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Let $C_c^\infty(\mathbb R^n)$ be the functions from $\mathbb R^n$ to $\mathbb R$ with compact support, further let $X$ be a separable Hilbert space with a fixed orthonormal basis $(e_n)_n$. Define the cylindrical functions:

\begin{align*} \text{Cyl}(X) := \{f : X \to &\mathbb R : \text{there exists } d \in \mathbb{N} \text{ and } \phi \in C_c^\infty(\mathbb R^n) \text{ such that }\\\ &f(x) = \phi(\langle x, e_1 \rangle, \ldots , \langle x, e_d \rangle ) \text{ for all } x \in X \} \end{align*}

Now, if $\langle . , . \rangle$ is the inner product on $X$ define $$\langle x, y \rangle_\omega := \sum_n \frac{1}{n^2} \langle x, e_n \rangle \langle e_n, y \rangle$$

Now, why is every $f \in \text{Cyl}(X)$ continuous with respect to $\langle . , . \rangle_\omega$? Sure, it is Lipschitz and continuous with respect to the weak topology (because it is with respect to the strong topology). Further I know that on bounded sets, the topology induced by $\langle . , . \rangle_\omega$ is the same as the weak topology. However, $f^{-1}(A)$ does not have to be bounded. What am I missing, I'm sure it is easy.

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I've usually seen cylinder functions defined by $f(x) = \phi(\langle x, u_1 \rangle, \dots, \langle x, u_d \rangle)$ where $u_1, \dots, u_d$ may be any elements of $X$, not just elements of the chosen orthonormal basis. My class of functions seems to be strictly larger than yours... –  Nate Eldredge Oct 27 '10 at 22:56
    
@Nate: Gradient Flows in Metric Spaces and in the Space of Probability Measures by Ambrosio et al uses the definition I give (pg 113). Do you have a reference where I can find your version? –  Jonas Teuwen Oct 27 '10 at 23:39
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For each $j$, the inequality $$|\langle x , e_j\rangle| = j \sqrt{\langle x , e_j\rangle\langle e_j , x\rangle\over j^2}\leq j\sqrt{\langle x , x\rangle_\omega}$$ shows that the map $x\mapsto \langle x , e_j\rangle$ is continuous from $(X , \langle\cdot ,\cdot\rangle_\omega)$ to $\mathbb R$. Thus, for fixed $d$ the map $x\mapsto (\langle x , e_1\rangle, \dots,\langle x , e_d\rangle)$ is continuous from $(X , \langle\cdot ,\cdot\rangle_\omega) $ into ${\mathbb R}^d$. Composing with the smooth map $\phi:{\mathbb R}^d\to {\mathbb R}$ gives you a continuous function from $(X , \langle\cdot ,\cdot\rangle_\omega)$ into $\mathbb R$ again.

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Thank you. I knew it was not so hard. –  Jonas Teuwen Oct 27 '10 at 21:12
    
You're welcome. When I was a student, I briefly thought that the $\langle \cdot , \cdot\rangle_\omega$ topology was the same as the weak topology. I'm glad to see that you didn't fall into that trap! –  Byron Schmuland Oct 27 '10 at 21:21
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