# Computing $e^{isD}$ for a differential operator D

I'm trying to understand functional calculus of unbounded operators and everywhere I see proofs of its existence, but it seems that no one ever dares to compute some easy example.

Lets take $D = i\tfrac{d}{dx}$ for example. I know that $e^{isD}$ is the operator on $L^2(\mathbb{R})$ given by translation by s but I'm completely lost if I try to show this. I tried writing $e^{isD} = e^{-s\tfrac{d}{dx}} = \sum_{k=0}^\infty \tfrac{1}{k!}(-s\tfrac{d}{dx})^k$ and then $(e^{isD}u)(y) = \sum_{k=0}^\infty \tfrac{1}{k!}(-s)^k(\tfrac{d^k}{dx^k}u)(y)$ but this does not give the right result. So I can't use continuity arguments in general to compute $e^{isD}$.

So, how does one compute $e^{isD}$ for some differential operator $D$ on $\mathbb{R}$?

Thanks.

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Meanwhile, I found the following thread on MathOverflow, which is a generalized version of my question here: mathoverflow.net/questions/87937 –  AlexE Nov 15 '12 at 9:59

Let us recall some basic ideas. Define

$$\hat{f}(\xi) = \int_{\Bbb{R}} f(x)e^{-2\pi i \xi x} \, dx \quad \text{and} \quad \check{f}(x) = \int_{\Bbb{R}} f(\xi)e^{2\pi i x \xi} \, d\xi.$$

Then for a function $f \in \mathcal{S}(\Bbb{R})$ of Schwartz class, we have

$$\left(\frac{d^n}{dx^n} f \right)^{\wedge}(\xi) = (2\pi i \xi)^n \hat{f} (\xi).$$

Thus if we put $D = \frac{1}{2\pi i} \frac{d}{dx}$, we have

$$D^n f(x) = \big(\xi^n \hat{f}(\xi)\big)^{\vee}(x).$$

That is, under Fourier transform, differentiation operator is indeed a multiplicative operator. This observation is the key ingredient to define a general class of differential operators. Indeed, for $g \in L^{\infty}(\Bbb{R})$ the differential operator

$$g(D) : L^2(\Bbb{R}) \to L^2(\Bbb{R})$$

is define as

$$g(D)f(x) = \big(g(\xi) \hat{f}(\xi)\big)^{\vee}(x), \qquad f \in L^2(\Bbb{R}).$$

This definition completely make sense since $g(\xi)\hat{f}(\xi) \in L^2(\Bbb{R})$ by Hölder's inequality. (If we change the condition of $g$, then the corresponding operator $g(D)$ no longer stands for an endomorphism on $L^2(\Bbb{R})$, but rather an operator between two Sobolev spaces.)

For example, we can formally write $e^{-s \frac{d}{dx}} = g(D)$ where $g(x) = e^{-2\pi i s x} \in L^{\infty}(\Bbb{R})$. Then by utilizing some identities, we have

$$e^{-s \frac{d}{dx}}f(x) = g(D)f(x) = \big(e^{-2\pi i s \xi} \hat{f}(\xi)\big)^{\vee}(x) = f(x-s).$$

If you want to replace $D$ by some other differential operators, you may instead change $g$ by some other function.

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Thanks! That is exactly the argument I was looking for! –  AlexE Nov 15 '12 at 9:57

Suppose that $u$ is analytic. Then, by Taylor, around $y$ $$u(y+t)=\sum_{k=0}^{\infty} \frac1{k!}t^k\cdot u^{(k)}(y),$$ so, following your calculations: $$(e^{isD}u)(y)=u(y-s).$$ And for the rest, only the density of analytic functions in $L^2$ is needed (and that $D$ is defined accordingly)..

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That's a nice argument for the case $D = i\tfrac{d}{dx}$! Thanks! –  AlexE Nov 15 '12 at 9:55