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For a real-valued loss function depending only on evaluations at $x_1,\dots ,x_n$, and a positive squared-norm penalty, any minimizing tuple in a sum of reproducing-kernel Hilbert spaces has the formwith the same coefficient vector for every component. The tuple must be the unique minimum-norm decomposition of its sum. Apply the representer theorem to that sum and note that the displayed tuple has total squared norm ${\sum }_j{\hat{\alpha }}^T{K}_j\hat{\alpha }={\hat{\alpha }}^T({\sum }_j{K}_j)\hat{\alpha }$, exactly the norm of the sum. This argument is conditional on existence of a minimizer; arbitrary loss functions need not attain their infimum.

A real positive-definite kernel is a symmetric function $k:\mathcal{X}\times \mathcal{X}\to \mathbb{R}$ such that, for every finite collection $x_1,\dots ,x_n$ and every $a\in {\mathbb{R}}^n$,Thus every kernel matrix is a positive semidefinite matrix; “positive definite” here does not require strict positivity. A Reproducing-kernel Hilbert space is a Hilbert space $\mathcal{H}$ of functions on $\mathcal{X}$ in which every point evaluation is a continuous linear functional. Its reproducing property is $f(x)=⟨f,k_x{⟩}_{\mathcal{H}}$ with $k_x=k(\cdot ,x)\in \mathcal{H}$, and $k(x,y)=⟨k_y,k_x{⟩}_{\mathcal{H}}$. For a sum of positive-definite kernels, the quadratic form of its kernel matrix is a sum of nonnegative quadratic forms. Symmetry also survives summation. Consequently $\boxed{k=\sum _j{k}_j\text{ is a positive-definite kernel}.}$

For the representer theorem, put $V=\mathrm{span}\{k_{x_1},\dots ,k_{x_n}\}$. This is a closed finite-dimensional vector space. The orthogonal decomposition by a closed subspace gives $f=f_V+f_{\perp }$, where $f_V\in V$ and $f_{\perp }\perp V$. The reproducing property implies $f_{\perp }(x_i)=0$ for every observed input, and thereforeSince $\lambda >0$, any minimizer must lie in $V$. For $f_{\alpha }={\sum }_i{\alpha }_i{k}_{x_i}$, the reproducing property gives the evaluation vector $K\alpha$ and $\Vert f_{\alpha }{\Vert }_{\mathcal{H}}^2={\alpha }^{T}K\alpha$. Hence $Q_1(f_{\alpha })=M(\alpha )$. If $\hat{f}$ minimizes $Q_1$, its representation yields a minimizing $\hat{\alpha }$. Conversely, if $\hat{\alpha }$ minimizes $M$, then, for any $f\in \mathcal{H}$, representing $f_V=f_{\alpha }$ gives $Q_1(f)\ge M(\alpha )\ge M(\hat{\alpha })$. ThusThis is an equivalence of minimizers, not an existence or uniqueness assertion: the loss function is arbitrary. A singular kernel matrix also allows nonunique coefficients, since differences in $\ker K$ represent the zero function. Strict convexity of the loss was not used.

Now use the stated minimum-norm description of a sum of reproducing-kernel Hilbert spaces. Let $\hat{f}={\sum }_j{\hat{f}}_j$. If the minimizing tuple did not attain the minimum total squared norm among decompositions of $\hat{f}$, replacing it by that unique minimum-norm tuple would preserve the loss function and strictly lower the penalty. This contradicts optimality. ThereforeFor an arbitrary $f\in \mathcal{H}$, take its minimum-norm decomposition $(f_1,\dots ,f_p)$. Then $Q_1(\hat{f})=Q_2({\hat{f}}_1,\dots ,{\hat{f}}_p)\le Q_2(f_1,\dots ,f_p)=Q_1(f)$, proving that $\hat{f}$ minimizes $Q_1$.

Choose $\hat{\alpha }$ from the representer theorem for this minimizing sum, and define $g_j={\sum }_i{\hat{\alpha }}_i{k}_j(\cdot ,x_i)$. Write $K_j$ for the component kernel matrices. ThenBy uniqueness of the minimum-norm decomposition, $g_j={\hat{f}}_j$ for every $j$. This proves the common representer coefficients for a sum of kernels conclusion:The same coefficient vector works simultaneously. Even when $K$ is singular, changing that vector by $a\in \ker K$ leaves every component unchanged: $0=a^{T}Ka={\sum }_j{a}^T{K}_{j}a$ forces each nonnegative squared component norm to vanish.