Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:37:27Z http://mathoverflow.net/feeds/question/60034 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60034/is-there-a-continuous-function-on-f-mathbbr-rightarrow-mathbbr-with-unco Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points? Kate 2011-03-30T05:07:24Z 2011-03-30T20:15:02Z <p>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.</p> <p>eg. $f(x)=sin(x)$ has countably infinite turning points as opposed to $f(x)=x^2$ which has one.</p> <p>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 . . . </p> http://mathoverflow.net/questions/60034/is-there-a-continuous-function-on-f-mathbbr-rightarrow-mathbbr-with-unco/60057#60057 Answer by ght for Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points? ght 2011-03-30T11:20:00Z 2011-03-30T11:20:00Z <p>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? </p> http://mathoverflow.net/questions/60034/is-there-a-continuous-function-on-f-mathbbr-rightarrow-mathbbr-with-unco/60061#60061 Answer by Jon Bannon for Is there a continuous function on $f:\mathbb{R} \rightarrow \mathbb{R}$ with uncountably infinite turning points? Jon Bannon 2011-03-30T12:34:18Z 2011-03-30T20:15:02Z <p>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.</p>