Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.