Prove that the function $\xi\in R \mapsto {e^{i\cdot \xi\cdot λ}-1\over i\cdot \xi}-λ$ is $C^{\infty}$

Question

Prove that the following function is $C^\infty$ in the point $\xi=0$: $$f:\Bbb R\to\Bbb C:\xi\mapsto {e^{i\cdot\xi\cdot λ}-1\over i\cdot\xi}-λ$$

Any ideas how to prove this? I am trying to think some ideas but I cannot find any way to prove it.

Answer 1

To elaborate on Daniel Fischer's comment: We have

Let $$\tag1 f\colon x\mapsto \sum_{n=0}^\infty a_nx^n$$ be a power series and $R=1/\limsup\sqrt[n]{|a_n|}>0$ its radius of convergence. Then the series converges absolutely for all $x$ with $|x|<R$ and we can take the derivative termwise, i.e., $f'(x)=\sum_{n=0}^\infty (n+1)a_{n+1}x^n$ where this series has the same radius of convergence. It follows that $f$ is $C^\infty$ on $(-R,R)$ (or on all of $\mathbb R$ if $R=\infty$).

Now we know (or in fact define) $e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$ and verify that this series has infinte radius of convergence. Then we can deduce that for all $x\ne 0$ $$\begin{align}\frac{e^{i\lambda x}-1}{i x}&=\frac{\sum_{n=0}^\infty \frac{(i\lambda)^n}{n!}x^n-1}{ix}\\ &=\frac{\sum_{n=1}^\infty \frac{(i\lambda)^n}{n!}x^n}{ix}\\ &=\frac{i\lambda x\sum_{n=0}^\infty \frac{(i\lambda)^n}{(n+1)!}x^n}{ix}\\ &=\lambda \sum_{n=0}^\infty \frac{(i\lambda)^n}{(n+1)!}x^n\\ \end{align} $$ Specifically, the right hand side evaluates to $\lambda $ at $x=0$ so that finally the function $$ x\mapsto\begin{cases}\frac{e^{i\lambda x}-1}{i x}-\lambda&\text{if $x\ne0$}\\ 0&\text{if $x=0$}\end{cases}$$ is $C^\infty$ on all of $\mathbb R$.