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$k,k_1$ and $k_2$ are kernels on $\mathcal{X}\times\mathcal{X}$, and $k=k_1+k_2$, then we have the following properties for the RKHS (Reproducing Kernel Hilbert Space) $\mathcal{H}$, $\mathcal{H}_1$ and $\mathcal{H}_2$:

$$\mathcal{H}=\mathcal{H}_1+\mathcal{H}_2=\left\{f_1+f_2:f_1\in \mathcal{H}_1, f_2\in\mathcal{H}_2\right\}(1)$$

and for any $f\in \mathcal{H}$,

$$ ||f||^2_{\mathcal{H}}=\min_{f_1,f_2}\left\{||f_1||^2_{\mathcal{H_1}}+||f_2||^2_{\mathcal{H_2}}:f_1+f_2=f\right\}(2)$$

Could someone help to prove (1) or (2)?

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So $\mathcal{H}$ is the RKHS of $k$, and we want to prove that $\mathcal{H}=\mathcal{H}_1+\mathcal{H}_2$ (1) and that the topology of $\mathcal{H}$ is the one defined by (2).

Preliminaries. First of all, let's notice that, for all $x \in \mathcal{X}$, $$k(x,\cdot)=k_1(x,\cdot)+k_2(x,\cdot) \in \mathcal{H}_1+\mathcal{H}_2.$$

Topology of $\mathcal{H}_1+\mathcal{H}_2$. To be able to prove the reproducing property, let's focus on the topology of $\mathcal{H}_1+\mathcal{H}_2$. We denote $E=\mathcal{H}_1 \times \mathcal{H}_2$. This set is a Hilbert space if we equip it with the norm $$\left \| \cdot \right \|_{E}:(f_1,f_2) \mapsto \sqrt{\left \| f_1 \right \|_{\mathcal{H}_1}^2+\left \| f_2 \right \|_{\mathcal{H}_2}^2},$$in fact, $E$ is the orthogonal direct sum of $\mathcal{H}_1$ and $\mathcal{H}_2$.

We want to compare the topologies of $\mathcal{H}_1+\mathcal{H}_2$ and $E$. A direct link between these spaces is the natural surjection $ s: \left| \begin{array}{rcl} E & \longrightarrow &\mathcal{H}_1+\mathcal{H}_2 \\ (f_1,f_2) & \longmapsto & f_1+f_2 \\ \end{array} \right.$. We are going to try to "make $s$ injective" (of course, "vanilla $s$" is not necessarily injective). In order to do so, let's consider $N=s^{-1}(\{0\})$. We will begin by proving that $N$ is a closed subset of $E$:

Let $(f_n,-f_n)$ be a sequence of elements of $N$ converging in $E$ to $(f,g)$. By definition of the $E$-norm, $(f_n)_{n\geq 1}$ converges in $\mathcal{H}_1$ to $f$ and $(-f_n)_{n\geq 1}$ converges in $\mathcal{H}_2$ to $g$. Since convergence in a RKHS implies ponctual convergence, we will have $f=-g$ an therefore $(f,g) \in N$. $N$ is therefore a closed subset of $E$.

Since $N$ is closed, $E$ is equal to the direct sum of $N$ and its orthogonal complement $O$. The restriction $\tilde s$ of $s$ to $O$ will therefore be a bijection.

Now that we have a linear bijection, we can equip $\mathcal{H}_1+\mathcal{H}_2$ with an Hilbertian structure inherited from $E$: with the norm $\left \| \cdot \right \|_{\mathcal{H}_1+\mathcal{H}_2}:f \mapsto \left \| \tilde{s}^{-1}(f) \right \|_{E}$, $\mathcal{H}_1+\mathcal{H}_2$ is indeed a Hilbert space.

Now we can prove (2). For every couple of functions $c=(f_1,f_2) \in E$, we can use the orthogonal decomposition $E=N \oplus O $ to write $c=(f_1^N,f_2^N)+\tilde{s}^{-1}(f_1+f_2)$ with $(f_1^N,f_2^N) \in N$. Therefore $$\left \| c \right \|_{E}^2 =\left \| (f_1^N,f_2^N) \right \|_{E}^2 + \left \| f_1+f_2 \right \|_{\mathcal{H}_1+\mathcal{H}_2}^2.$$ Therefore, for every function $f=f_1+f_2$ of $\mathcal{H}_1+\mathcal{H}_2$, we have $\left \| f \right \|_{\mathcal{H}_1+\mathcal{H}_2} \leq \sqrt{\left \| f_1 \right \|_{\mathcal{H}_1}^2+\left \| f_2 \right \|_{\mathcal{H}_2}^2}$ with equality if, and only if $(f_1,f_2) \in O$. We can then deduce (2) $$\left \| f \right \|_{\mathcal{H}_1+\mathcal{H}_2} = \min_{(f_1,f_2)\in E, f_1+f_2=f} \sqrt{\left \| f_1 \right \|_{\mathcal{H}_1}^2+\left \| f_2 \right \|_{\mathcal{H}_2}^2}.$$

Reproducing property. In fact, to prove that $\mathcal{H}_1+\mathcal{H}_2$ endowed with the norm we just defined is the RKHS of $k_1+k_2$, we still need to prove the reproducing property: let $x \in \mathcal{X}$ and $f \in \mathcal{H}_1+\mathcal{H}_2$. We can write $f=\tilde{s}(f_1,f_2)$ and $k(x,\cdot)=\tilde{s}(A(x,\cdot),B(x,\cdot))$ where $(f_1,f_2)$ and $(A_x,B_x)$ live in $O$. Thus, $$\langle f,K_x \rangle_{\mathcal{H}_1+\mathcal{H}_2} = \langle(f_1,f_2) , (A(x,\cdot),B(x,\cdot)) \rangle_{E}=\langle(f_1,f_2) , \left(k_1(x,\cdot),k_2(x,\cdot)\right)+(A(x,\cdot)-k_1(x,\cdot),B(x,\cdot)-k_2(x,\cdot))\rangle_E$$ but, since $ A(x,\cdot)-k_1(x,\cdot)+B(x,\cdot)-k_2(x,\cdot)=0,$ the vector $(A_x-k_1(x,\cdot),B_x-k_2(x,\cdot))$ is orthogonal to every element in $O$, and in particular to $(f_1,f_2)$. Consequently, $$\langle f,K_x \rangle_{\mathcal{H}_1+\mathcal{H}_2} = \langle(f_1,f_2) , \left(k_1(x,\cdot),k_2(x,\cdot)\right) \rangle_E = f_1(x)+f_2(x)=f(x)$$ and the reproducing property is true.

$\mathcal{H}_1+\mathcal{H}_2$ is therefore the RKHS of $k_1+k_2$.

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    $\begingroup$ Nice proof. Nitpicking: wouldn't $N$ be $s^{-1}(\{0\})$ (extra braces)? $\endgroup$ – P. Camilleri Mar 9 '15 at 14:44
  • $\begingroup$ @M.Massias : yes! thanks! $\endgroup$ – PAM Mar 9 '15 at 17:27

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