We denote by $E$ the vector space of continuous functions defined on $[0,1]$ and taking values in $\mathbb{R}$ equipped with the inner product defined by $$\langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ We consider a function $A : [0,1] \times [0,1] \rightarrow \mathbb{R}$ continuous. We are interested in the application $T : E \rightarrow E$ defined by $$T(f)(x) = \int_0^1 A(x,t) f(t) \, \mathrm{d}t$$ We suppose that $\ker T$ is finite-dimensional. We denote by $S$ the inverse bijection of the isomorphism induced by $T$ from $(\ker T)^\perp$ onto $\operatorname{Im} T$. We define the inner product $\varphi$ on $\operatorname{Im} T$ by setting, for all $(f,g) \in (\operatorname{Im} T)^2$, $$\varphi(f,g) = \langle S(f), S(g) \rangle$$ We consider the application $K$ defined on $[0,1]^2$ by $$K(x,y) = \int_0^1 A(x,t) A(y,t) \, \mathrm{d}t$$ Show that $(\operatorname{Im} T, \varphi)$ is a reproducing kernel Hilbert space, with kernel $K$.
We denote by $E$ the vector space of continuous functions defined on $[0,1]$ and taking values in $\mathbb{R}$ equipped with the inner product defined by
$$\langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$
We consider a function $A : [0,1] \times [0,1] \rightarrow \mathbb{R}$ continuous. We are interested in the application $T : E \rightarrow E$ defined by
$$T(f)(x) = \int_0^1 A(x,t) f(t) \, \mathrm{d}t$$
We suppose that $\ker T$ is finite-dimensional. We denote by $S$ the inverse bijection of the isomorphism induced by $T$ from $(\ker T)^\perp$ onto $\operatorname{Im} T$.\\
We define the inner product $\varphi$ on $\operatorname{Im} T$ by setting, for all $(f,g) \in (\operatorname{Im} T)^2$,
$$\varphi(f,g) = \langle S(f), S(g) \rangle$$
We consider the application $K$ defined on $[0,1]^2$ by
$$K(x,y) = \int_0^1 A(x,t) A(y,t) \, \mathrm{d}t$$
Show that $(\operatorname{Im} T, \varphi)$ is a reproducing kernel Hilbert space, with kernel $K$.