Let $B$ be a Banach space. Let $x\in B.$ Then by Hahn-Banach there exists $f\in B^\ast$ such that $\|f\| =1$ and $f(x) = \|x\|.$
Question: can we strengthen this result so that now, for each $\epsilon>0$ we can find $f$ satisfying the following inequality: for all $y\in B$ such that $\|y\|=\|x\|,$ whenever $f(y)=\|x\|,$ we have $\|y-x\|<\epsilon$? This will simplify the proof of many results.
Edit: I just realised that strict convexity needs to be assumed.