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Suppose $X$ and $Y$ are Banach spaces, and let $A$ be a bounded operator from $X$ to $Y$.

Does there exist $r\gt0$ such that for every $B\in L(X,Y)$ and $\left\|A-B\right\|\lt r$:

  1. $A$ is surjective implies $B$ is surjective, or
  2. $A$ is injective implies $B$ is injective?

The following is my current work under the surjective condition. From open mapping theorem, we only need to prove that $\operatorname{ran}A$ is a second category set. As $Y$ is a Banach space, by Baire category theorem it is Baire space, so we only need to prove that $\operatorname{ran}A$ has nonempty interior. From open mapping theorem, we know that $A$ is an open map because it is a closed operator, so there exists $r_1$ such that $O(0,r_1)\subseteq Y$ is contained in $A(O(0,1)).$ I guess that if we choose $r$ enough small, then we can prove that $0$ is in the interior of $B(O(0,1)),$ but I cannot prove it.

And for the injective condition, I guess that there exist some counterexample. I try to let $X=C[0,1]$ and $A()=\int^_0().$ Is this a proper counterexample?

Homer
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    "A is injective implies B is surjective": don't you rather want A is injective implies B is injective? – Anne Bauval Nov 01 '23 at 17:03
  • For a counterexample for the injective condition: maybe try $X = C(\mathbb{R})$ with the $L^\infty$ norm, $A$ is multiplication by $e^{-x^2}$. – Daniel Schepler Nov 01 '23 at 17:42
  • Injectivity is not preserved. Any compact operator in an infinite dimensional Hilbert space is approximated by finite dimensional operators, which are obviously not injective. There are injective compact operators.

    On the other hand surjectivity is preserved. The proof can be performed along the lines of the Banach open mapping theorem.

     – Ryszard Szwarc Nov 01 '23 at 17:47
  • Yes,I made a mistake.I have corrected it. – Homer Nov 04 '23 at 00:49

1 Answers1

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The problem concerning surjectivity is solved here.

In my opinion the simplest counterexample for injectivity is the following. Let $X=Y=\ell^2(\mathbb{N})$ and $$A(x_1,x_2,\ldots,x_n,\ldots )\\ =(\lambda x_1,\lambda^2x_2,\ldots, \lambda^nx_n,\ldots)$$ where $0<|\lambda|<1.$ Then $A$ is injective. Let $B_n$ act by $$B_n(x_1,x_2,\ldots,x_n,\ldots )\\ =(\lambda x_1,\lambda^2x_2,\ldots, \lambda^nx_n,0,0,\ldots)$$ Then $\|A-B_n\|=|\lambda|^{n+1}\to 0.$ The operators $B_n$ are not injective as $B_ne_{n+1}=0,$ where $\{e_k\}_{k=1}^\infty $ denotes the standard basis of $\ell^2(\mathbb{N}).$

Concerning the operator $A$ mentioned at the end of OP let $$(B_nf)(x)=\begin{cases} 0& 0\le x\le {1\over n}\\ \int\limits_{1\over n}^x f(t)\,dt& {1\over n}<x\le 1 \end{cases}$$ Then $\|A-B_n\|={1\over n}.$ Every operator $B_n$ is not injective as it vanishes on functions supported in $[0,{1\over n}].$

Ryszard Szwarc
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