation of any point on any subspace exists and is unique, which is the orthogonal projection of that point.
Theorem (Riesz representation):
Theorem (Lax-Milgram):
Linear operator is an operator between two vector spaces that is compatible with their linear structures. Bounded linear operator (the image of any bounded set in $X$ is bounded in $Y$). Compact linear operator (the image of any bounded set in $X$ is pre-compact in $Y$).
Continuous operator (topological vector spaces). Weakly continuous operator. Strongly continuous operator. Compact operator, aka completely-continuous operator.
For a Hilbert space $(H, \langle \cdot , \cdot \rangle)$ of scalar functions on $X$, a scalar function $K(x,y)$ on $X \times X$ is a reproducing kernel of the Hilbert space if: let $K_y(x) \equiv K(x,y)$,
Reproducing kernels are symmetric and positive definite.
Reproducing kernel Hilbert space (RKHS) is a Hilbert space with a reproducing kernel [@Aronszajn1950].
Theorem (equivalent definition of RKHS): A Hilbert space of functions on a set $X$ is a reproducing kernel Hilbert space iff $f(y) \le c(y) |f|, \forall y \in X$, where $c(y) \equiv |K_y|$.
A reproducing kernel Hilbert space uniquely defines a reproducing kernel which is symmetric and positive definite.
Theorem: For a symmetric positive definite kernel $K$ on a set $X$, there is a unique Hilbert space of functions on $X$ for which $K$ is a reproducing kernel.
Therefore, reproducing kernel Hilbert spaces of functions on a domain are in one-to-one correspondence with positive definite kernels on the domain. We can denote any reproducing kernel Hilbert space as $H_K$, where $K$ is the unique reproducing kernel of the Hilbert space and $H_K$ is the unique Hilbert space generated by the symmetric positive definite kernel $K$.
If the space $H_K$ is sufficiently rich, then the reproducing kernel $K$ is positive definite.
Let $X$ has measure $\mu$, which defines a Hilbert space $(L^2_\mu, \langle \cdot , \cdot \rangle_\mu)$. Define integral operator $L_K$ on $H_K$ as $(L_K f)(x) = \langle K_x, f \rangle_\mu$. If $X$ is compact, then $L_K$ is compact and self-adjoint w.r.t $L^2_\mu$, so its eigenfunctions $\{e_i\}_{i \in \mathbb{N}}$ form an orthonormal basis of $L^2_\mu$, its eigenvalues $\{\lambda_i\}_{i \in \mathbb{N}}$ have finite multiplicities and converge to zero, and $K(x,y) = \sum_{i \in \mathbb{N}} \lambda_i e_i(x) e_i(y)$. If $f(x) = \sum_{i \in \mathbb{N}} a_i e_i(x)$, then $L_K f = \sum_{i \in \mathbb{N}} \lambda_i a_i e_i(x)$. It can be shown that the eigenfunctions are in $H_K$, so $\langle e_i, e_j \rangle = \delta_{ij} / \lambda_i$, and $f(x) = \sum_{i \in \mathbb{N}} a_i e_i(x) \in H_K$ iff $\sum_{i \in \mathbb{N}} a_i^2 / \lambda_i < \infty$. Let $L_K^{1/2}$ be the only positive definite self-adjoint operator that satifies $L_K^{1/2} \circ L_K^{1/2} = L_K$, then $L_K^{1/2}$ is an isomorphism from $L^2_\mu$ to $H_K$.