If I'm reading it correctly, your definition of cpd would include any kernel where $k(x, x) > 0$ for any point $x$, by just considering the subset $\{ x \}$.
One way to phrase a matrix being positive definite is that $c^T A c > 0$, for all vectors $c$; a kernel is positive definite if that holds for all Gram matrices $A$. The conditioning in cpd refers to reducing the "for all vectors $c$" part, not the "for all Gram matrices $A$" part. Specifically, the most common definition is reducing it to "for all vectors $c$ with $1^T c = 0$.
These are useful because, just as pd kernels correspond to inner products in some Hilbert space, cpd kernels are closely connected to distances in some Hilbert space. SVMs and kernel PCA both work with any cpd kernel, though a particular optimization algorithm might not.
Here's the full definition from Schölkopf and Smola's Learning with Kernels (2002). Here $\mathbb K$ refers to either $\mathbb R$ or $\mathbb C$.
Definition 2.20 (Conditionally Positive Definite Matrix) A symmetric $m \times m$ matrix $K$ ($m \ge 2$) taking values in $\mathbb K$ and satisfying
$$ \sum_{i, j = 1}^m c_i \bar{c}_j K_{ij} \ge 0 \text{ for all } c_i \in \mathbb K, \text{ with } \sum_{i=1}^m c_i = 0$$
is called conditionally positive definite (cpd).
Definition 2.21 (Conditionally Positive Definite Kernel) Let $\mathcal X$ be a nonempty set. A function $k : \mathcal X \times \mathcal X \to \mathbb K$ which for all $m \ge 2, x_1, \dots, x_m \in \mathcal X$ gives rise to a conditionally positive definite Gram matrix is called a conditionally positive definite (cpd) kernel.
I don't know a simple eigenvalue-based definition.
To prove a kernel is cpd, you can either prove it directly from the definition above (usually difficult), or you can try to build it out of other cpd kernels using various properties that preserve cpd-ness.
Here are some useful connections to pd kernels from that book:
Proposition 2.22 (Constructing PD Kernels from CPD Kernels [42, Lemma 2.1, p. 74]) Let $x_0 \in \mathcal X$, and let $k$ be a symmetric kernel on $\mathcal X \times \mathcal X$. Then
$$\tilde k(x, x') := \tfrac12 \left( k(x, x') - k(x, x_0) - k(x_0, x') + k(x_0, x_0) \right)$$
is positive definite if and only if $k$ is conditionally positive definite.
[42] is Berg, Christensen, and Ressel's Harmonic Analysis on Semigroups (1984).
Proposition 2.28 (Connection PD – CPD [465, Thm. 1, p. 527]) A kernel $k$ is conditionally positive definite if and only if $\exp(t k)$ is positive definite for all $t > 0$.
[465] is the classic paper Metric spaces and positive definite functions of Schoenberg (Transactions of the American Mathematical Society 44:522-536, 1938).
Yes, it is. The proof is as follows,
$$K(x, y) = \min(f(x)g(y), f(y)g(x))=\min(\frac{f(x)}{g(x)}, \frac{f(y)}{g(y)})g(x)g(y)$$
$$\min(\frac{f(x)}{g(x)}, \frac{f(y)}{g(y)})=\int\mathbb{1}_{[0, \frac{f(x)}{g(x)}]}
\mathbb{1}_{[0, \frac{f(y)}{g(y)}]} = \langle \mathbb{1}_{[0, \frac{f(x)}{g(x)}]}, \mathbb{1}_{[0, \frac{f(y)}{g(y)}]} \rangle = \langle \phi(x), \phi(y) \rangle$$
By Aronszajn theorem using $\phi(x) = \mathbb{1}_{[0, \frac{f(x)}{g(x)}]}$ , $\min(\frac{f(x)}{g(x)}, \frac{f(y)}{g(y)})$ is positive definite kernel.
$g(x)g(y)$ is trivially a positive kernel as it is a product.
$K(x, y)$ is thus the product of two positive definite kernels, so it is a positive definite kernel.
Best Answer
To show that $K(x, y)$ is semi-positive definite, it is equivalent to show that for any $(x, y) \in [0, 1]^2$, the Gram matrix \begin{align} M(x,y) = \begin{bmatrix} \min(x, x) - x^2 & \min(x, y) - xy \\ \min(x, y) - xy & \min(y, y) - y^2 \end{bmatrix} = \begin{bmatrix} x - x^2 & \min(x, y) - xy \\ \min(x, y) - xy & y - y^2 \end{bmatrix} \end{align} is a semi-positive definite matrix. Since it is trivial that $x - x^2 \geq 0$ when $x \in [0, 1]$, it is sufficient to verify that the determinant of $M(x, y)$ is non-negative.
For simplicity, denote $\min(x, y)$ by $z$. Straightforward calculation shows that \begin{align} \det(M(x, y)) & = (x - x^2)(y - y^2) - (z - xy)^2 = xy - xy^2 - x^2y -z^2 +2zxy \\ & = \begin{cases} xy - xy^2 + x^2y - x^2 = x(1 - y)(y - x) & \text{ if } x \leq y \\ xy + xy^2 - x^2y - y^2 = y(1 - x)(x - y) & \text{ if } x > y \end{cases} \\ & \geq 0. \end{align}
This completes the proof.