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Let Y be a closed subspace of a normed linear space X. Show that Y* is isometrically isomorhpic with $X^*/Y^\perp$, where $Y^\perp$ is the set of functionals $\ell$ that vanish on Y.

I have a little problem understanding what I am supposed to do. Do I need to find an bijection between the sets that is group and distance preserving?

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You have to find a linear bijective $X^*/Y^\bot \to Y^*$ that is distance preserving. Note that linear maps $X^*/Y^\bot \to Y^*$ correspond to linear maps $X^* \to Y^*$, which's kernel contains $Y^\bot$. There is a very simple map $X^* \to Y^*$ (keep in mind that $Y \subseteq X$) that will work here.

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    Im not sure I follow. Can we define som map $T: X^* \rightarrow Y^*$ as the projection of $X^*$ to $Y^*$? would this give kernel $Y^\perp$?2012-11-26
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    Elements of $X^*$ are linear maps $x^*\colon X \to \mathbb K$, right? Recall $Y \subseteq X$, how can you get a linear map $Tx^*\colon Y \to \mathbb K$ from $x^*$?2012-11-26
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    So $T\ell(y) = \ell(y) \Rightarrow Kerr(T) =Y^\perp$ and $T_2: X^*/Y^\perp \rightarrow Y^* \Rightarrow T([\ell]) = T_2(\ell)$2012-11-26
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    exactly.${}{}{}$2012-11-26