Let $\mathcal{R} : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ denote the reproducing kernel of a Hilbert space $\mathcal{H}$ of functions on $[0,1]$ endowed with inner product $ \langle \cdot, \cdot \rangle_{\mathcal{H}}$. For my particular setting I am able to show that
$$ \sup_{x \in [0,1]} ||\mathcal{R}(x, \cdot)||_{\mathcal{H}} \leq C,$$
for some $C>0$.
Now let us imagine that we have a set of $n$ values $t_j$ in $[0,1]$, from which we evaluate $\mathcal{R}(t_i, t_j)$ and form the corresponding matrix $\mathbf{R}$.
What I am wondering is, if we can also establish a similar bound for the largest eigenvalue of this positive-semidefinite matrix. That is, if for a vector $\mathbf{a} \in \mathbb
{R}^n$ we can have
$$ \sum_j^n \sum_i^n a_i a_j \mathcal{R}(t_i, t_j) \leq B C \sum_{j=1}^n a_j^2,$$
for another constant $B$.
All help is greatly appreciated, thank you.
Best Answer
Using a property of reproducing kernels, and convexity of the squared norm:
\begin{aligned} \sum_i^n \sum_j^n a_i a_j \mathcal{R}(t_i, t_j) &= \sum_i^n \sum_j^n a_i a_j <\mathcal{R}(t_i, \cdot), \mathcal{R}(t_j, \cdot)>_\mathcal{H} \\ &= <\sum_i^n a_i \mathcal{R}(t_i, \cdot), \sum_j^n a_j \mathcal{R}(t_j, \cdot)>_\mathcal{H}\\ &= \left\Vert a \right\Vert_1^2 \left\Vert \sum_i^n \frac{a_i}{\left\Vert a \right\Vert_1} \mathcal{R}(t_i, \cdot) \right\Vert^2_\mathcal{H}\\ &\leq \left\Vert a \right\Vert_1^2 \sum_i^n \frac{a_i}{\left\Vert a \right\Vert_1} \left\Vert \mathcal{R}(t_i, \cdot) \right\Vert^2_\mathcal{H} \\ & \leq \left\Vert a \right\Vert_1^2 C^2 \\ & \leq \frac{\left\Vert a \right\Vert_2^2}{(\min_{\Vert x \Vert_1=1} \Vert x \Vert_2)^2} C^2 \end{aligned}