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I have been trying to prove this fact, but I am having trouble. I cannot even understand it intuitively. What makes it impossible to make the image grow until it is the whole space?

Francisco Maion
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  • True! Not very helpful to me though. Why should this be? What you stated is basically the definition of an open set – Francisco Maion Feb 28 '18 at 18:46
  • It is more intuitive when $A,B$ are Hilbert spaces. Choose orthonormal basis $e_1,e_2,...$ for $B$, if $T$ is not onto then one of the orthonormal basis elements is not in it's range, say $e_1$. Then it follows that $T(A)$ is orthogonal to $e_1$ and therefore the distance would be at least $1$. But $e_1$ is in the image of $S$ making $|T-S|_{op}\geq 1$ – Yanko Feb 28 '18 at 18:58
  • did you learn about 'almost orthogonal vectors'? – Yanko Feb 28 '18 at 19:14
  • I did not! Can you link a reference? – Francisco Maion Feb 28 '18 at 23:27
  • Couldn't find a reference (maybe this is not the right term) anyway I post an answer to your question. The lemma in the answer is exactly this 'almost orthogonality' I was talking about. – Yanko Mar 01 '18 at 16:35

1 Answers1

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I claim that if $S$ is onto, and $\|T-S\|_{op}\leq \frac{1}{2\cdot \|S\|_{op}}$ then $T$ is onto (This means that there exists an open subset around $S$ of onto operators).

Before we prove this claim I need a lemma

Lemma (existance of almost orthogonal vectors): Let $X$ be a normed space and $M\subseteq X$ be a closed subspace, then for every $1>\varepsilon>0$ there is some $x_\varepsilon\in X$ with $\|x_\varepsilon\| = 1$ and $d(x_\varepsilon,M)>1-\varepsilon$

Proof of Lemma: Let $u\in X\backslash M$, then $d(u,M)=d>0$. By definition, there exists $v\in M$ with $\|u-v\|<\frac{d}{1-\varepsilon}$. Observe that $d(u-v,M) = d(u,M)$ (because $M$ is a subspace). Let $x_\varepsilon = \frac{u-v}{\|u-v\|}$ we have that $d(x_\varepsilon,M)$ = $\frac{1}{\|u-v\|}d(u-v,M) = \frac{d}{\|u-v\|}>1-\varepsilon$. This proves the lemma.

Proof of the main claim: Suppose by contradiction that $T$ is not onto, then $T(A)\subseteq B$ is a closed vector space. Therefore, by the lemma for $\varepsilon=1/2$ there exists an element $b\in B$ with $\|b\|_B=1$ and $d_B(b,T(A))>1/2$.

Now $S$ is onto, so there exists $a$ such that $S(a)=b$. Since $\|b\|_B=1$ we have that $\|a\|_A\geq 1/\|S\|_{op}$.

It follows that $\|S(a)-T(a)\|_B > 1/2$ and therefore $\|S-T\|_{op}\geq 1/2\cdot \|a\| \geq 1/(2\|S\|_{op})$

Yanko
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