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I was studying on Follan's Real Analysis book and in Chapter 5.3, I saw the Open Mapping Theorem which stated (with some adjustment) as follows:

Let $X,Y$ be Banach spaces and $T:X\to Y$ be linear and bounded operator. If $T$ is surjective, then $T$ maps open sets to open set.

Then, based on the expression above, I expect if $T$ is nonlinear then the conclusion must fail. So I consider an example as follows: Let $X=Y=\mathbb{R}$ be (simplest?) Banach spaces and then take a mapping $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = x^2$. Then consider open interval $U:=(-1,1)$ then $f(U) = [0,1)$ which is not open. Is my thinking correct?

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    $\begingroup$ I don't "do" functional analysis, but that is the standard good example in topology. (Though not surjective.) $\endgroup$
    – Randall
    Sep 24, 2018 at 17:03
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    $\begingroup$ But $f$ isn't surjective, so you don't have a counterexample. Consider instead a cubic polynomial. $\endgroup$
    – Aweygan
    Sep 24, 2018 at 18:14
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    $\begingroup$ I think I might be shocked to learn that $[0,1]$ is a Banach space, but then again, I don't know anything about them. $\endgroup$
    – Randall
    Sep 24, 2018 at 19:44
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    $\begingroup$ Anyway, does $f: \mathbb{R} \to \mathbb{R}$ by $f(x)=x^3 -3x^2$ work for the same reason as your proposed example? It's bounded (in the operator sense), onto, but look at $f(0,3)$. $\endgroup$
    – Randall
    Sep 24, 2018 at 19:51
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    $\begingroup$ @Randall. There was a crank on this site who, among other odd things, said "Would you be shocked when I tell you that $1$ is a prime number?" $\endgroup$
    – DanielWainfleet
    Sep 25, 2018 at 3:41

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Consider $f: \mathbb{R} \to \mathbb{R}$ by $f(x)=x^3 -3x^2$. This is bounded (as it's continuous) and surjective. However, the open set $(0,3)$ in the domain maps to $[-4,0)$, which isn't open.

(Aweygan gets the assist.)

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    $\begingroup$ Now I don’t like this answer because my intuition on what bounded means is totally shot without having linearity to rely on. Is there a right notion of bounded nonlinear operator? $\endgroup$
    – Randall
    Sep 25, 2018 at 11:23

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