Let $X$ a Hilbert space, $U$ be a basic weak neighbourhood of $0$ in $X$, and P an orthogonal projection. I want to prove that there exists $\varepsilon >0$ such that for any $x\in X$ with $\|x\| \leq 1$, if $\|Px\|\geq 1-\varepsilon$, then $(I-P)x \in U$.
I thought I would start with something simpler, like proving it for $U=\{x\in X: |\langle x,y\rangle < \delta\}$. In this case $(I-P)x \in U$ is equivalent to $|\langle (I-P)x,y\rangle| < \delta$. But I have not managed to show such an $\varepsilon$ exists even for this case.
I guess that this will probably be used at some point: $\|x-Px\|^2 = \|x\|^2 - \|Px\|^2 \leq 1 - (1-\varepsilon)^2 \leq 2\varepsilon - \epsilon^2$.
I would really appreciate some help. Thanks :)