Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
What is $\mathbb K$? –  tomasz Jun 25 '12 at 13:39
1  
Probably either $\mathbb{R}$ or $\mathbb{C}$. –  Siminore Jun 25 '12 at 13:56
    
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. –  FranckStudiesCommEng Jun 25 '12 at 17:29

1 Answer 1

up vote 2 down vote accepted

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 $k_t(s)=\alpha_t+\beta_t s$.

share|improve this answer
    
Thank you Leonid. For the record, I obtain $k_{t}(s)=4-6\overline{t}+\left(12\overline{t}-6\right)s$. –  FranckStudiesCommEng Jun 25 '12 at 19:17
2  
@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. –  user31373 Jun 25 '12 at 20:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.