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Let $X$ be a normed vector space and $M$ a linear subspace of $X$. Using the Hahn-Banach Theorem, I managed to prove that

For each $g\in X^*$ there exists $h\in M^\perp$ such that $||g-h||=d(g,M^\perp).$

Here, $X^*$ denotes the dual space of $X$ and $M^\perp=\{f\in X^*:f(x)=0~\forall x\in M\}$. Now, I'm trying to prove:

If for each functional $f\in M^*$ there exists a unique functional $F\in X^*$ such that $||F||_{X^*}=||f||_{M^*}$, then for each $g\in X^*$ there exists a unique functional $h\in M^\perp$ such that $||g-h||=d(g,M^\perp)$.

What I've tried so far is letting $f=g|_M$, the restriction of $g$ to $M$. Clearly $f\in M^*$, then there exists a unique functional $||F||_{X^*}=||f||_{M^*}$. I defined the function $h=g-F$, and tried to prove that $h\in M^\perp$. But my hypothesis doesn't say $F$ extends $f$, just preserves the norm. I'm not sure how to use the uniqueness part, there might be somehow a way to use that in order to show that $F|_M=f=g|_M$, or maybe I'm in the wrong direction. Any help would be appreciated.

Edit: that cannot be unique since $||F||=||-F||$, if I add $F|_M=f$ maybe everything works. I'm not sure if there is a weaker hypthesis we can add than that one. Just in case, this is an introductory Functional Analysis problem.

J. W. Tanner
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Fabrizio G
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1 Answers1

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The condition "for each functional $f\in M^*$ there exists a unique functional $F\in X^*$ such that $\|F\|_{X^{*}} = \|f\|_{M^{*}}$" is not suitable for this problem, because if $f\in M^{*}$ and $f \neq 0$ then the set \begin{equation} \{F\in X^{*} : \|F\|_{X^{*}} = \|f\|_{M^{*}}\} \end{equation} contains strictly more than one element. It is instead the condition "for each functional $f\in M^*$ there exists a unique functional $F\in X^*$ such that $\|F\|_{X^{*}} = \|f\|_{M^{*}}$ and $F\vert_{M} = f$" that is more appropriate for this problem. Hence we consider the following problem as suggested near the end of the question.

Let $X$ be a normed space and $M$ a linear subspace of $X$. Suppose that for each $f\in M^{*}$ there exists a unique $F\in X^{*}$ such that $F\vert_{M} = f$ and $\|F\|_{X^{*}} = \|f\|_{M^{*}}$. Then for each $g\in X^{*}$ there exists a unique $h\in M^{\perp}$ such that $\|g - h\|_{X^{*}} = d(g, M^{\perp})$, where $d(g, M^{\perp}) = \inf\{\|g - f\| : f\in M^{\perp}\}$.

Fix $g\in X^{*}$ throughout. We first establish existence. Using this result we have $g\vert_{M} \in M^{*}$ and $\|g\vert_{M}\|_{M^{*}} = d(g, M^{\perp})$. By a Hahn-Banach extension theorem there exists $\tilde{g} \in X^{*}$ such that $\tilde{g}\vert_{M} = g\vert_{M}$ and $\|\tilde{g}\|_{X^{*}} = \|g\vert_{M}\|_{M^{*}}$. Define $h := g - \tilde{g}$. Then $h\in M^{\perp}$ and \begin{equation} \|g - h\|_{X^{*}} = \|\tilde{g}\|_{X^{*}} = \|g\vert_{M}\|_{M^{*}} = d(g, M^{\perp}) . \end{equation} Hence $h$ has the desired properties. This shows existence.

We now show uniqueness. Assume there are $h_{1}, h_{2}\in M^{\perp}$ such that $\|g - h_{1}\|_{X^{*}} = d(g, M^{\perp}) = \|g - h_{2}\|_{X^{*}}$. Since $h_{1}, h_{2}\in M^{\perp}$ and $\|g\vert_{M}\|_{M^{*}} = d(g, M^{\perp})$, we deduce the equalities $(g - h_{1})\vert_{M} = g\vert_{M} = (g - h_{2})\vert_{M}$, $\|(g - h_{1})\vert_{M}\|_{M^{*}} = \|g - h_{1}\|_{X^{*}}$ and $\|(g - h_{2})\vert_{M}\|_{M^{*}} = \|g - h_{2}\|_{X^{*}}$. Hence $g - h_{1}$ and $g - h_{2}$ are both norm-preserving linear extensions of $g$, and as such extensions are unique by assumption, we obtain $g - h_{1} = g - h_{2}$ and therefore $h_{1} = h_{2}$. This shows uniqueness.

Dean Miller
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