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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.

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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}

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  • $\begingroup$ How exactly do you get the last inequality? I think it is wrong. $\endgroup$ – JohnK Oct 9 '20 at 10:55
  • $\begingroup$ By subadditivity property of the norm, and the scalability. See points 1 and 2 in the definition here: wiki. $\endgroup$ – karamel-kitty Oct 9 '20 at 13:09
  • $\begingroup$ But this is a square-norm so this is not correct. $\endgroup$ – JohnK Oct 9 '20 at 13:26
  • $\begingroup$ I have updated my answer. It's not the 2-norm as you wanted but the 1-norm that appears naturally. $\endgroup$ – karamel-kitty Oct 11 '20 at 7:16
  • $\begingroup$ While this fixes the previous mistake, this is not what the question is about. $\endgroup$ – JohnK Oct 11 '20 at 8:58

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