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Continuing with the previous question, I want to ask one more question which seems true but I don't have any idea again. If $A\colon X\to \mathbb{C}^{n}$ is a finite rank bounded linear operator, then does the following holds? $$\overline{\{Ax:\|x\|\leq 1\}}\subset \{Ax:\|x\|\leq 2\}$$ Without any further assumptions on $X$, I think this might be false, but I can't find any counter examples. It seems that $X=c_{0}$ given in the answer to the previous question can't be a counterexample.

Gibbs
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Seewoo Lee
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  • This is true if ${Ax:||x||\leq 2}$ is repleced by ${Ax:||x||\leq N}$ for a suitable integer $N$ depending on $A$. – Kavi Rama Murthy Jun 01 '18 at 08:58
  • @KaviRamaMurthy Can you give any hints? Also, if $||A||$ is large, is the statement false in general? Actually, I tried to show the same thing for $(1, 2)$ replaced by arbitrary $(a, b)$ with $0<a<b$ but I failed. – Seewoo Lee Jun 01 '18 at 09:24
  • I do not know the complete answer to the question yet but as far as my statement is concerned you can do the following: consider $A$ as an operator from $X$ onto its range. The range is finite dimensional, hence complete. This implies that $A$ is an open map. So the image of the open unit ball contains some ball around the origin in the range. From this my claim follows easily. – Kavi Rama Murthy Jun 01 '18 at 09:28

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The statement is true, and it is even true if you would replace the $2$ with $1+\varepsilon$.

First, we can consider $A$ as an operator from $X$ onto its range, which is again finite-dimensional. So, without loss of generality we can assume that $A$ is surjective. We define the set $$ D=\{Ax : \|x\|\leq 1\}. $$ Using linearity, we can see that the original claim is equivalent to $$ \overline D \subset 2D. $$

Note that it can be shown that $D$ is convex, symmetric, and $0$ is in the interior of $D$.

Let $\delta>0$ be such that $\overline{B_\delta(0)}\subset D$. Consider a sequence $z_n\in D$ such that $z_n\to z$, but $z\not\in D$. After some calculations, it can be seen that $$ \gamma_n^{-1} z = (1-\beta_n) z_n + \beta_n\delta \|z-z_n\|^{-1} (z-z_n) $$ with a suitable $\beta_n\in (0,1)$ and where $\gamma_n= 1+\delta^{-1} \|z-z_n\|$. Since this is a convex combination, this implies $\gamma_n^{-1}z \in D$, or $z\in \gamma_n D$.

Since $\gamma_n\to 1$, this means that $$ \overline D \subset (1+\varepsilon) D $$ for every $\varepsilon>0$.

This is only a sketch of a proof and maybe there are less complicated ways to prove your claim.

harfe
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