Characterisation of $\left(-\frac{d^2}{dx^2}\right)^{1/2}$ Let $T:=-\frac{d^2}{dx^2}$ on $L^2([0,1])$ (let's say with Dirichlet boundary conditions). This is a positive operator and therefore has a positive square-root $\left(-\frac{d^2}{dx^2}\right)^{1/2}$
My question: Is there a nice characterisation of this square-root? I know that in $n$ dimensions the operator $(-\Delta)^{1/2}$ is quite nasty, but I suspected that it might be better in one dimension.
It would be particularly nice to have some relation between $i\frac{d}{dx}$ and $\left(-\frac{d^2}{dx^2}\right)^{1/2}$. Does anyone know whether such a thing exists?
Maybe a more precise way to pose this question is the following: By the polar decomposition, there is a unitary operator $U$ such that 
$$i\frac{d}{dx} = U\left(-\frac{d^2}{dx^2}\right)^{1/2}.$$
What's $U$?
 A: This problem probably is best solved in terms of the Fourier-series. If you want Dirichlet conditions on $[0,1]$, you should actually look at the Fourier series of 2-periodic odd functions on $[-1,1]$, however the length of the interval will just give you constants anyway. So for sake of simplicity I'll do periodic boundary conditions on $[0,2\pi]$ instead.
So if $c_k$ are the Fourier coefficients of $u$ in the basis $e^{ikx}$, then the operator $\frac{d}{dx}$ simply translates to $c_k \mapsto ik c_k$.
In the same way, $-\frac{d^2}{dx^2}$ on the Fourier side just is the multiplier $k^2$. But then it is not hard to see that $\left(-\frac{d^2}{dx^2}\right)^{1/2}$ translates to $c_k \mapsto |k| c_k$. (The eigenvalues are of the form $k^2$ with eigenfunctions $e^{ikx}$ and $e^{-ikx}$ each) If you compare this to $c_k \mapsto -k c_k$, which is $i\frac{d}{dx}$ on the Fourier side, you will see what your operator $U$ is in terms of the Fourier-coefficients, as it simply flips their sign for all positive $k$.
A: The operator $T$ has a complete orthonormal basis of eigenfunctions:
$$
              s_n(x) = \sqrt{2}\sin(n\pi x/L),\;\; n=1,2,3,\cdots.
$$
These functions are in the domain of $T$, with
$$
       Ts_n = \lambda_n s_n,\;\;\; \lambda_n=\frac{n^2\pi^2}{L^2},\;\; n=1,2,3,\cdots.
$$
And
$$
   \mathcal{D}(T) = \{ f \in L^2[0,1] : \sum_{n=1}^{\infty}\lambda_n^2|\langle f,s_n\rangle|^2 < \infty \} \\
         Tf = \sum_{n=1}^{\infty}\lambda_n \langle f,s_n\rangle s_n.
$$
The unique positive square root $\sqrt{T}$ of $T$ is characterized as
$$
      \mathcal{D}(T) = \{ f \in L^2 : \sum_{n=1}^{\infty}\lambda_n|\langle f,s_n\rangle|^2 < \infty\} \\
         \sqrt{T}f = \sum_{n=1}^{\infty}\sqrt{\lambda_n} \langle f,s_n\rangle s_n.
$$
Another orthonormal basis for $L^2[0,1]$ consists of the functions $c_n$ given by
$$
               c_0=1,c_1=\sqrt{2}\cos(\pi x/L),c_2=\sqrt{2}\cos(2\pi x/L),\cdots.
$$
The derivative operator $i\frac{d}{dx}$ with $0$ endpoint conditions is defined on $\mathcal{D}(\sqrt{T})$ and is given by
$$
                    i\frac{d}{dx}f = i\sum_{n=1}^{\infty}\sqrt{\lambda_n}\langle f,s_n\rangle c_n
$$
These differentiation operators are related by an isometry $V$ that is not unitary:
$$
                 i\frac{d}{dx} = V\sqrt{T} \\
                  Vf = \sum_{n=1}^{\infty}\langle f,s_n\rangle c_n.
$$
These differentiation operators cannot be related by a unitary $V$ because $\sqrt{T}$ is selfadjoint on $\mathcal{D}(\sqrt{T})$ with a dense range, while the range of $i\frac{d}{dx}$ restricted to $\mathcal{D}(\sqrt{T})$ is orthogonal to the constant function $1$.
