# A simple counter example for Open Mapping Theorem when Linearity Assumption is Dropped

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?

• I don't "do" functional analysis, but that is the standard good example in topology. (Though not surjective.) – Randall Sep 24 '18 at 17:03
• But $f$ isn't surjective, so you don't have a counterexample. Consider instead a cubic polynomial. – Aweygan Sep 24 '18 at 18:14
• 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. – Randall Sep 24 '18 at 19:44
• 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 $f(0,3)$. – Randall Sep 24 '18 at 19:51
• @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?" – DanielWainfleet Sep 25 '18 at 3:41

## 1 Answer

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.)

• 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? – Randall Sep 25 '18 at 11:23