If $f(x)$ is close to $0$ then necessarily $x$ is close to $\ker f$?

Question

Suppose that $X$ is a real Banach space and $f:X \to \mathbb{R}$ is a continuous linear functional. Is it true that for any $\varepsilon>0$ there is a $\delta>0$ such that for any $x \in X$ we have: $$|f(x)|<\delta \Rightarrow (\exists z \in \ker f) (\Vert x-z \Vert <\varepsilon).$$

This is certainly true if $X=\mathbb{R}^n$. I suspect this might be false for infinite-dimensional $X$, but I haven´t been able to find a counterexample.

Thanks!

Answer 1

If $f$ is identically $0$ there is nothing to show. Otherwise pick $x_0$ such that $f(x_0)\neq 0$ and write $x$ as $$x=x-\frac{f(x)}{f(x_0)}x_0+\frac{f(x)}{f(x_0)}x_0.$$ If we define $z(x):=x-\frac{f(x)}{f(x_0)}x_0$, then $z\in\ker f$ and $\lVert x-z(x)\rVert=\frac{|f(x)|}{|f(x_0)]}\lVert x_0\rVert$, hence $\delta:=\frac{\varepsilon}{\lVert x_0\rVert}|f(x_0)|$ does the job.