Confusion about inner product between two functions

functional-analysisfunctionsinner-productsreproducing-kernel-hilbert-spaces

In this video, the professor defined a reproducing kernel function $K(x, \cdot)=\sum_{j=1}^{P}\phi_j(x)\phi_j$, and a function $f=\sum_{j=1}^Pw_j\phi_j$ and claim

$$
\langle f,K(x,\cdot)\rangle= \sum_{j=1}^Pw_j\phi_j(x)=f(x) \tag 1
$$

He explained here

if there is one-to-one correspondence between the functions and their coefficients, we can compute the inner product between these two functions by taking the inner product of the coefficients.

I don't understand why this is the case. More precisely, where is the $\phi_j$ in $K(x,\cdot)$ and $f$ gone? I've searched online, finding that the definition of the inner product between two functions $f$ and $g$ are

$$
\langle f,g\rangle=\int_a^b f(x)g(x)dx \tag 2
$$

Based on the definition of Equation 2, shouldn't $\langle f,K(x,\cdot)\rangle$ be computed as

$$
\langle f,K(x,\cdot)\rangle= \sum_{j=1}^Pw_j\phi_j(x)\langle\phi_j,\phi_j\rangle
$$

Why would Equation 1 omit $\langle\phi_j,\phi_j\rangle$?

Best Answer

I think it is assumed that the $\phi_j$'s form an orthonormal basis. Therefore $\langle \phi_j,\phi_k \rangle=0$ if $j\neq k$ and $\langle \phi_j,\phi_j \rangle=1$.