Hilbert Space
Definition 1 (Norm) Let $\mathcal{F}$ be a vector space over $\mathbb{R}$ (For example $\mathcal{F}=\mathbb{R}^n$ is a vector space). A function $||\cdot||_{\mathcal{F}}: \mathcal{F} \to [0, \inf)$ is said to be a norm on $\mathcal{F}$ if ($||\cdot||_{\mathcal{F}}$ 是一个有效norm算子要满足以下条件):
- For $f \in \mathcal{F}$, $||f||_{\mathcal{F}}=0$ if and only if $f=0$. (norm separates points)
- $|\lambda f|_{\mathcal{F}}=|\lambda||f|_{\mathcal{F}}$, $\forall \lambda \in \mathbb{R}, \forall f \in \mathcal{F}$ (positive homogeneity).
- $|f+g|_{\mathcal{F}} \leq|f|_{\mathcal{F}}+|g|_{\mathcal{F}}, \forall f, g \in \mathcal{F}$ (triangle inequality).
向$||\cdot||_{\mathcal{F}}$中输入任意一个向量，只要满足以上条件，那么$||\cdot||_{\mathcal{F}}$是一个valid norm operator.
Inner Product
An inner product takes two elements of a vector space $\mathcal{X}$ and outputs a number. An inner product could be a usual dot product: $\langle\mathbf{u}, \mathbf{v}\rangle=\mathbf{u}^{\prime} \mathbf{v}=\sum_{i} u^{(i)} v^{(i)}$ (Inner Product can be Dot Product). Or the inner product could be something fancier（即内积不一定表示为点积的形式）. If an Inner Product $\langle \cdot,\cdot \rangle$ is valid, it MUST satisfy the following conditions:
Symmetry $$\langle u, v\rangle=\langle v, u\rangle \quad \forall u, v \in \mathcal{X}$$
Bilinearity $$\langle\alpha u+\beta v, w\rangle=\alpha\langle u, w\rangle+\beta\langle v, w\rangle \quad \forall u, v, w \in \mathcal{X}, \forall \alpha, \beta \in \mathbf{R}$$
Strict Positive Definiteness $$ \begin{gathered} \langle u, u\rangle \geq 0 \forall x \in \mathcal{X} \\ \langle u, u\rangle=0 \Longleftrightarrow u=0 \end{gathered}$$
An inner product space (or pre-Hilbert space) is a vector space together with an inner product. （包含内积运算的向量空间称为 内积空间，即可以定义内积运算的向量空间）。
Kernel is a kind of Inner Product. For example, the Gaussian kernel is defined as: $$ \begin{equation} \langle u, v \rangle = k(u,v) = \exp({-\frac{||u-v||^2}{2\sigma}}) \tag{1} \end{equation} $$
Hilbert Space
Definition 2 (Hilbert Space) A Hilbert Space is an Inner Product space that is complete and separable with respect to the norm defined by the inner product.
‘Complete’ means sequences converge to elements of the space - there aren’t any “holes” in the space.
Finite States
Given finite input space ${x_1, x_2, \cdots x_m }$. I want to be able to take inner products between any two of them using my function $k$ as the inner product ($k$ is customized and satisfy three conditions. For example, $k$ is a Gaussian inner product as Eq.(1)). Inner products by definition are symmetric, so $k(x_i, x_j)=k(x_j, x_i)$ , which yields a symmetric matrix $\mathbf{K}$.
Since $\mathbf{K}$ is real symmetric, and this means we can diagonalize it （实对称阵可以对角化，即特征分解）, and the eigendecomposition takes this form: $$ \begin{equation} \begin{aligned} \mathbf{K} &=\mathbf{V} \Lambda \mathbf{V}^T \\ &= \mathbf{V} \begin{bmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \cdots &\ & & &\lambda_m \end{bmatrix} \mathbf{V}^T \\ &= \begin{bmatrix} v_1 & v_2 & \cdots v_m \end{bmatrix} \begin{bmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \cdots &\\ & & &\lambda_m \end{bmatrix} \begin{bmatrix} v_1^T\\ v_2^T\\ \cdots \\ v_m^T \end{bmatrix}\\ &=v_1\lambda_1 v_1^T + \cdots + v_m\lambda_m v_m^T = \sum_{t=1}^m v_t\lambda_tv_t^T \end{aligned} \tag{2} \end{equation} $$ Let the $i$-th element of vector $v$ as $v^{(i)}$, then $$ \begin{equation} \begin{aligned} \mathbf{K}_{ij} = k(x_i, x_j) &= [\sum_{t=1}^m v_t\lambda_tv_t^T]_{ij}\\ &=\sum^m_{t=1} v_t^{(i)} \lambda_t v_t^{(j)} \end{aligned} \tag{3} \end{equation} $$ If $\mathbf{K}$ is a positive semi-definite (PSD) matrix, then $\lambda_1, \cdots \lambda_m \geq 0$.
Assumption 1. All $\lambda_t$ are nonnegative.
