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Let $X$ be a Banach Space and $Y$ be a closed subspace of $X$. Define the quotient space of $X$ as collection $\left\{ \left\{ x+y:y\in Y\right\} :x\in X\right\} $. Can we find a a surjective linear map $T:X/Y\rightarrow X$ such that $\left\Vert T\left( \left\{ x+y:y\in Y\right\} \right) \right\Vert _{X}=\inf \left\{ \left\Vert x-y\right\Vert _{X}:y\in Y\right\} $?

First, I just assume that $X$ is Hilbert Space. Then, define $T\left( \left\{ x+y:y\in Y\right\} \right) =x-P_{Y}\left( x\right) $ where $P_{Y}$ is orthogonal projection on $Y$, and I get the result immediately from definition of projection. But, I can't extend this to Banach space since the norm doesn't come from inner product so I can't define orthogonal projection like in Hilbert space. Maybe one could simplify $X=l^p, 1\leq p \leq \infty $ first.


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But the map you give in the Hilbert space case isn't surjective! – martini Nov 18 '12 at 7:22
@martini: Is that projection surjective? Could you give me the surjective one? – beginner Nov 18 '12 at 7:31
up vote 1 down vote accepted

Let $X$ be a Banach space with $\mathrm{dim}(X)>1$, then take arbitrary $f\in X^*\setminus\{0\}$ and set $Y=\mathrm{Ker}(f)$. In this case $\mathrm{dim}(X/Y)=1<\mathrm{dim}(X)$. If there exist a surjective isometry between $T:X/Y\to X$ in particular it would be a linear isomorphism. This will imply $\mathrm{dim}(X/Y)=\mathrm{dim}(X)$. Contradiction, so in general it is impossible, to construct such an isometry.

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