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The problem is:

Let E be a Banach space and $F\subset E$ be a closed linear subspace. Prove that for every $x \in E$ there exists $y \in F$ such that $\left\Vert x-y\right\Vert =\inf\left\{ \left\Vert x-z\right\Vert :\, z\in F\right\} =\left\Vert x+F\right\Vert _{E/F}$.

My efforts:

By definition, we know that $\left\Vert x+F\right\Vert _{E/F}:=\inf\left\{ \left\Vert x+z\right\Vert :\, z\in F\right\} $, and since $F$ is a linear subspace, we have that $\inf\left\{ \left\Vert x+z\right\Vert :\, z\in F\right\} =\inf\left\{ \left\Vert x-z\right\Vert :\, z\in F\right\} $.

My idea:

We know that for every bounded sequence in a reflexive Banach space, there exists a weakly convergent subsequence.

My question:

How can I now use the weak convergence to prove the first equality of my statement? Have I done any mistake so far?

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up vote 4 down vote accepted

Sketch of proof. We can find a sequence $\{y_n'\}\subset F$ such that $\lVert x-y_n'\rVert_E\leq \lVert x+F\rVert_{E/F}+\frac 1n$. We have $$\lVert y_n'\rVert\leq \lVert x-y_n'\rVert_E+\lVert x\rVert_E\leq \lVert x\rVert_E+F\rVert_{E/F}+\frac 1n\leq \lVert x\rVert_E+F\rVert_{E/F}+1,$$ hence the sequence $\{y_n'\}$ bounded. Since $F$ is a closed subspace of a reflexive subspace, it's itself reflexive hence we can extract a weakly converging subsequence (in $F$) $\{y_n\}$ to some $y\in F$. We have that $x-y_n$ converges weakly to $x-y$ in $E$ and we know that if $u_n\rightharpoonup u$ then $\lVert u\rVert\leq \liminf_n\lVert u_n\rVert$. This inequality shows that $y$ does the job.

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So we have that $\left\Vert x-y\right\Vert \leq\liminf\limits _{n}$$\left\{ \left\Vert x-y_{n}\right\Vert \right\} =\inf\left\{ \left\Vert x-z\right\Vert :\, z\in F\right\} =\left\Vert x+F\right\Vert _{E/F}$. But how can I now prove that the first inequality is actually an equality? – franckysnow May 28 '12 at 11:26
Or, showing this would be enough:$\left\Vert x+F\right\Vert _{E/F}=\inf\left\{ \left\Vert x-z\right\Vert :\, z\in F\right\} \leq\left\Vert x-y\right\Vert \qquad\forall y\in F$ ? – franckysnow May 28 '12 at 11:50
Yes, it's by definition of infimum. – Davide Giraudo May 28 '12 at 12:26

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