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I wanted to show that $X' / U^\perp \cong U'$, for $U$ being a closed subspace of the Banach space $X$.

Therefore I looked at $l: X' / U^\perp \cong U' , x' + U^\perp=[x'] \mapsto x'|_U$.

It is clear that this map is onto, as we just take a $u' \in U'$. Then I can use Hahn-Banach to get an appropriate $x'$ such that $x'|_U = u'$ and in particular $||x'||=||u'||$.

Now I need to show that is isometric: since we always have $\|[x']\|\le \|x'\|$ and by Hahn-Banach we have $\|x'\|=\|x'_{|U}\|$, this shows $\|[x′]\|\le \|x'_{|D}\|$ and it should be always true that $\forall y' \in U^\perp: \|x'-y'\| \ge \|x'_{|D}\|$, hence $\|[x']\|\ge \|x'_{|D}\|$. Is this reasoning correct? The isometric property follows.

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  • $\begingroup$ $x'-x'_{| U}$ does not make sense. You can't subtract functions which have different domains. Suggestion: split the proof of the equality of norms in two parts: $\le$ and $\ge$. Let us know which part you have trouble with. $\endgroup$
    – user127096
    Apr 5 '14 at 17:31
  • $\begingroup$ Consider the restriction map $r : X^\prime \to U^\prime$ which sends a functional on $X$ to its restriction to $U$. Check that $r$ is a contraction and that $\ker r = U^\perp$, use Hahn-Banach to show that $r$ is surjective, and use the open mapping theorem to conclude that $r$ induces an isomorphism $X^\prime/U^\perp \to U^\prime$. Now, think about how you used Hahn-Banach to prove surjectivity. How does this also imply that the induced map $X^\prime/U^\perp \to U^\prime$ is isometric? $\endgroup$ Apr 5 '14 at 17:39
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    $\begingroup$ @cheapeffectivedietpills since we always have $||[x']||\le ||x'||$ and by Hahn-Banach we have $||x'|| = ||x'|_D||$, this shows $||[x']||\le ||x'|_D||$, right? $\endgroup$
    – user66906
    Apr 5 '14 at 18:45
  • $\begingroup$ @cheapeffectivedietpills and it should be always true that $\forall y' \in U^\perp: ||x'-y'|| \ge ||x'|_D||$, hence $||[x']||\ge ||x'|_D||$ $\endgroup$
    – user66906
    Apr 5 '14 at 18:51
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    $\begingroup$ I tried to fix the formulas you pasted in without LaTeX and with mismatched notation. It was not an enjoyable thing to do. $\endgroup$
    – user127096
    Apr 5 '14 at 19:06
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It is mostly correct now, but could be written better.

Consider $l:X'/U^\perp\to U'$, which sends each coset of $U^\perp$ to its restriction to $U$. This map is well-defined because all elements of the coset agree on $U$.

The map is surjective, because every functional $f$ on $U$ can be extended to a functional $g$ on $X$ (by Hahn-Banach), and $l([g])=f$. Moreover, we can choose $g$ so that $\|g\| = \|f\|$. Then $\|[g]\|\le \|g\|= \|f\|$. We have shown that $l$ does not decrease the norm. Since the norm of restriction is bounded by the norm of original function, $l$ does not increase the norm either. Thus, $l$ is an isometry.

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