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.

  • $\begingroup$ 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... $\endgroup$ – Nate Eldredge Oct 27 '10 at 22:56
  • $\begingroup$ @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? $\endgroup$ – Jonas Teuwen Oct 27 '10 at 23:39

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.

| cite | improve this answer | |
  • $\begingroup$ 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! $\endgroup$ – user6096 Oct 27 '10 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.