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Let $f \colon \mathbb{R} \to \mathbb{R}$ be a continuous functions that is nowhere differentiable. From this question (Does there exist a nowhere differentiable, everywhere continous, monotone somewhere function?) , I know that it follows that $f$ is monotone on no interval.

Let $x$ be a real number. We say that $f$ is non-decreasing at $x$ if there is a neighborhood of $x$, $N_x$, such that $\frac{f(y)-f(x)}{y-x} \ge 0$ if $y \in N_x-\{x\}$.

If a function is continuous everywhere and differentiable nowhere, does it follow that it is monotonic at no point? If this is not the case can you please give a counterexample?

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  • $\begingroup$ @Brian In the question I have written the definition of being non-decreasing at a point. The definition for non-increasing at a point is obtained by substituting $\ge$ by $\le$. And a function is monotone at $x$ if it is non-decreasing at $x$ or non-increasing at $x$. This does not imply monotonicity in a neighborhood, at least for general functions. $\endgroup$ – Student Jul 27 '14 at 1:36
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This does not follow: here's a cheap way to modify a given $f$ so that it becomes monotone at a point. Consider a local minimum $x_0$ of $f$. Let's assume that $x_0=f(x_0)=0$. Then $f(x)\ge 0$ in a neighborhood of $0$, so $$ g(x) = \begin{cases} f(x) & x\ge 0\\ -f(x) & x<0 \end{cases} $$ is monotone at $0$. If I'm spectacularly unlucky here, then $g$ is now differentiable at $0$, but then I can simply redefine it as $2f(x)$ for $x\ge 0$.

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  • $\begingroup$ Thanks, this is clever. Must every function that is nowhere differentiable have a local minimum? But most importantly, could you show that your redefined $g$ is indeed non-differentiable at $0$? For example, the function $f(x)=x/2$ if $x\ge 0$ , $f(x)=-x$ if $x<0$ is continuous but not differentiable at $0$ but $g(x)=2f(x)=x$ if $x\ge 0$ and $g(x)=x$ if $x<0$ is differentiable at $0$. $\endgroup$ – Student Jul 27 '14 at 2:40
  • $\begingroup$ (1) I was a little sloppy here (I was secretly thinking about the Weierstrass function which has local minima), but it's easy to make precise: consider the min of $f$ on $[0,1]$. Say it's at $x=0$ (if it's not an endpoint, then it's a local min). Now consider the min on $[-1,0]$. Again we have our local min unless this is $-1$. In this case, look at the max on $[-1,0]$. Done if this is $-1$ ($f$ constant on $[-1,0]$) or not an endpoint, and if it's $0$, then $f$ is increasing at $x=0$ without modification. $\endgroup$ – user138530 Jul 27 '14 at 2:49
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    $\begingroup$ (2) Usually $g$ as originally defined will not be differentiable at $0$ (I normally don't change from $f$ to $2f$, that is probably the misunderstanding here). However, it might be, but only in this very special situation: the one-sided derivatives $(Df)_-$, $(Df)_+$ of $f$ at zero exist and $-(Df)_-=(Df)_+$. We also know that these are $\not=0$ (because otherwise $f$ would be differentiable). So changing $f$ to $cf$ (say $2f$) on one side will get rid of this unintended differentiability. $\endgroup$ – user138530 Jul 27 '14 at 2:53
  • $\begingroup$ Oh I see, so if the original $g$ is differentiable at $0$ then the redefined $g$ won't be. Thanks! $\endgroup$ – Student Jul 27 '14 at 2:59

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