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Let $X$ and $Y$ be two Banach spaces and $A:X\rightarrow Y$ a linear, continuous map. Let $M\subseteq X$ be a closed, convex subset of the unit sphere in $X$. When is $A(M)\subseteq Y$ closed?

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Compactness of $M$ would be sufficient and no reasonable other condition comes to mind. Are you allowed to assume further properties on $A$? – t.b. May 1 '12 at 9:30
For the operator norm of $A$ we have $\Vert A\Vert\leq 1$... – Andy Teich May 1 '12 at 9:34
Okay, but you can always reduce to that case by scaling. I was more thinking of $A$ being a compact operator or $A$ being bounded away from zero or something like that. – t.b. May 1 '12 at 9:36
up vote 0 down vote accepted

Even in finite dimensions this can fail: let $X=\mathbb{R}^2$, $Y = \mathbb{R}$, $A$ be projection onto the first coordinate, and $M = \{ (x,y) : x > 0, y \ge 1/x\}$. In particular compactness of $A$ isn't sufficient.

Compactness of $M$ is obviously sufficient. Boundedness of $M$ is sufficient in finite dimensions thanks to Heine-Borel, but not in infinite dimensions: take $X = C([0,1])$, $Y = L^1([0,1])$, $A$ the inclusion map, and $M$ the closed unit ball of $X$; you can approximate, say, $1_{[0,1/2]}$ in $L^1$ norm by continuous functions bounded by 1.

If $A$ is bounded away from 0 (i.e. there is a constant $c$ with $\|Ax\| \ge c\|x\|$), then $A$ is a homeomorphism and so $A(M)$ is closed. Likewise, if $A$ is bijective, then it follows from the open mapping theorem that it is again a homeomorphism.

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