Exercise 12.15: Sobolev Kernels and Norms
Follows directly from Proposition 12.27, which allow us to add the norms of $\Hb_1$ and $\Hb_2$. The norm of $\Hb_2$ follows from Example 12.17, and we now identify the norm of $\Hb_1$.
For $\Hb_1$, we show that the inner product is given by
\begin{equation} \label{eq:inner-product} \lra{f, g}_{\Hb_1} = \sum_{\ell=0}^{\alpha - 1} f^{(\ell)}(0) g^{(\ell)}(0) \end{equation}
by showing that the reproducing property is true for the kernel
\begin{equation} k_1(x, z) = \sum_{\ell = 0}^{\alpha - 1} \frac{x^\ell}{\ell!}\frac{z^\ell}{\ell!}. \end{equation}
Since there is only one Hilbert space for which the reproducing property is true for the kernel $k_1$ (Moore–Aronszajn theorem), \eqref{eq:inner-product} must indeed be the inner product for $\Hb_1$, which means that the norm of $\Hb_1$ is given by
\begin{equation} \norm{f}_{\Hb_1} = \lra{f, f}_{\Hb_1} = \sum_{\ell=0}^{\alpha - 1} (f^{(\ell)}(0))^2. \end{equation}
Consider a polynomial of degree $\alpha - 1$:
\begin{equation} f(x) = \sum_{\ell=0}^{\alpha - 1} a_i x^\ell. \end{equation}
Then, by a Taylor expansion of $f$ around $x = 0$,
\begin{equation} f(x) = \sum_{\ell=0}^{\alpha - 1} \frac{x^{\ell}}{\ell!} f^{(\ell)}(0) \overset{(*)}{=} \lra{k(x, \vardot), f}_{\Hb_1} \end{equation}
where equality $(*)$ follows from the observation that
\begin{equation} \left.\left(\frac{\sd}{\sd z}\right)^\ell k_1(x, z)\right|_{z = 0} = \frac{x^{\ell}}{\ell!}. \end{equation}