let, $H=id/dx$ be the operator define on its maximal domain $D(H)=\{f\in L^2(\mathbb{R}):f'\in L^2(\mathbb{R})\}\subseteq L^2(\mathbb{R})$. Now, define $\exp(-H)=\sum_{n=0}^\infty(-1)^n\frac{H^n}{n!}$. I am unable to find the domain $D(exp(-H))$.
My observations:
We know $\exp(-H)(f)=\sum_{n=0}^\infty(-i)^n\frac{f^{(n)}}{n!}$ should be in $L^2(\mathbb{R})$. So, $f$ should be almost everywhere smooth and all it's derivative $f^{(n)}$ should be in $L^2(\mathbb{R})$ .
At first I misunderstood that $C_0^\infty(\mathbb{R})$ will be a subset of $D(\exp(-H))$, which is not true because if I take a smooth bump function, a compactly supported smooth function gives value $1$ in a certain interval and multiply with $cos(x)$ then that function does't belongs to $D(\exp(H))$.
we know , if f is a holomorphic function on a ball $B(a,r)$ then $f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(z-a)^{n}$ for all $z\in B(a,r)$ and $f(a+1)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}$ when $r>1$. Absolute Convergence of that series is same as the convergence of the series of $\exp(-H)(f)(a)$. Hence if our function $f$ can be extend as a holomorphic function on a neighborhood of $\{z=x+iy\in \mathbb{C}: y\in[-1,1]\}$ then $\exp(-H)(f)$ is defined but I don't know if it is $L^2(\mathbb{R})$.
That's all.
Any suggestion or help will be appreciated.