We consider this feature map: $$ \begin{equation} \Phi\left(x_{i}\right)=\left[\sqrt{\lambda_{1}} v_{1}^{(i)}, \ldots, \sqrt{\lambda_{t}} v_{t}^{(i)}, \ldots, \sqrt{\lambda_{m}} v_{m}^{(i)}\right] \in \mathbb{R}^m \tag{4} \end{equation} $$ (writing it for $x_j$ too): $$ \begin{equation} \boldsymbol{\Phi}\left(x_{j}\right)=\left[\sqrt{\lambda_{1}} v_{1}^{(j)}, \ldots, \sqrt{\lambda_{t}} v_{t}^{(j)}, \ldots, \sqrt{\lambda_{m}} v_{m}^{(j)}\right] \in \mathbb{R}^m \tag{5} \end{equation} $$ 即 $\Phi: \mathcal{X} \to \mathbb{R}^m$ 将$x\in \mathcal{X}$映射到$m$维向量空间$\mathbb{R}^m$中的一个点。
With this choice, the inner product $k$ is just defined as a dot product in $\mathbb{R}^m$: $$ \begin{equation} \left\langle\Phi\left(x_{i}\right), \Phi\left(x_{j}\right)\right\rangle_{\mathbf{R}^{m}}=\sum_{t=1}^{m} \lambda_{t} v_{t}^{(i)} v_{t}^{(j)}=\left(\mathbf{V} \Lambda \mathbf{V}^{\prime}\right)_{i j}=K_{i j}=k\left(x_{i}, x_{j}\right) \tag{6} \end{equation} $$ If there exists an eigenvalue $\lambda_s <0$ (即$\sqrt{\lambda_s} = \sqrt{|\lambda_s|} i$). $\lambda_s$对应的特征向量$v_s$。用$v_s \in \mathbb{R}^m$的$m$个元素$v_s = [v_s^{(1)},\cdots, v_s^{(m)}]$, 来对$\Phi(x_1),\cdots, \Phi(x_m)$做线性组合： $$ \begin{equation} \mathbf{z}=\sum_{i=1}^{m} v_{s}^{(i)} \boldsymbol{\Phi}\left(x_{i}\right) \tag{7} \end{equation} $$
It is obvious that $\mathbf{z} \in \mathbb{R}^m$. Then calculate $$ \begin{equation} \begin{aligned} |\mathbf{z}|_{2}^{2} &=\langle\mathbf{z}, \mathbf{z}\rangle_{\mathbf{R}^{m}}=\sum_{i} \sum_{j} v_{s}^{(i)} \boldsymbol{\Phi}\left(x_{i}\right)^{T} \boldsymbol{\Phi}\left(x_{j}\right) v_{s}^{(j)}=\sum_{i} \sum_{j} v_{s}^{(i)} K_{i j} v_{s}^{(j)} \\ &=\mathbf{v}_{s}^{T} \mathbf{K} \mathbf{v}_{s}=\lambda_{s}<0 \end{aligned} \tag{8} \end{equation} $$ which conflicts with the geometry of the feature space.
如果$\mathbf{K}$不是半正定，那么feature space $\mathbb{R}^m$存在小于0的值。所以假设Assumption不成立。即，若$k$表示有限集的内积，那么它的Gram Matrix一定半正定(PSD)，否则无法保证该空间中的norm大于0。
有效的内积对应的Gram Matrix 必定PSD.
Kernel
Definition 3. (Kernel) A function $k: \mathcal{X} \times \mathcal{X} \to \mathbb{R}$ is a kernel if
- $k$ is symmetric: $k(x,y) = k(y,x)$.
- $k$ gives rise to a positive semi-definite “Gram matrix,” i.e., for any $m\in \mathbb{N}$ and any $x_1,\cdots,x_m$ chosen from $X$, the Gram matrix $\mathbf{K}$ defined by $\mathbf{K}_{ij} = k(x_i,x_j)$ is positive semi-definite.
Another way to show that a matrix $\mathbf{K}$ is positive semi-definite is to show that $$ \begin{equation} \forall \mathbf{c} \in \mathbf{R}^{m}, \mathbf{c}^{T} \mathbf{K} \mathbf{c} \geq 0 \tag{9} \end{equation} $$ Here are some nice properties of $k$:
- $k(u,u) \geq 0$ (Think about the Gram matrix of $m = 1$.)
- $k(u, v) \leq \sqrt{k(u, u) k(v, v)}$ (This is the Cauchy-Schwarz inequality.)
