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It's true that in a normed space $X$ we have $d(x,\ker(f))=\frac{|f(x)|}{\|f\|}$ for any $x\in X$ and $f \in X^*$?

Otto Holzmann
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  • Hi and welcome! Can you please add some context (e.g. definitions) to your question? – MattAllegro Feb 23 '15 at 19:34
  • My guess is that this works if and only if your space is reflexive, where we can represent $f$ with some kind of inner product. – Ben Grossmann Feb 23 '15 at 19:42
  • I agree, it seems that in this case, writing $f(x) = \langle x,y\rangle$), $d(x,\textrm{ker}(f)) = |\textrm{proj}_y x| = \left| \dfrac{\langle x, y\rangle}{|y|^2}y \right| = \dfrac{|\langle x,y \rangle |}{|y|}$.

    I couldn't quickly come up with a counterexample for other spaces, but a guess out of nowhere is that $f = \delta$ and $X = C^\infty$ might work.

     – BaronVT Feb 23 '15 at 19:54

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Obviously you must assume $f \ne 0$ for your equation to be defined.

For any $y \in \text{ker}(f)$ we have $\|x - y\| \ge |f(x)|/\|f\|$. In the other direction, for any $\epsilon > 0$ there exists $z \in X$ with $f(z) = 1$ and $\|z\| \le (1+\epsilon)/\|f\|$. Then $y = x - f(x) z \in \ker(f)$ with $\|x - y\| = |f(x)| \|z\| \le (1 + \epsilon) |f(x)|/\|f\|$. Take $\epsilon \to 0+$.

Robert Israel
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  • Can you please tell how did you deduced that $y\in ker(f) $ , we have $||x-y||\geq |f(x)| / ||f||$ ? –  Jan 04 '23 at 18:28
  • By definition of the norm on $X^*$, $|f(x-y)| \le |f| |x-y|$, but $f(x-y) = f(x)-f(y) = f(x)$. – Robert Israel Jan 04 '23 at 21:18