The following definition is given as the Fourier transform of a Borel probability measure $\mu$ on $E$, a Banach Space (Real):

$\hat{\mu}: E^*\rightarrow \mathbb{C}$ defined by

$\hat{\mu}(x^*):=\int_E \; \exp(-i(x,x^*))d\mu(x),$

where $(\cdot,\cdot)$ is the duality pairing.

One defines the Fourier transform a random variable $X: \Omega\rightarrow E$ as the Fourier transform of its distribution $\mu_X$, where $(\Omega, \mathcal{F}, \mathbf{P})$ is a probability triple. Note that a $E$-valued random variable is defined to be strongly $\mathbf{P}$ measurable, so as to ensure that $X$ is separable valued (Petti).

Is it true that $E^*\cong \hat{E}$, where $\hat{E}$ is the Pontryagin Dual of $(E,+)$. That is, $\hat{E}=Hom_{cont}((E,+),S^1)$. We know this to be true for $\mathbb{R}$, where

$\mathbb{R}\cong\mathbb{R}^*\cong\hat{\mathbb{R}}$. Can I extend the proof of $\mathbb{R}\cong\hat{\mathbb{R}}$ to the general case? Below is the proof for $\mathbb{R}$:

Let $\phi\in\hat{\mathbb{R}}$. Since $\phi$ is a continuous map from $\mathbb{R}$ to $S^1$, then there exists an $a\in\mathbb{R}^+$ such that
$A=\int_0^a \phi(t)dt\ne 0$ . Since $\phi$ is a homomorphism, then
$\phi(x)A=\int_0^a \phi(t+x)dt=\int_x^{a+x} \phi(t)dt$. Applying the Fundamental Theorem of Calculus, we obtain the differential equation.

$\phi'(x)=A^{-1}(\phi(a+x)-\phi(x))=A^{-1}\phi(x)(\phi(a)-1).$ Thus, we obtain
$\phi(x)=C \exp(A^{-1}\left(\phi(a)-1\right)x),$ for some constant $C$. Since $\phi(0)=1$, then $C=1$. Furthermore, since $|\phi(x)|=1$, then $A^{-1}(\phi(a)-1)=2\pi i \xi$ for some $\xi\in\mathbb{R}$.

This completes the proof. One obtains the result for $\mathbb{R}^n$ using induction and the result: $\hat{G_1\times G_2}\cong \hat{G_1}\times \hat{G_2}$. To extend this proof to a general Banach Space, we would need Change of Variable and some form of Fundamental Theorem of Calculus.

Thank you in advance for your help.