Reproducing Kernel Hilbert Space (RKHS)
给定一个kernel $k(\cdot, \cdot): \mathcal{X} \times \mathcal{X} \to \mathbb{R}$. 定义一个函数空间（space of functions）$\mathbf{R}^{\mathcal{X}}:={f: \mathcal{X} \rightarrow \mathbb{R}}$. $\mathbf{R}^{\mathcal{X}}$ 是一个 Hilbert Space， 该空间中的每个元素是一个$\mathcal{X}$映射到$\mathbb{R}$的函数。
令$k_x(\cdot) = k(x, \cdot)$, 假设$x$是一个定值（Constant），自变量（输入）用$\cdot$表示。那么$k(x, \cdot)$ 也是$\mathbf{R}^{\mathcal{X}}$空间中的一个函数。
每个函数$k_x(\cdot)$ 都与一个特定的$x \in \mathcal{X}$有关，即每个$x$对应于一个函数$k_x(\cdot) = k(\cdot, x)$. 这种对应关系表示为$\Phi(x) = k_x(\cdot) = k(x,\cdot)$, 即： $$ \begin{equation} \Phi: x \longmapsto k(\cdot, x) \tag{10} \end{equation} $$ 即 $\Phi$的输入为$x\in \mathcal{X}$, 输出一个函数, 输出的函数属于$\mathbf{R}^{\mathcal{X}}$空间。
在连续空间$\mathcal{X}$中，$x \in \mathcal{X}$ 有无穷多种情况，那么$\Phi(x)=k_x(\cdot)=k(x, \cdot)$也有无穷多种情况，即无穷多种函数。 这些函数可以span 一个Hilbert Space: $$ \begin{equation} \mathcal{H}_k = \operatorname{span}({\Phi(x): x \in \mathcal{X}})=\left\{f(\cdot)=\sum_{i=1}^{m} \alpha_{i} k\left(\cdot, x_{i}\right): m \in \mathbf{N}, x_{i} \in \mathcal{X}, \alpha_{i} \in \mathbf{R}\right\} \tag{11} \end{equation} $$ 其中$k(x,\cdot)=\Phi(x)$可以理解为将$x$映射为一个函数（or vector）。上述Hilbert Space是由任意$k(x, \cdot)$线性组合而成的函数空间，该空间中的每个元素可以表示为 $$ \begin{equation} f(\cdot)=\sum_{i=1}^{m} \alpha_{i} k\left(\cdot, x_{i}\right) \tag{12} \end{equation} $$ 所以$\mathcal{H}$可以看作是kernel $k$对应的一个Hilbert Space。
给定$\mathcal{H}$中的任意两个函数$f(\cdot)=\sum_{i=1}^{m} \alpha_{i} k\left(\cdot, x_{i}\right)$, $g(\cdot)=\sum_{j=1}^{m^{\prime}} \beta_{j} k\left(\cdot, x_{j}^{\prime}\right)$。注意$f(\cdot)$和$g(\cdot)$可以表示$\mathcal{H}$中任意两个元素。我们将$\mathcal{H}$上的内积定义为： $$ \begin{equation} \langle f, g\rangle_{\mathcal{H}_{k}}=\sum_{i=1}^{m} \sum_{j=1}^{m^{\prime}} \alpha_{i} \beta_{j} k\left(x_{i}, x_{j}^{\prime}\right) \tag{13} \end{equation} $$ 由Proof证明了该内积符合三个条件，顾上式是$\mathcal{H}$空间中一个有效的内积算子。注：$\mathcal{H}_k$表示该Hilbert Space是由函数 $k(x,\cdot)$ span而成的，与Kernel $k$有关.
$k(x,\cdot)$也是$\mathcal{H}_k$中的一个函数，那么它与 $f$的内积为： $$ \begin{equation} \langle k(\cdot, x), f\rangle_{\mathcal{H}_{k}}= \sum_{i=1}^m \alpha_i k(x,x_i) =f(x) \tag{14} \end{equation} $$ Theorem 1. $k(\cdot, \cdot)$ is a reproducing kernel of a Hilbert space $\mathcal{H}_k$ if $f(x)=\langle k(x, \cdot), f(\cdot)\rangle$.
$\mathcal{H}_k$ 为$k(\cdot, \cdot)$的再生核希尔伯特空间。
同理，$k(\cdot, x_i)$, $k(\cdot, x_j)$都为$\mathcal{H}_k$中的函数， 计算他们的内积: $$ \begin{equation} \left\langle k(\cdot, x_i), k\left(\cdot, x_j\right)\right\rangle_{\mathcal{H}_{k}}=k\left(x_i, x_j\right) \tag{15} \end{equation} $$ 因为$ k(\cdot, x_i) = \Phi(x_i)$, $ k(\cdot, x_j) = \Phi(x_j)$, 所以 $$ \begin{equation} k\left(x_i, x_j\right) = \left\langle \Phi(x_i), \Phi(x_j)\right\rangle_{\mathcal{H}_{k}} \tag{16} \end{equation} $$ 表示将$x_i$和$x_j$ 映射成$\mathcal{H}_k$中的函数（向量）后再做内积。
参考文献
[1] https://people.eecs.berkeley.edu/~bartlett/courses/281b-sp08/7.pdf
[2] http://www.stats.ox.ac.uk/~sejdinov/teaching/atml14/Theory_2014.pdf
我把本文整理成了PDF