How to make sense of $(1-e^{tD})f$? I'm sophomore student in college. Recently, I'm thinking about series expansion of operators. When I supposed that f is an $C^\infty$-function and D is the differential operator d/dt. According to integation by parts,
$$ \begin{align} \int f'(t) dt &=tf'-\int tf''(t) dt \\ &=tf'-\frac{1}{2}t^2f''+\int \frac{t^2}{2}f'''dt \\ &= \left (\frac{tD}{1!}-\frac{(tD)^2}{2!}+\frac{(tD)^3}{3!}- \dots \right ) f \\ &=\left ( 1-e^{-tD} \right )f \end{align} $$
seems to make sense.
But I don't know about 'functional analysis' or 'operator theory', etc.
I want to know how can I make a proof of things like that.
What should I study to proof that?
 A: Functional analysis and operator theory are precisely the areas where this sort of idea is rigorously studied. In particular, since you're working with infinite-dimensional spaces such as $C^\infty$ and a linear (though unbounded) operator $d/dt$ defined on an infinite-dimensional space, there's no way you can avoid dealing with functional analytic details.
However, the concept of the exponential of a linear operator can be defined in finite dimensions, where the full methods of functional analysis are not strictly required. This is something you sometimes run across in undergraduate ODEs: if $A:\mathbb{C}^n\to\mathbb{C}^n$ is a linear operator, then its matrix exponential is given by the series
$$
e^A = \sum_{n=0}^\infty \frac{A^n}{n!},
$$
where $A^n$ is interpreted as the $n$-fold composition of $A$ with itself. The sum converges in the sense that the partial sums
$$
\sum_{n=0}^N \frac{A^n}{n!}
$$
converge entrywise to a $n\times n$ matrix, which is precisely what we denote by $e^A$. The series also converges in the sense of operator norm (or any other norm equivalent to it, like the $L^p$ matrix norms or the finite-dimensional Hilbert-Schmidt norm). This can be proved with the knowledge from a typical undergraduate analysis sequence.
More generally, if
$$
f(z) = \sum_{n=0}^\infty a_nz^n
$$
is a convergent complex power series with radius of convergence $R>0$, then one can define the operator
$$
f(A) = \sum_{n=0}^\infty a_nA^n,
$$
and this series converges to a linear operator provided that the eigenvalues of $A$ lie inside the disc of convergence. This can be proved in a similar manner to the same claim for $e^A$, which is the special case where $f(z) = e^z$.
This last idea generalizes to the setting of bounded linear operators on a Banach space. One needs to be a bit careful with definitions, because the spectral theory of Banach space maps is more complicated in infinite dimensions, but basically one can say that if the spectrum of $A$ is contained in the disc of convergence of $f$, then the series definition of $f(A)$ makes sense and gives you a bounded linear operator. This is known as the holomorphic functional calculus, and it gives us a rigorous foundation for working with objects like $e^{tD}$ when $D$ is a bounded operator on a suitable Banach space.
Sadly, in your example with $C^\infty$ and $D = d/dt$, the operator $D$ can fail to be bounded, and $C^\infty$ can be hard to present as a Banach space. (A Frechet space structure is much more natural for $C^\infty$.) But on $C^1(K)$, $K$ a compact set, we have a natural norm giving a Banach space and $D$ is indeed bounded in that norm. In that situation one can consider the holomorphic functional calculus of $D$ as an operator on $C^1(K)$, and $e^{tD}$ can be well-defined using the series definition with little hassle.
All of this can be studied superficially after a pretty basic introduction to the basic objects of functional analysis, like Banach spaces and bounded linear transformations. A deeper study would entail a thorough study of spectral theory and extension to the continuous and Borel functional calculi.
A: Let $D$ be differentiation with respect to $x$. If $f$ has a Taylor series,
$$
      e^{tD}f=\sum_{n=0}^{\infty}\frac{t^{n}D^{n}}{n!}f(x)=\sum_{n=0}^{\infty}\frac{t^{n}f^{(n)}(x)}{n!} = f(x+t).
$$
This basically agrees with more advanced definitions. This is the so-called translation semigroup. You can see that $(T(t)f)(x)=f(x+t)$ defines an operator $T$ with the exponential property:
$$
              T(t')T(t)f=T(t')f(x+t)=f(x+t+t')=T(t'+t)f \\
                         T(t')T(t)=T(t'+t).
$$
For yours, $(1-e^{tD})f=f(x)-f(x+t)$.
