Differential operators with arbitrary functions? By Taylor expansion, one has $$f(x+t) = \sum_{k=0}^∞ \frac{D^k}{k!}f(x)([x+t]-x)^k = \sum_{k=0}^∞ \frac{(Dt)^k}{k!}f(x)$$ and hence one could say $e^{Dt}$ is translation by $t$. But this isn't a "differential operator", not by Wikipedia's definition. My questions: in what sense is this a truly convergent "differential operator"? I have seen similar ones e.g. replace $D$ with the Laplacian; what does this mean, and why do we care about them?
 A: *

*Starting with the familiar
$$
e^t=\sum_{n=0}^\infty\frac{t^n}{n!}.
$$
We know that $e^{t+s}=e^te^s$.

*For a matrix $A\in M_{n\times n}$
$$
e^{At}=\sum_{n=0}^\infty\frac{A^nt^n}{n!}.
$$
What we know is the same $e^{A(t+s)}=e^{At}e^{As}$ and that it is the fundamental solution to $\dot x=Ax$. If we denote the fundamental solution by $T(t)=e^{At}$, the former property can be written as 
$$
T(t+s)=T(t)T(s),\qquad T(0)=I.
$$
Thus, the mapping $T(t)=e^{At}$ defines a $C_0$ semigroup. Here, you can interpret $T(t)$ as the shift along the trajectories of $\dot x=Ax$. 

*For an abstract one-parametric $C_0$ semigroup on a Banach space $X$ we define the generator $A$ as (in the derivative manner) 
$$
Ax=\lim_{t\to 0^+}\frac{1}{t}(T(t)-I)x,\qquad\forall x\in X.
$$
If a $C_0$ semigroup is uniformly continuous, which means that $t\mapsto T(t)$ is continuos wrt the uniform operator topology, then there exists a bounded operator $A$ such that $T(t)=e^{At}$. It means that 


all uniformly continuous semigroups look like exponentials of some bounded operator.


*Consider a left translation semigroup 
$$
T(t)f(s)=f(s+t).
$$
It can be shown to be not uniformly continuous, unfortunately, but it is still quite nice in the sense that it is strongly continuous semigroup (i.e. wrt the strong operator topology) on $L^p$, $1\le p<\infty$. For strongly continuous semigroup it is still possible to define the generator $A$, however it is no longer bounded, but a closed operator. For the left shift semigroup the generator is the differential operator
$$
Af(s)=f'(s)
$$
defined on the Sobolev space (the space with the graph metric) $W^{1,p}$. So $A=D$, and it is tempting to write $T(t)=e^{Dt}$, but it is no longer possible, since the semigroup is not uniformly bounded. However, there are certain similarities with $e^{Dt}$ as we have seen above.


P.S. Note that all the differential operators in Wikipedia have a finite sum definition. It is precise the problem here: to give interpretation to the infinte series we have to go beyond the exponential.
