= Solution
A real <positive-semidefinite kernel> is a symmetric function $k$ such that every finite Gram matrix $K_{ij}=k(x_i,x_j)$ is positive semidefinite. The <representer theorem> says that any minimizer in a <Reproducing-kernel Hilbert space> of an objective depending on $f$ only through $f(x_1),\ldots,f(x_n)$ and a strictly increasing function of $\lVert f\rVert_{\mathcal H}$ lies in
$$
\operatorname{span}\{k(x_1,\mathord\cdot),\ldots,k(x_n,\mathord\cdot)\}.
$$
Indeed, write $f=f_\parallel+f_\perp$ relative to this span. The reproducing property gives $f_\perp(x_i)=0$ for every $i$, while the <Pythagorean theorem in an inner-product space> gives $\lVert f\rVert^2=\lVert f_\parallel\rVert^2+\lVert f_\perp\rVert^2$. Removing a nonzero perpendicular component preserves all data values and strictly decreases the penalty, proving the theorem.
Apply this decomposition to both optimizers and write $f=\sum_i\alpha_i k(X_i,\cdot)$ and $g=\sum_i\beta_i l(Y_i,\cdot)$. If $K$ and $L$ are the two <Gram matrices>, then
$$
\sum_i f(X_i)g(Y_i)=\alpha^TKL\beta,
\qquad \lVert f\rVert_{\mathcal H}^2=\alpha^TK\alpha,
\qquad \lVert g\rVert_{\mathcal G}^2=\beta^TL\beta.
$$
Writing $u=K^{1/2}\alpha$ and $v=L^{1/2}\beta$, with pseudoinverses on the respective ranges, turns the supremum into
$$
\sup_{\lVert u\rVert_2,\lVert v\rVert_2\leq1}u^TK^{1/2}L^{1/2}v
=\sigma_{\max}(K^{1/2}L^{1/2}).
$$
Solved by gpt-5.6-sol high.
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