According to the Theorem of Aronszajn at the top of p. 351 HERE, if $K(x,y)$ is a reproducing kernel on $E\times E$ for a Hilbert space $\cal H$ of functions with inner product $\langle \,\, ,\,\, \rangle$, then for $E_1 \subset E$, the restriction of $K(x,y)$ to $E_1\times E_1$ is a reproducing kernel for the restriction of functions in $\cal H$ to $E_1$ with norm inherited from $\cal H$ as indicated in the link.
Yet when I try this with a simple example, with $K(x,y)\equiv \frac{1}{\pi} \cos(x-y)$ on $[0,2\pi]\times[0,2\pi]$ with $\cal H\equiv$ span $\{\cos x, \sin x\}$ with inner product $\langle f, g \rangle \equiv \int_0^{2\pi} f(x)g(x)dx$, I can show that $K(x, y)$ is a reproducing kernel for $\cal H$ but when I then try to restrict the kernel and the function space to be defined on $E_1 = [0,\frac{\pi}{2}]$ for example, the new restricted kernel does not reproduce the restricted functions.
Example: On the full space, $\int_0^{2\pi} K(x,y) \cos x dx = \cos y$ as one would expect. On the restricted space, $\int_0^{\frac{\pi}{2}} K(x, y) \cos x dx = \frac{1}{4} \cos y + \frac{1}{2\pi} \sin y \neq \cos y$ on $[0,\frac{\pi}{2}]$.
I'm sure I am misinterpreting Aronszajn's theorem somehow, but I am not able to see why or how. I would be grateful for any assistance.
Edit: Perhaps this will yield some insight.
The Theorem in Aronszajn is a bit misleading in my opinion. Now that I have studied it further, I see that the inner product on the space of restricted functions is nothing more than the original inner product applied to the restricted functions extended back to the original space. In other words, for the example given, if the restricted functions are defined on $[0, \frac{\pi}{2}]$ as $K_1$, $\cos x$, then the inner product $\langle \,\,, \,\,\rangle_1$ of Aronszajn's Theorem would be for fixed $y\in[0,\frac{\pi}{2}]$, $\langle \cos x , K_1(x, y) \rangle_1 = \langle \cos x, K(x,y) \rangle |_{[0,\frac{\pi}{2}]}$, which is a kind of obvious and empty result in this setting, or any other in which the projection of the original Hilbert space onto the subspace $F_0$ of functions that are everywhere $0$ on $E_1$, has $F_0=\{0\}$.