Consider a reproducing kernel Hilbert space $\mathcal{H}$, whose elements are functions from $X \rightarrow \mathbb{R}$.
Let $t \in X$. Since $\mathcal{H}$ is a RHKS, there exists a unique function $K_t \in \mathcal{H}$ (called the representer of $t$) such that $\langle K_t, f \rangle = f(t)$ for all $f \in \mathcal{H}$. (my ultimate goal will be to find concrete examples of $K_t$ and $f$)
We define $K(t, x) := K_t(x) = \langle K_t, K_x \rangle$ to be the reproducing kernel of $\mathcal{H}$.
One popular kernel is the linear kernel $K:X \times X \rightarrow \mathbb{R}$ defined by $K(x, y) = x \cdot y$.
Let's take $X = \mathbb{R}$. So if we take $3 \in X$, then $K(3, x) = K_3(x) = 3 \cdot x$ is the representer of $3$ in $\mathcal{H}$.
Thus, $\langle K_3, f \rangle = f(3)$ for any $f \in \mathcal{H}$. Say $f = K_3$, then $\langle K_3, K_3 \rangle = \int_{-\infty}^{\infty} K_3(x) K_3(x) dx = \int_{-\infty}^{\infty} 9x^2 dx = 0 \neq 9 = K_3(3) = f(3)$?
Similarly, what if we take $K$ to be the RBF kernel?
I'm certainly misunderstanding something here...any help would be appreciated!
$\langle 3x,3x\rangle=3x$ doesn't make sense, as the left-hand side is a number and the right-hand side is a function. Perhaps some of the problem is the notation.
The correct evaluation is $\langle K_3,K_3\rangle =K_3(3) = 9$.
Note that the RKHS in this case is one dimensional, all of its elements being of the form $f(x) = cx =K_c(x)$ for some $c\in\mathbb R$.
I am not familiar with the RBF kernel, but: What is your question about it?
Added, after edits to the question 2016-07-07 23:09:18Z:
You have made an incorrect assumption about an integral representation of the inner product. You have defined the kernel function such that, by definition, $\langle K_x,K_y\rangle = xy$. Furthermore, to elaborate on the remark above, all of your kernel functions are scalar multiples of $K_1$, hence your space is the one dimensional span of $K_1$, i.e., the functions all have the form $x\mapsto cx$ for some $c\in \mathbb R$. If you would like a representation of your inner product independent of kernel functions, it is $\langle f,g\rangle=f(1)g(1)$. This is an integral in a way, with counting measure on a one point set.