I am slightly puzzled by Exercise 22 in Berlinet, Thomas-Agnan. It states
Prove that a function $f$ belongs to $\mathcal{H}_K$ if and only if there exists a constant $C$ such that for all $s$ and $t$ in $E$, $f(s)f(t)\leq C^2 K(s,t)$ and the minimum of such $C$ coincides with $|| f||_K$.
What is wrong with the following simple counterexample?
Let $E=\{1, 2\}$, $\mathcal{H}_K$ be $\mathbb{R}^2=\{f\,|f:E\rightarrow \mathbb{R}\}$ with $K(i,j)=\delta_{ij}$ the standard scalar product. Choose the function such that $f(1)=f(2)=1$. Then $f\in\mathcal{H}_K$ but since $f(1)f(2)=1$ and $K(1,2)=<e_1,e_2>=0$ the required inequality cannot hold!?
The inequality should be taken as a kernel inequality, that is, there exist a constant $C > 0$ such that $C^2K(s, t) - f(s)(t)$ is a kernel.
In your example $K$ would be the $2\times2$ identity matrix and $f(s)f(t)$ would generate the matrix $$ \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}. $$ Taking, for example, $C = \sqrt2$, we obtain $$ C^2K(s, t) - f(s)f(t) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}. $$ This is positive semi-definite since its sub-determinants are $1$ and $0$. Moreover this is the least such $C$ therefore the norm of $f$ is $\sqrt2$, which can also be computed directly.