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Definition of the problem

Let $\mathcal{H}$ be a Hilbert space which consists of functions, defined on a set $S$. Let $k:S\times S \rightarrow \mathbb{K}$ is our reproducing kernel for $\mathcal{H}$. Now, let $\mathcal{H}$ be the two-dimensional subspace of $L^2(0,1)$, consisting of the functions $f(t)=a+bt,\ a,b\in \mathbb{K}$. I am asked to find the reproducing kernel for $\mathcal{H}$.

My efforts

I have to show that $f(t)=\left\langle f,k_{t}\right\rangle $. Given that $f(t)=a+bt$, we have to find $k_t$ such that $a+bt= \left\langle a+bt,k_{t}\right\rangle$. And in $L^2(0,1)$, we know that $\left\langle f,k_{t}\right\rangle =\int f\left(t\right)\overline{k_{t}}dt=f\left(t\right)$.

My question

How could I find from there the reproducing kernel of $\mathcal H$? Should I integrate over $dt$, and between $(0,1)$ ?

Thank you, Franck!

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  • $\begingroup$ What is $\mathbb K$? $\endgroup$
    – tomasz
    Jun 25 '12 at 13:39
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    $\begingroup$ Probably either $\mathbb{R}$ or $\mathbb{C}$. $\endgroup$
    – Siminore
    Jun 25 '12 at 13:56
  • $\begingroup$ Yes, $\mathbb K$ is either $\mathbb R$ or $\mathbb C$. I think in some other books, it can also be denoted by $\mathbb F$. Sorry for the ambiguity. $\endgroup$ Jun 25 '12 at 17:29
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First, the formula $a+bt= \left\langle a+bt,k_{t}\right\rangle$ is a mess: two different variables are denoted by the same letter $t$. It should be $a+bt= \int_0^1 (a+bs)\overline{k_{t}(s)}\,ds$. Now if this is to hold for all $a$ and for all $b$, then the coefficients of $a$ and $b$ must be the same on both sides. $1=\int_0^1 \overline{k_{t}(s)}\,ds$ and $t=\int_0^1 s\overline{k_{t}(s)}\,ds$. This gives you two equations for two unknowns in $k_t(s)=\alpha_t+\beta_t s$.

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  • $\begingroup$ Thank you Leonid. For the record, I obtain $k_{t}(s)=4-6\overline{t}+\left(12\overline{t}-6\right)s$. $\endgroup$ Jun 25 '12 at 19:17
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    $\begingroup$ @FranckStudiesCommEng Since $t$ and $s$ belong to $[0,1]$, the complex conjugate is unnecessary. In a more symmetric form, $k_t(s)=4-6(t+s)+12ts$. Notice that $k_t(s)=k_s(t)$, and this is not a coincidence. $\endgroup$
    – user31373
    Jun 25 '12 at 20:01

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