How to justify to pull the Fourier transform operator out of an exponential function

Question

Let's assume we have a finite measure $\nu$ on $(\mathbb R,\mathcal B(\mathbb R))$ (particulary I am interesting in the Lévy measure of a compound Poisson process), that means we have $\nu(\mathbb R)=\lambda<\infty$.

I am looking for an argument to justify the following operation (while $\mathcal F$ stands for the Fourier transform and $\nu^{*k}$ for the $k$-th convolution):

\begin{align} \exp\left(\mathcal F[\nu](u) \right)=&\sum_{k=0}^{\infty}\frac{1}{k!}\mathcal F^k[\nu](u) \tag1\\ =&\mathcal F\left( \sum_{k=0}^{\infty}\frac{1}{k!}\nu^{*k}\right)(u) \tag2\\ =& \mathcal F\left(e^{\nu} \right)(u) \tag3 \end{align} while $(1)$ seems justified - just by plugging in - I have problems with the second equality. $(3)$ should then just be understood in usual exponential way with the convention that $\nu^0\equiv \delta_0$. Ideas and references are highly welcomed.

I know that for example for $h_n,h\in L^2(\mathbb R)$ we have something similar like $$ \lVert h_n - h \rVert_{L^2(\mathbb R)} \to 0 \implies\lVert \mathcal Fh_n -\mathcal F h \rVert_{L^2(\mathbb R)}\to 0 $$

Addendum:

The Fourier transform of a finite measure $\nu$ on $\mathbb R$ is defined as $$ u\in \mathbb R:\mathcal F[\nu](u)=\int_{\mathbb R}e^{iux}\nu(dx) $$

Context:

Originally I was looking for a derivation of the following expression \begin{align} \mathcal{F}^{-1}[\frac1\varphi(-\bullet)]=\sum_{k=0}^{\infty}\frac{e^{\lambda}{(-1)}^k}{k!}\bar{\nu}^{* k} \end{align} with $\varphi$ being the characteristic function of the compound Poisson process and $\bar{\nu}(A):=\nu(-A)$.

It actually started out quite nice, I began to derive this from scratch, so first the characteristic function, then the reciprocal and then I ended up doing this hand-wavy argument above for which I actually have no justification. In the underlying paper they mentioned actually a different method to derive expressions for $\displaystyle \mathcal{F}^{-1}[\frac1\varphi(-\bullet)]$ but I was wondering whether it would actually also work along the way I thought it might.

Answer 1

Once you know that the Fourier transform of a convolution is the pointwise product of the Fourier transforms of the factors,

$$\mathcal{F}[\mu \ast \nu](u) = \mathcal{F}[\mu](u)\cdot \mathcal{F}[\nu](u),$$

the second equality is just the linearity and continuity of the Fourier transform. By linearity, we have

$$\sum_{k = 0}^{N} \frac{1}{k!}\bigl(\mathcal{F}[\nu](u)\bigr)^k = \mathcal{F}\Biggl[\sum_{k = 0}^{N} \frac{1}{k!} \nu^{\ast k}\Biggr](u).$$

Since the exponential series

$$\sum_{k = 0}^{\infty} \frac{1}{k!} \nu^{\ast k}$$

converges in norm - it even converges absolutely because $\lVert \mu \ast \nu\rVert \leqslant \lVert\mu\rVert\cdot \lVert\nu\rVert$ - the continuity of the Fourier transform with respect to the norm on the space of finite measures yields the second equality.

The continuity of $\mathcal{F}$ follows from

$$\Biggl\lvert\int_{\mathbb{R}} e^{iux} \mu(dx)\Biggr\rvert \leqslant \int_{\mathbb{R}} \lvert e^{iux}\rvert\, \lvert\mu\rvert(dx) = \lvert\mu\rvert(\mathbb{R}) = \lVert\mu\rVert.$$

The norm on the space of finite measures corresponds to the $L^1$-norm for measurable functions, not to the $L^2$-norm, and the relevant inequality for Fourier transforms of integrable functions is

$$\lVert \mathcal{F}[f]\rVert_{C_0(\mathbb{R})} \leqslant \lVert f\rVert_{L^1(\mathbb{R})},$$

where the norm on $C_0(\mathbb{R})$ - the space of continuous functions vanishing at infinity - is the maximum norm.