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.



Fourier Transform of measure on Banach Space (a question about Pontryagin Duality)
Yes. Let $\phi:(E,+)\rightarrow S^1$ be a continuous homomorphism. For each $x\in E$, the map $\mathbb R\rightarrow S^1; t\mapsto \phi(tx)$ is continuous and a homomorphism, so there is some $\mu(x)\in\mathbb R$ with $$\phi(tx) = \exp(i t \mu(x)) \qquad (x\in E,t\in\mathbb R).$$ Then, for $x,y\in E$ and $\lambda\in\mathbb R$,
$$ \exp(it\mu(\lambda x+y)) = \phi(t(\lambda x+y)) = \phi(t\lambda x)\phi(ty)
= \exp\big(it(\lambda\mu(x)+\mu(y)\big). $$
Letting $t\rightarrow 0$, we conclude that
$$\mu(\lambda x+y) = \lambda\mu(x) + \mu(y).$$
So $\mu$ is a linear map. If $x\_n\rightarrow 0$ and $\mu(x\_n)\rightarrow\mu$ then $$ \exp(it\mu) = \lim\_n \exp(it\mu(x\_n)) = \lim\_n \phi(tx\_n) = \phi(tx) = \exp(it\mu(x)), $$
for all $t$, so again, $\mu=\mu(x)$, and we conclude that $\mu$ is continuous.


So $\mu\in E^\*$ and $\phi(x) = e^{i\mu(x)}$ as you want.


