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Let real power series $f(t)={\sum }_{k\ge 0}{f}_k{t}^k$ and $g(t)={\sum }_{k\ge 0}{g}_k{t}^k$ converge on $[-1,1]$, with $f_k\ge |g_k|$ for every index including zero. If $X⪰0$ has entries there, applying $f$ within two diagonal blocks and $g$ across them preserves positive semidefiniteness. Indeed the coefficient matrix $H_k$ is a bipartite block-constant positive semidefinite matrix. Each $H_k\circ X^{\circ k}$ is positive semidefinite by the Schur product theorem, and their sum converges to the transformed matrix. Closedness of the positive semidefinite cone completes the proof. Convergence at one gives $\sum f_k<\infty$, so coefficient domination ensures absolute convergence of both series throughout the interval. Without the zero-index condition, $f=-1$, $g=0$ and $X=0$ give a counterexample.

For integer $k\ge 1$, the entrywise power has entries $X_{ij}^k$ and equals the repeated Hadamard product. The zero power is the all-ones matrix, even at zero entries, because it represents the constant function one. Every nonnegative integer Hadamard power of a positive semidefinite matrix is positive semidefinite by the Schur product theorem. It is not an ordinary matrix power.

The matrix with entries $(P\circ Q{)}_{ij}=P_{ij}{Q}_{ij}$. It differs from ordinary matrix multiplication. The Schur product theorem states that it preserves positive semidefiniteness when both factors are positive semidefinite matrices.

For each coefficient index $k\ge 0$, let $H_k$ have constant diagonal blocks $f_{k}J$ and cross blocks $g_{k}J$. The preceding block-constant argument gives $H_k⪰0$. Define the Hadamard powers $X^{\circ k}$, withThis is the constant entrywise power, including at zero entries; it is not the identity matrix. Repeated use of the Schur product theorem shows $X^{\circ k}⪰0$ for every $k\ge 0$.

Applying the Schur product theorem again makes every summand $H_k\circ X^{\circ k}$ positive semidefinite. The partial sums are positive semidefinite, and their entrywise limit is exactly $Y$. In finite dimension this is a matrix-norm limit; the positive semidefinite cone is closed. ConsequentlyEndpoint convergence follows from the stated expansions on the full interval: $f_k\ge 0$ and $\sum f_k=f(1)<\infty$, while $\sum |g_k|\le \sum f_k$. Thus the expansions converge absolutely at every matrix entry. This is coefficient-dominated entrywise positivity.

The coefficient condition must include index zero. We interpret the PDF's $\mathbb{N}$ accordingly. If it means only positive integers and no condition is imposed on $f_0,g_0$, the assertion is false: take $n=m=1$, $X=0$, $f=-1$ and $g=0$. All positive-index inequalities hold, but $Y=-I_2$.

For a finite sample, the Gram matrix of $k_1{k}_2$ is the entrywise product of the two positive-semidefinite Gram matrices. The Schur product theorem makes it positive semidefinite, proving the product closure.

The Gaussian kernel with bandwidth ${\sigma }^2$ isFactor it asThe linear kernel $x^T{x}^{\prime }$ is positive semidefinite; products and nonnegative scalar multiples preserve positivity, as does multiplication by $a(x)a(x^{\prime })$. The partial sums are therefore kernels, and pointwise-limit closure proves that the Gaussian kernel is positive semidefinite.

PutThe numerator is the linear kernel, while the reciprocal of the denominator is the kernel from part c applied to $\Vert x{\Vert }^2$ and $\Vert y{\Vert }^2$. 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 $|a(x,y)|\le 1/2$, soEach 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, thenThe feature-space norm gives symmetry and the triangle inequality. Finally, $d(x,y)=0$ implies $k_2(x,y)=1$, hence $a(x,y)=1/2$ andwhich is equivalent to $\Vert x-y{\Vert }^2=0$. Thus $d$ is a metric rather than merely a pseudometric.

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.