I am trying to prove:

If M is a subspace of a normed space $X$, that $\overline{M}=\bigcap\{\ker(\phi):\phi|_{M} = 0 \}$

It is really easy to see that $\overline{M} \subset \bigcap\{\ker(\phi):\phi|_{M} = 0 \}$. However, I don't know how to use the Hahn-Banach theorem to prove the other inclusion.

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    $\begingroup$ Consider a point $x\notin\overline M$, try to build some functional $\phi$ … ? $\endgroup$ – Harald Hanche-Olsen Dec 16 '14 at 16:43
  • $\begingroup$ If $x\notin \overline{M}$, the there is a functional $\phi \in X^*$, $\phi|_{M}=0$ and $\phi(x)=d(x,\overline{M})$. But how can I conclude the assertion? $\endgroup$ – Victor Ronchim Dec 16 '14 at 16:48
  • $\begingroup$ By looking carefully at the meaning of $\bigcap\{\ker(\phi):\phi|_{M} = 0 \}$. Does $x$ belong, or does it not? $\endgroup$ – Harald Hanche-Olsen Dec 16 '14 at 16:55
  • $\begingroup$ Thank you @HaraldHanche-Olsen, if I consider $\phi_x$ for all $x\notin \overline{M}$ as I described above, then the proof is done. $\endgroup$ – Victor Ronchim Dec 16 '14 at 16:57
  • $\begingroup$ @VictorRonchim, you should post your own solution as an answer if you managed to do it. Might help anyone who sees this in the future :) $\endgroup$ – Ivo Terek Dec 16 '14 at 16:58

For each $x\notin \overline{M}$, as a consequence of Hahn-Banach Theorem, there is $\phi_x \in X^*$, $\|\phi_x \|=1$, $\phi_{x}|_{M}=0$, $\phi_x(x)=d(x,\overline{M})$.

If $y\in \bigcap\{\ker(\phi): \phi|_M=0\} \subset \bigcap\{\ker(\phi_x)\}$, then $\phi_x(y) = 0 \quad \forall x \notin \overline{M}$. Thus $y\in \overline{M}.$

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    $\begingroup$ Could you clarify why does $\phi_x|_{M}$ have norm 1? For my understanding, I could only show it has a norm smaller or equal to 1. $\endgroup$ – Wenkui Liu Dec 11 '18 at 15:46
  • $\begingroup$ @WenkuiLiu it follows from the definition of $d(x,\overline{M})$. The argument is in the 1.9.7 Corollary of the book An Introduction to Banach Space Theory, from Megginson. There is an alternate proof where one can define a functional $\varphi (z +\alpha x) = \alpha. d(x,\overline{M})$ over the subspace $\overline{M} + [x]$. Both proofs are essentially the same. $\endgroup$ – Victor Ronchim Jan 28 at 2:51

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