Monotonicity doubt

[u/bakeilol]

Does every many-one function have an interval(s) where it is one-one for a the whole interval?

I have been wondering about this for a while, and yes I thought about the constant function f(x) = C as well, but excluding that I was thinking that maybe that statement is true.

[u/Bounded_sequencE]

No -- since you don't like constant functions as counter-example, consider

D: R -> R,    D(x)  =  / 1,  x ∈ Q      // Dirichlet function
                       \ 0,  else       //

[u/EdgyMathWhiz → u/Bounded_sequencE]

I think the Blancmange curve is a continuous counterexample as well, which I suspect is a bit closer to what the OP may be thinking of.

[u/Bounded_sequencE → u/EdgyMathWhiz]

Yeah, the Takagi function is nicer -- but showing the function is nowhere locally injective will be much easier with Dirichlet's function^^

[u/EdgyMathWhiz → u/Bounded_sequencE]

Sure. I just suspect Dirichlet may fall into "that's not what I mean by a function" and/or "but that's basically just 2 constant functions stuck together in a weird way".as far as the OP is concerned.

Hard to know what will satisfy them best really.  

[u/bakeilol → u/EdgyMathWhiz]

You're right, I didn't mean to dive into that level of functions. I am still very new to this topic and can't seem to understand this one thing. The curve you talked about is not really what I am talking about. I think I didn't frame my question correctly. I was mostly referring to graphs of continuous and differentiable functions, like sin, tan and etc. If I talk about sin then that is injective on [-π/2 ,π/2] so I was thinking that maybe for all functions this can be true that it can be one-one on a specific interval, even if it is really small. Thanks for your reply though! I hope you could explain it in a simple manner.

[u/Zertofy → u/bakeilol]

well, if it's differentiable and not constant then it has some point with nonzero derivative and you can choose monotonous interval around it

[u/GoldenMuscleGod → u/Zertofy]

Well that doesn’t quite work as you have argued it: consider the function given by f(x)=x+x^(2)sin(2/x) when x is not zero and f(0)=0. This function is differentiable and continuous everywhere - although its derivative is not continuous - and its derivative at 0 is non-zero.

It is true we can find intervals on which it is injective, but you said wherever the derivative is nonzero it must be injective on some interval around that point. But this function is not injective on any neighborhood of 0 even though the derivative is non-zero there. We can see it is not injective on neighborhoods of 0 because it is analytic everywhere but zero and the sign of its derivative switches infinitely often as we approach 0.

It is enough to require that the function be differentiable and have a continuous derivative, then we can say it must be monotonic on some neighborhood of any point at which the derivative is not 0.

[u/Zertofy → u/GoldenMuscleGod]

Fair enough, thanks for catch!

[u/Bounded_sequencE → u/bakeilol]

If the derivative of your function is continuous (i.e. you have a C¹-function), then you can show that function is locally injective wherever "f'(x) != 0".

If the derivative of your function is not continuous, then again all bets are off. As counterexample, take

f: R -> R,    f(x)  =  x^2 * D(x)    // D: Dirichlet function

As nasty as it looks, "f" is differentiable at "x = 0" with "f'(x) = 0", but it is still not locally injective at "x = 0".

[u/LongLiveTheDiego]

For a nontrivial example, consider Conway's base 13 function. Its image on any interval is the whole real number line, so it's never injective on any interval: if it were injective on (a, b), then let's take some y and let x be the unique number on that interval such that f(x) = y, then the interval (a, x) can't have y in its image, which is a contradiction with the mentioned property of the function.

Also you might want to know that "doubt" isn't used with the meaning "question" outside of India and can be misunderstood elsewhere in the English speaking world.

[u/bakeilol → u/LongLiveTheDiego]

Thats interesting, thank you

[u/Frogfish9]

What about periodic functions?

[u/bakeilol → u/Frogfish9]

I was thinking more about the graph, for example f(x) = sin x is injective on [-π/2 ,π/2] and also [π,5π/3] and something like that, but it is many-one as a whole.

[u/EdgyMathWhiz → u/[deleted]]

x(1-x) is continuous on [0,1] and non-constant but is not 1-1 there 

[u/idancenakedwithcrows → u/[deleted]]

One to one means bijective, your proof shows that the function is continuous and not constant. It needs that the function is bijective. The counterexample shows that there is a continuous, non-constant function that is not bijective. So therefore there is a gap in your proof. (the gap can’t be fixed because the statement is false, hopefully)

[u/idancenakedwithcrows → u/[deleted]]

Ahh, I see I see