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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 . . .

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    $\begingroup$ You need to be more specific. The (graph of the) real-valued function f(x,y) = x - x^2 has uncountably many points (x,y) with a partial derivative of 0 and second partial negative. It is likely there are 2-D versions of Brownian motion which might come closer to what you actually intend to visualize. Gerhard "Ask Me About System Design" Paseman, 2011.03.29 $\endgroup$
    – Gerhard Paseman
    Commented Mar 30, 2011 at 5:55
  • $\begingroup$ Please use the "edit" link below the question, and describe the definition of "turning point" that you are using. $\endgroup$
    – S. Carnahan ♦
    Commented Mar 30, 2011 at 11:11
  • $\begingroup$ Sorry for being so late to edit and for being vague, I think the definition of turning point I use above makes it impossible. $\endgroup$
    – Kate
    Commented Mar 30, 2011 at 12:20
  • $\begingroup$ Still confusing. You ask about functions on ${\bf R}^2$ but your examples are of functions on $\bf R$ - only the graph is in ${\bf R}^2$. So what do you mean? $\endgroup$
    – Gerry Myerson
    Commented Mar 30, 2011 at 12:24
  • $\begingroup$ Ah - you answered my question while I was typing it in. Thanks. $\endgroup$
    – Gerry Myerson
    Commented Mar 30, 2011 at 12:25

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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.

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  • $\begingroup$ Thanks, I will check that out. I realized this when I defined turning points more precisely. Thanks for helping despite the problem not being quite up to scratch! $\endgroup$
    – Kate
    Commented Mar 30, 2011 at 12:43
  • $\begingroup$ I do not follow this argument. You can choose the neighborhoods to be rational intervals, but how do you know that such a neighborhood contains only one turning point? $\endgroup$
    – Michael Renardy
    Commented Mar 30, 2011 at 13:13
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    $\begingroup$ @Michael: You don't know that such a neighborhood N contains only one turning point, but you do know that at most one point attains the maximum of f in N and at most one attains the minimum of f in N. Any other turning points, even if they're in N, will have had other neighborhoods N' assigned to them --- neighborhoods in which they achieve the maximum or minimum value of f. $\endgroup$
    – Andreas Blass
    Commented Mar 30, 2011 at 13:48
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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?

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