Mercer's Theorem and Reproducing Kernels

The kernel is bounded on the compact square $[a,b]^2$, so $\int\!\!\int K^2<\infty$, which makes $T_K$ a Hilbert-Schmidt operator. Fix an orthonormal basis $(\psi_i)$ of $L^2$. The products $(\psi_i(s)\psi_j(t))_{i,j}$ form an orthonormal basis of $L^2([a,b]^2)$, so $K=\sum_{i,j} c_{ij}\psi_i(s)\psi_j(t)$ with $\sum_{i,j}c_{ij}^2=\int\!\!\int K^2<\infty$. Truncating the expansion to $i,j\le N$ gives a kernel whose operator $F_N$ has finite-dimensional range, and $\norm{T_K-F_N}^2\le \int\!\!\int(K-K_N)^2=\sum_{i\text{ or }j>N}c_{ij}^2\to 0$, the operator norm being bounded by the Hilbert-Schmidt norm. So $T_K$ is a norm limit of finite-rank operators, hence compact. Self-adjointness is $\ip{T_K f}{g}=\int\!\!\int K(s,t)f(t)g(s)=\ip{f}{T_K g}$ by symmetry and Fubini, and positivity holds by hypothesis on $K$.

The kernel is uniformly continuous on the compact square, so for $\varepsilon>0$ there is $\delta>0$ with $\abs{K(s,t)-K(s',t)}<\varepsilon$ whenever $\abs{s-s'}<\delta$, uniformly in $t$. Then by Cauchy-Schwarz

so $T_K f$ is continuous. Since $\lambda_n\varphi_n=T_K\varphi_n$ lies in the range of $T_K$, the function $\lambda_n^{-1}T_K\varphi_n$ is a continuous representative of $\varphi_n$ when $\lambda_n>0$.

Write $S_N(s,t)=\sum_{n\le N}\lambda_n\varphi_n(s)\varphi_n(t)$ for the partial sums.

Convergence in $L^2$. By the spectral theorem the closure of the range of $T_K$ is the closed span of $\{\varphi_n:\lambda_n>0\}$, and self-adjointness gives $(\overline{\operatorname{ran}T_K})^\perp=\ker T_K$, so any direction orthogonal to all $\varphi_n$ lies in $\ker T_K$. Extend $\{\varphi_n:\lambda_n>0\}$ by an orthonormal basis of $\ker T_K$ to a complete orthonormal basis $(u_i)$ of $L^2$; then $(u_i(s)u_j(t))$ is a complete orthonormal basis of $L^2([a,b]^2)$. The coefficient $\ip{K}{u_i\otimes u_j}=\ip{T_K u_j}{u_i}$ vanishes unless $u_j=\varphi_n$ with $\lambda_n>0$, in which case it equals $\lambda_n\delta_{ij}$. Hence $K=\sum_{n:\lambda_n>0}\lambda_n\varphi_n\otimes\varphi_n$ in $L^2([a,b]^2)$, so $S_N\to K$ in $L^2$.

The diagonal is bounded. The remainder $K-S_N$ has integral operator $T_K-\sum_{n\le N}\lambda_n \varphi_n\otimes\varphi_n$, which is positive because $\ip{(T_K-\sum_{n\le N}\lambda_n\varphi_n\otimes \varphi_n)f}{f}=\sum_{n>N}\lambda_n\ip{f}{\varphi_n}^2\ge 0$. A continuous positive kernel is nonnegative on the diagonal, since a strictly negative value $K_N(s_0,s_0)<0$ would persist on a square $[s_0-\eta,s_0+\eta]^2$ by continuity and, with $g=\mathbf 1_{[s_0-\eta,s_0+\eta]}$, make $\ip{T_{K_N}g}{g}=\int\!\!\int_{[s_0-\eta,s_0+\eta]^2}K_N(s,t)\,ds\,dt<0$. Thus $\sum_{n\le N}\lambda_n\varphi_n(s)^2\le K(s,s)$ for all $s$ and $N$, and the increasing sums converge pointwise to a limit $g(s)\le K(s,s)$.

