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Having X , a normed vector space and X* = B(X,R) its dual space ( R is the real numbers). Show that for all f contained in X*, we have that Ker f included in X is a closed subspace.

Knowing that X* is a Banach space, since R is a complete space, does that make X a Banach space too ? Needing a little help with this one.

Ben Grossmann
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mimi
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  • It is not necessary for $X$ to be a Banach space. Indeed, however, $X^*$ will be a Banach space whether or not $X$ is complete (see this post for instance) – Ben Grossmann Nov 25 '18 at 19:51
  • This answer really comes down to applying the definitions. In particular: what does $\ker f$ mean, and what does it mean for a subspace to be closed? – Ben Grossmann Nov 25 '18 at 19:52

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That $\ker f$ is a linear subspace is evident. To prove that it is closed, notice that $\ker f = f^{-1}(\{0\})$, which is the preimage of the closed set $\{0\}$ under the continuous map $f$.

This has nothing to do with $X$ being or not being a Banach space.

MSDG
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