A reproducing kernel Hilbert Space is a Hilbert space , $H$, of functions , $f$ on a set $E$, so that
i) For any $f$, there is $x\in E$ with $f(x)\not=0$
ii) For any $x\in E$, there is $f\in H$ so $f(x)\not=0$
iii) For any $x,y\in E$, there is $f$ in $H$ so $f(x)\not=f(y)$
iv) For any $x\in E$, there is $C_x$ so that $|f(x)|\leq C_x|f|_H$
a) Prove that for any $x\in E$, there is $k_x\in H$, so that for any $f\in H$, $f(x)=<k_x,f>$.
Sol. I have this. For any $x\in E$, $l_x:H\to\mathbb{C}$ such that $l_x(f)=f(x)$ (evaluation function) By Riesz Representation, there $k_x\in H$ such that $l_x(f)=<k_x,f>$ we are done.
b) Prove that each $k_x\not=0$ and $x\not=y \Rightarrow k_x\not=k_y$
c) Prove that finite linear combinations of $\left\{k_x\right\}_{x\in E}$ are dense in $H$. (Hint: If $S$ is the closure of the span of the $\left\{k_x\right\}_{x\in E}$ and $f\in S^{\perp}$, whatcan you say about $f(x)$?)
Sol: Let $S=\overline{Span(\left\{k_x\right\}_{x\in E})}$. Then $H=S \oplus S^{\perp}$. Let $f\in H$, if $f=0$, then $f\in S$ obvious. If $f\not=0$, and $f\in S$ we are done. If $f\not=0$ and $f\in S^{\perp}$ then $f(x)=<k_x,f>=0$ then $f(x)=0$ a contradiction.
d)The Reproducing kernel of $H$ is the function $K$ on $E\times E$ given by $K(x,y)=<k_y,k_x>=k_x(y)$. Prove that $K(y,x)=\overline{K(x,y)}.$
sol: $K(y,x)=<k_x,k_y>=\overline{<k_y,k_x>}=\overline{K(x,y)}$.
e) Prove for any $x_1,\ldots, x_n \in E$, and $\lambda \in \mathbb{C}^n,$ we have $\sum_{i,j=1}^{n}\overline{\lambda}_i\lambda_jK(x_i,x_j)\geq 0.$
How prove that b) and e)? For e) I should be induction?
The assertions in (b) follow immediately from (ii) and (iii) of the definition.
For (e), the idea should be to show that the desired sum equals $\langle k, k \rangle$ where $k = \sum_{i=1}^n \lambda_i k_{x_i}$. (I am not sure in your setup if the inner product is supposed to be conjugate linear in the first variable or the second, so perhaps it should be $k= \sum_{i=1}^n \overline{\lambda_i} k_{x_i}$ instead.)