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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!

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  • $\begingroup$ Do you know quotient spaces and the quotient norm? $\endgroup$
    – Daniel Fischer
    Nov 11, 2014 at 16:16
  • $\begingroup$ Suppose that $f$ is not zero, let $u$ (fixed) such that $f(u)=1$, use $z=x-f(x)u$. $\endgroup$
    – Kelenner
    Nov 11, 2014 at 16:38

1 Answer 1

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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.

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  • $\begingroup$ Thank you very much Davide, I guess you meant $\delta:=\frac{\varepsilon}{\Vert x_0 \Vert}|f(x_0)|$. $\endgroup$
    – Ramiro de la Vega
    Nov 11, 2014 at 16:51
  • $\begingroup$ Yes, fixed now. $\endgroup$
    – Davide Giraudo
    Nov 11, 2014 at 16:55

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