Let $f\in E'$ (dual space of $E$), $f\neq 0$, $M=\{f=0\}$. Prove that $\operatorname{dist}(x,M)=\dfrac {|f(x)|}{\|f\|}$.
My idea: $\operatorname{dist}(x,M)=\inf\limits_{v\in M}\|x-v\|$, $|f(x-v)|=|f(x)|\leq \|f\|\cdot\|x-v\|\implies \|x-v\|\geq\dfrac {|f(x)|}{\|f\|}$. But I can't prove the opposite inequality. Can anyone help me? Thank you.
The case $x \in M$ is trivial.
Assume $x \not\in M$. Then we can show $E = \{\lambda \, x + v: \lambda \in \mathbb K, v \in M\}$ (how?). Let $\varepsilon > 0$ be given. Since $$\| f \| = \sup_{e \in E, \|e\| \le 1} |f(e)|,$$ there is $e \in E$ with $\|e\| \le 1$ and $\| f \| \le |f(e)| + \varepsilon$. Now, write this $e$ as $\lambda \, x + v$ and conclude.
