Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points? - MathOverflow

Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points?

2011-03-30T05:07:24Z

I was thinking about the statement "if f is continuous on the interval I, there is not necessarily an interval J in I on which f is monotone." and this led me to the question "does there exist a continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ that has uncountably infinite turning points?" when I say turning point I'm talking about a point (x,f(x)) s.t there exists an open ball around that point where f(x) is either the highest or lowest value within that ball.

eg. $f(x)=sin(x)$ has countably infinite turning points as opposed to $f(x)=x^2$ which has one.

I cant think of a reason that convinces me that its impossible yet I can conceptualize a function that does this. Is it impossible? or does there exist such a function? I certainly get the impression this is impossible . . .

Answer by ght for Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points?

2011-03-30T11:20:00Z

It seems to me that two dimensional Brownian motion is the example you are looking for. Can you please be more precise about what do you mean by turning point?

Answer by Jon Bannon for Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points?

2011-03-30T12:34:18Z

Of course you want to rule out the constant function, so you probably mean that there is a unique highest and lowest point in the neighborhood. Assuming this, with your new definition of turning point, you can choose your neighborhoods to be intervals with rational endpoints. This will force the number of turning points to be countable.