Convergence in $t$ for fixed $s$. For $m>N$, Cauchy-Schwarz on the increment gives

where $S_N^{\mathrm d}(s)=\sum_{n\le N}\lambda_n\varphi_n(s)^2$ and the second factor uses the diagonal bound at $t$. Since $S_N^{\mathrm d}(s)\to g(s)$, the sums $S_N(s,\cdot)$ are uniformly Cauchy in $t$, converging uniformly to a continuous function $F_s$ with $F_s(t)=\sum_n\lambda_n\varphi_n(s)\varphi_n(t)$.

A pointwise inequality. The remainder $K_N=K-S_N$ is a continuous kernel with positive operator, so its values obey $K_N(s,t)^2\le K_N(s,s)K_N(t,t)$ for all $s,t$. To see this, evaluate the positivity $\ip{T_{K_N}f}{f}\ge 0$ on $f=\alpha\mathbf 1_I/\abs I+\beta\mathbf 1_J/\abs J$ for intervals $I\ni s$, $J\ni t$ shrinking to their centres. Continuity sends the quadratic form to $\alpha^2 K_N(s,s)+2\alpha \beta K_N(s,t)+\beta^2 K_N(t,t)\ge 0$ for all real $\alpha,\beta$, and a positive semidefinite binary form has nonnegative discriminant, which is the inequality.

Identification. For each $s$ the uniform limit gives $K_N(s,\cdot)=K(s,\cdot)-S_N(s,\cdot)\to K(s,\cdot) -F_s=:h_s$ uniformly in $t$, with $h_s$ continuous. Separately, $S_N\to K$ in $L^2([a,b]^2)$ forces, along a subsequence, $S_N(s,\cdot)\to K(s,\cdot)$ in $L^2(t)$ for almost every $s$, so $h_s=0$ on a set $G$ of full measure, which is dense. For $s'\in G$ the a.e.-$t$ equality $F_{s'}=K(s',\cdot)$ between continuous functions holds everywhere, so at $t=s'$ we get $g(s')=F_{s'}(s')=K(s',s')$, whence the diagonal $K_N(s',s')=K(s',s')-S_N^{\mathrm d}(s')\to K(s',s')-g(s')=0$. Fixing any $s_0$ and any $s'\in G$, the diagonal bound $0\le K_N(s_0,s_0)\le K(s_0,s_0)$ keeps the first factor bounded, so the pointwise inequality gives $K_N(s_0,s')^2\le K_N(s_0,s_0)K_N(s',s')\le K(s_0,s_0)K_N(s',s')\to 0$, so $h_{s_0}(s')=0$. The continuous $h_{s_0}$ thus vanishes on the dense $G$, hence everywhere, giving $K(s_0,\cdot)=F_{s_0}$ for every $s_0$. So $K(s,t)=\sum_n\lambda_n\varphi_n(s)\varphi_n(t)$ for all $(s,t)$, and in particular $g(s)=K(s,s)$.

Uniformity. The continuous increasing sums $S_N^{\mathrm d}(s)$ now converge to the continuous limit $K(s,s)$ on the compact interval $[a,b]$, so by Dini's theorem the convergence is uniform in $s$. The bound Equation (4) then has its first factor uniformly small and its second bounded by $\max_t K(t,t)^{1/2}$, making $S_N\to K$ uniformly on $[a,b]^2$. The same bound with absolute values term by term gives absolute convergence at each point.

Integrate $K(s,s)=\sum_n\lambda_n\varphi_n(s)^2$ over $[a,b]$. Uniform convergence permits term-by-term integration, and $\int\varphi_n^2=1$ leaves $\sum_n\lambda_n$.

The coordinates of $K_s$ are $\lambda_n\varphi_n(s)$, so $\ip{f}{K_s}_{\mathcal H_K}=\sum_{n:\lambda_n>0} a_n\cdot\lambda_n\varphi_n(s)/\lambda_n=\sum_{n:\lambda_n>0} a_n\varphi_n(s)=f(s)$, the last equality being the expansion of $f$ evaluated at $s$, which converges absolutely since $\sum_{n:\lambda_n>0}\abs{a_n \varphi_n(s)}\le\norm f_{\mathcal H_K}K(s,s)^{1/2}$ by Cauchy-Schwarz.