How many cusps could a convex function have?

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Let $f:mathbbR to mathbbR$ be a convex function. How many cusps (sharp corners) could it have? Are them numerable or not?



I could say that they are infinite and numerable, thinking to a function like $f(x)=arcsinbigg(fracsinxsqrt2 - cosxbigg)$, which is not convex. If I am right, how to prove it?




real-analysis functions




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  • 2




    what do you mean by sharp corner (specifically: what does sharp mean)? Would you call the behavior of the graph of $f(x) =frac1n |x|$, $n >1$ at $x=0$ a sharp corner, or would you require a slope $>1$, for example?
    – Thomas
    Jul 20 at 15:51






  • 1




    The $f(x)$ you have defined is good
    – F.inc
    Jul 20 at 16:23










  • why would they be infinite?
    – zhw.
    Jul 21 at 1:31














up vote
2
down vote

favorite
3












Let $f:mathbbR to mathbbR$ be a convex function. How many cusps (sharp corners) could it have? Are them numerable or not?



I could say that they are infinite and numerable, thinking to a function like $f(x)=arcsinbigg(fracsinxsqrt2 - cosxbigg)$, which is not convex. If I am right, how to prove it?




real-analysis functions




share|cite|improve this question















  • 2




    what do you mean by sharp corner (specifically: what does sharp mean)? Would you call the behavior of the graph of $f(x) =frac1n |x|$, $n >1$ at $x=0$ a sharp corner, or would you require a slope $>1$, for example?
    – Thomas
    Jul 20 at 15:51






  • 1




    The $f(x)$ you have defined is good
    – F.inc
    Jul 20 at 16:23










  • why would they be infinite?
    – zhw.
    Jul 21 at 1:31












up vote
2
down vote

favorite
3









up vote
2
down vote

favorite
3






3





Let $f:mathbbR to mathbbR$ be a convex function. How many cusps (sharp corners) could it have? Are them numerable or not?



I could say that they are infinite and numerable, thinking to a function like $f(x)=arcsinbigg(fracsinxsqrt2 - cosxbigg)$, which is not convex. If I am right, how to prove it?




real-analysis functions




share|cite|improve this question











Let $f:mathbbR to mathbbR$ be a convex function. How many cusps (sharp corners) could it have? Are them numerable or not?



I could say that they are infinite and numerable, thinking to a function like $f(x)=arcsinbigg(fracsinxsqrt2 - cosxbigg)$, which is not convex. If I am right, how to prove it?





real-analysis functions





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 20 at 15:26









F.inc

2688




2688







  • 2




    what do you mean by sharp corner (specifically: what does sharp mean)? Would you call the behavior of the graph of $f(x) =frac1n |x|$, $n >1$ at $x=0$ a sharp corner, or would you require a slope $>1$, for example?
    – Thomas
    Jul 20 at 15:51






  • 1




    The $f(x)$ you have defined is good
    – F.inc
    Jul 20 at 16:23










  • why would they be infinite?
    – zhw.
    Jul 21 at 1:31












  • 2




    what do you mean by sharp corner (specifically: what does sharp mean)? Would you call the behavior of the graph of $f(x) =frac1n |x|$, $n >1$ at $x=0$ a sharp corner, or would you require a slope $>1$, for example?
    – Thomas
    Jul 20 at 15:51






  • 1




    The $f(x)$ you have defined is good
    – F.inc
    Jul 20 at 16:23










  • why would they be infinite?
    – zhw.
    Jul 21 at 1:31







2




2




what do you mean by sharp corner (specifically: what does sharp mean)? Would you call the behavior of the graph of $f(x) =frac1n |x|$, $n >1$ at $x=0$ a sharp corner, or would you require a slope $>1$, for example?
– Thomas
Jul 20 at 15:51




what do you mean by sharp corner (specifically: what does sharp mean)? Would you call the behavior of the graph of $f(x) =frac1n |x|$, $n >1$ at $x=0$ a sharp corner, or would you require a slope $>1$, for example?
– Thomas
Jul 20 at 15:51




1




1




The $f(x)$ you have defined is good
– F.inc
Jul 20 at 16:23




The $f(x)$ you have defined is good
– F.inc
Jul 20 at 16:23












why would they be infinite?
– zhw.
Jul 21 at 1:31




why would they be infinite?
– zhw.
Jul 21 at 1:31















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