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.



Continuity of cylindrical functions.
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. 


