Reproducing Kernels of $H^1([0, 1])$

In this posting, we compute the reproducing kernels of the Sobolev space $H^1([0,1])$ equipped with the inner product

$$ \langle f, g \rangle = \int_{0}^{1} [ \overline{f(x)}g(x) + \overline{f'(x)}g'(x)] \,\mathrm{d}x. $$

In other words, we aim to find a function $K_a \in H^1([0, 1])$, for each $a \in [0, 1]$, such that $\langle K_a, f \rangle = f(a)$ for all $f \in H^1([0, 1])$.

 

Assume there exists a real-valued function $G_a \in H^2([0, 1])$ such that $G_a' = K_a$ and it solves the ODE

$$ G_a''(x) = G_a(x) - \mathbf{1}_{[a,1]}(x), \quad G_a(0) = 0, \quad G_a(1) = 1. \tag{*} $$

If this is the case, then we get

\begin{align*} & \int_{0}^{1} [f(x) K_a(x) + f'(x) K_a'(x)] \, \mathrm{d}x \\ &= \int_{0}^{1} [f(x) G_a'(x) + f'(x)(G_a(x) - \mathbf{1}_{[a,1]}(x)) ] \, \mathrm{d}x \\ &= f(1)G_a(1) - f(0)G_a(0) - f(1) + f(a) \\ &= f(a). \end{align*}

Hence $K_a = G_a'$ will be the desired reproducing kernel. The ODE $\text{(*)}$ can be solved in a fairly standard way, yielding

$$ K_a(x) = \frac{e^{-a-x}}{2 (e^2 - 1)} (e^{2 \max (a,x)} + e^2 ) (e^{2 \min (a,x)} + 1). $$