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By convex/cave I mean by the definition for an interval $(x,y)$ of $f$ is convex iff $f(\frac{a+b}{2})\geq\frac{f(a)+f(b)}{2}$ and is concave if $f(\frac{a+b}{2})\leq\frac{f(a)+f(b)}{2}$ where $a,b\in(x,y)$. This is merely so that the function does not have to be differentiable.

If it's not possible to have a function be continuous but not convex or concave on any nonempty interval, is it possible to construct a function which is not convex or concave on any nonempty interval? Has one already been created?

If it is possible to have a function be continuous but not convex or concave on any nonempty interval, is it possible to construct it? Has one already been created?

Is it possible to have one of these functions be differentiable once?

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    $\begingroup$ Presumably you want not convex/concave on any finite interval of positive length. The Weierstrass function should fit this bill. upload.wikimedia.org/wikipedia/commons/6/60/… (article at en.wikipedia.org/wiki/Weierstrass_function) $\endgroup$
    – Noam D. Elkies
    Commented Oct 5, 2017 at 4:01
  • $\begingroup$ Sorry, I worded that wrong. I meant nonempty, not infinite. $\endgroup$
    – Zetapology
    Commented Oct 5, 2017 at 4:02
  • $\begingroup$ Also, can there be such a function with a first derivative? $\endgroup$
    – Zetapology
    Commented Oct 5, 2017 at 4:04
  • $\begingroup$ @NoamD.Elkies the Weierstrass function doesn't work because it has local extrema; take $a$ and $b$ at equal, sufficiently small, distances to the left and right of such a local extremum, e.g. a maximum, then we have $(f(a)+f(b))/2 < f((a+b)/2)$. As the definition of convexity or concavity is based on the comparison of two reals, the outcome can only be less, equal or greater; quartum non datur. What would be needed, is a defintion based on limits. $\endgroup$
    – Manfred Weis
    Commented Oct 5, 2017 at 5:29
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    $\begingroup$ @ManfredWeis The question clearly speaks of concavity/convexity in an interval; your example doesn't imply that the condition holds for all $(a,b)$ in a small interval around even that local extremum. $\endgroup$
    – Steven Stadnicki
    Commented Oct 5, 2017 at 6:01

2 Answers 2

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It is well known that almost every path of a Brownian motion is nowhere monotone (i.e., not monotone on any interval). Hence the primitive (antiderivative) of almost every path is continuously differentiable but nowhere convex or concave.

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    $\begingroup$ It is also not hard to rig up a nowhere monotone function by an inductive construction, which will be a little more elementary - take a piecewise linear function, and repeatedly add a new vertex near an existing vertex either slightly above or below, then form the piecewise function with the new vertices. Also add vertices in the middle of any big gaps. By choosing the vertices close enough, the sequence is uniformly convergent, hence the limit is continuous, and we can choose the vertices sequentially to kill monotonicity on every intereval. $\endgroup$
    – Will Sawin
    Commented Oct 5, 2017 at 12:13
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Any convex (or concave) function on a convex set is locally Lipschitz, thus in particular a function $f$ which is convex or concave on an interval is absolutely continuous. Now if you construct a function which is continuous on $[a,b]$ but not absolutely continuous on any subinterval, this should fit the bill.

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