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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.

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 Hermitian matrix with unitary diagonalization $A=U\mathrm{diag}({\lambda }_i)U^{\ast }$, define $f(A)=U\mathrm{diag}(f({\lambda }_i))U^{\ast }$ for a function on its finite spectrum. This acts on eigenvalues, unlike an entrywise matrix function $f[A]$. For example, the spectral square of a matrix is its ordinary matrix product with itself, whereas its entrywise square is a Hadamard power.