# Reproducing Kernel of subspace of $L^2(0,1)$

### 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!

• What is $\mathbb K$? Jun 25 '12 at 13:39
• Probably either $\mathbb{R}$ or $\mathbb{C}$. 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. Jun 25 '12 at 17:29

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 $k_t(s)=\alpha_t+\beta_t s$.
• Thank you Leonid. For the record, I obtain $k_{t}(s)=4-6\overline{t}+\left(12\overline{t}-6\right)s$. Jun 25 '12 at 19:17
• @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.