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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.

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up vote 3 down vote accepted

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.

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I strongly, strongly suspect that this must be very well known (assuming I haven't made a stupid mistake, and it's really false). But I wouldn't know where to look for a reference... – Matthew Daws May 16 '11 at 15:07
Perfect! Thank you Matt. – user2048 May 16 '11 at 15:19
Seems good to me. Alternatively, as $E$ is locally path connected and path connected, we know that $\phi$ has the lift $\mu_n$ to $\mathbb{R}$. – George Lowther May 16 '11 at 16:40

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