Neumann series of the derivative operator

Question

Consider the Nemuann series of the $D=\frac{d}{dx}$ operator: $T = 1 + D+ D^2+...$

Applying T on $x$ yields $x+1$ (i.e., $Tx=x+1$) for $x \in R$.

I would appreciate if someone helps me to resolve the following confusion.

Since the sum above converges, we can rewrite $T$ as $\frac{1}{1-D}$. Now, lets compute $Tx$ as follows:

$Tx=(1 + D+ D^2+...)x=\frac{1}{1-D}x\equiv y \implies y - Dy = x \implies y-y^{'}=x,$

which is a first order differential equation with $y(x)=x+1-ce^{x}$ as the general solution, where $c$ is a constant.

The question: why two different solutions? I understand that the first one is a specific case of the second general solution (with $c=0$). My confusion is that I was expecting to get one solution no matter which method I use. This may sound very basic but I cannot wrap my head around it. Thanks.

Answer 1

When you rewrite $T$ as $T = (I-D)^{-1}$, you are assuming that $I-D$ is invertible. That statements require

(1) a space $X$ where $D$ is defined on, so that it makes sense to say that $I-D:X\to X$ is bijective, and

(2) a reason, after choosing $X$, why $I-D$ is bijective at all.

If $D : X\to X$ is just any bounded linear operator $D:X\to X$ between some Banach space $X$ with operator norm $\|D\|<1$, then the series

$$ T = I+D+D^2 + \cdots,$$ as a series in the space of bounded linear operator on $X$, converges to the inverse of $I-D$.

When $D = \frac{d}{dx}$, I guess one might choose $X$ to be any finite dimensional vector space spanned by $e^{\lambda_i x}$ with $|\lambda_i|<1$, but it might not be very interesting.

In another direction, for each $n\in \mathbb N$, your operator $T$ is defined as a linear operator on $X=P_n$, the space of polynomials in one variables of degree $\le n$. One can easily check that $I-D$ is invertible (e.g. using the standard basis $\{1, x, \cdots, x^n\}$ on $P_n$, the matrix representation of $D$ is strictly upper triangular)

Also, since $D^m = 0$ on $P_n$ whenever $m>n$,

\begin{align} (I-D) T &= (I-D) (I + D +D^2 + \cdots +) \\ &= (I-D)(I + D + \cdots + D^n) \\ &= I- D^{n+1} = I \end{align} and thus $T = (I-D)^{-1}$.

Let $P$ be the space of polynomial in one variables. Since $P = \cup_n P_n$, one checks that $I-D$ is also invertible on $P$ and $T = (I-D)^{-1}$.

Now go back to your calculation, for any polynomial $p\in P$, $y=Tp$ implies

$$\tag {1} y-y' = p,$$

but you have required $y\in P$ in order that $I-D$ is invertible. So what you have proved is just that for any polynomial $p$ there is only one polynomial $y$ which satisfies $(1)$.

Answer 2

The sum does not converge for all $y(x)$. In particular if the sum converges, then $D^ny(x)$ goes to $0$ when $n\to\infty$. This is only the case when $c=0$ in your example.