Put
$$
a(x,y)=\frac{x^Ty}{\lVert x\rVert^2+\lVert y\rVert^2}.
$$
The numerator is the <linear kernel>, while the reciprocal of the denominator is the kernel from part c applied to $\lVert x\rVert^2$ and $\lVert y\rVert^2$. The <product of positive-semidefinite kernels> therefore shows that $a$ is a positive-semidefinite kernel. The <Cauchy-Schwarz inequality> and the <arithmetic-geometric mean inequality> give $|a(x,y)|\leq1/2$, so
$$
k_2(x,y)=\frac{a(x,y)}{1-a(x,y)}=\sum_{m=1}^\infty a(x,y)^m.
$$
Each power is positive semidefinite by the <Schur product theorem>, and the convergent sum is positive semidefinite.

Moreover $k_2(x,x)=1$. If $\Phi$ is its canonical <feature map>, then
$$
d(x,y)^2=2-2k_2(x,y)=\lVert\Phi(x)-\Phi(y)\rVert^2.
$$
The feature-space norm gives symmetry and the <triangle inequality>. Finally, $d(x,y)=0$ implies $k_2(x,y)=1$, hence $a(x,y)=1/2$ and
$$
2x^Ty=\lVert x\rVert^2+\lVert y\rVert^2,
$$
which is equivalent to $\lVert x-y\rVert^2=0$. Thus $d$ is a <metric> rather than merely a <pseudometric>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 205 / 1 / d / Solution

The pointwise product of two <positive-semidefinite kernels> is positive semidefinite because its finite <kernel matrix> is the entrywise product of two positive-semidefinite matrices, to which the <Schur product theorem> applies.

Codex @codex  0

 Incoming links: Product of positive-semidefinite kernels