Prove that $K(x,y)=0.5\exp(-|x-y|)$ is a reproducing kernel for $W^{1,2}(\mathbb{R})$, i.e. that $K(x,y)\in W^{1,2}(\mathbb{R})$ and for the continuous representative $\hat{f}$ of $f\in W^{1,2}(\mathbb{R})$ prove the reproducing property.
My attempt:
Proving that $K(x,y)\in W^{1,2}(\mathbb{R})$ is easy. Writing down the reproducing property $\left<\hat{f}, K(x,\cdot)\right>=f(x)$ for all $x\in\mathbb{R}$:
$$\left<\hat{f}, K(x,\cdot)\right>=\int_\mathbb{R}\hat{f}(y)K(x,y)dy=\int_\mathbb{R}\hat{f}(y)K(x,y)dy-\int_{\mathbb{R}\setminus{(x-\delta,x+\delta)}}\hat{f}(y)\partial_{yy}K(x,y)dy-\int_{(x-\delta,x+\delta)}\hat{f}(y)\partial_{yy}K(x,y)dy$$
by integration by parts and the definition of weak derivative. Now
the second derivative with respect to y is $$\partial_{yy}K(x,y)=0.5\dfrac{\exp{(-|x-y|)}((x-y)^2+(x-y)^2 |x-y|-|x-y|^2)}{|x-y|^3}$$ So the middle and left term cancel in the limit $\delta\to 0$. How do I prove that the right term goes to $-\hat{f}(x)$?
The formula you have for the second derivative is too complicated, and more importantly, it misses a crucial component: Dirac delta at $x$. For any fixed $x$, $$ \partial_y K(x,y) = \frac12 \operatorname{sign}(y-x) e^{-|x-y|} $$ which jumps from $-1/2$ to $1/2$ at the point $x=y$. Therefore, the second derivative is $$ \partial_{yy} K(x,y) =-\delta_x + \frac12 e^{-|x-y|} $$ bringing us to the reason this $K$ is considered: $$ K(x,y)-\partial_{yy} K(x,y) = \delta_x $$ Hence, the reproducing property: $$ \langle f, K(x,\cdot)\rangle = \int (f(y)K(x,y)+f'(y)\partial_y K(x,y))\,dy= \int f(y)(K(x,y)-\partial_{yy} K(x,y))\,dy = f(x) $$
One could avoid the language of distributions by splitting the range of integration to $y>x$ and $y<x$, and dealing with the boundary terms at $x$. But avoiding distributions when considering Sobolev spaces would be counterproductive.