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I'm a bit embarrassed to admit that:

a) This is a rather unmotivated question.

b) I can't remember whether or not I've asked this before, but searching doesn't seem to turn anything up so ...

Consider some "shape" function $\phi: \mathbf{R} \to \mathbf{R}$. Then given some function $f: \mathbf{R} \to \mathbf{R}$, one can ask whether the "difference quotient",

$\lim_{y\to x} \frac{f(y)-f(x)}{\phi(y-x)}$,

exists at various points $x$. Letting $\phi(x) = x$ corresponds to taking normal derivatives, and intuitively when the limit exists this means that near $x$, the function $f$ "looks like" $\phi$ does near 0.

However, if the ratio $\phi(x)/x$ is not bounded above or away from 0 as $x\to 0$ (I'm mostly thinking of the case when it is neither, so that $\phi$ is "wildly oscillating" in some sense), then anywhere the above limit exists and is nonzero, the function $f$ is necessarily non-differentiable.

My question: If $\phi$ is some wildly oscillating function as described above (pick your favorite), can there be an $f$ for which this limit exists everywhere?

(Edit: I suppose I really want $\phi$ and $f$ to be continuous functions.)

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    $\begingroup$ Do you want the limit to be nonzero almost everywhere? $\endgroup$
    – Douglas Zare
    Commented Mar 15, 2010 at 22:50
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    $\begingroup$ How about taking $\phi$ to be a discontinuous additive function and $f=\phi$? $\endgroup$
    – Jonas Meyer
    Commented Mar 15, 2010 at 22:51
  • $\begingroup$ @Douglas Zare Not necessarily. Is there an easy (non-constant) example if it isn't required? @Jonas Meyer Good point. I guess I really want $\phi$ to be continuous, and $f$ to be a continuous nowhere-differentiable function. I really should have put the adjective continuous in a lot of places. $\endgroup$
    – Mike Hall
    Commented Mar 16, 2010 at 0:32
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    $\begingroup$ @Mike, Any $C^{0,\alpha}$-function $f$ and $\phi(x)=|x|^\beta$ for $0<\beta<\alpha<1$ will do. $\endgroup$
    – Anton Petrunin
    Commented Mar 16, 2010 at 1:12
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    $\begingroup$ But Anton, I don't think that meets the "wildly oscillating" criterion. $\endgroup$
    – Jonas Meyer
    Commented Mar 16, 2010 at 1:18

1 Answer 1

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Assume WLOG that $\phi(x)>0$ when $x>0$. Since the limit described exists for all $x$ in the source of $f$. We get for any $x$ the bound:

$f(x+\delta)-f(x) \leq C\phi(\delta)$

for $0 < \delta < \delta_0$ for some $C,\delta_0>0$ which may depend on $x$.

diving by $\delta$ we get by the assumptions on $\phi$ that

$\underline{\lim}_{\delta \to 0} ( \frac{f(x+\delta) - f(x)}{\delta}) \leq 0$

This is one the four derivatives of $f$, and proposition 2 chapter 5 in Real Analysis by H.L. Royden states that if $f$ is continuous then it is (non-strictly) decreasing. Similar for increasing. So $f$ is constant.

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  • $\begingroup$ “Assume WLOG that $\phi(x)>0$ when $x>0$.” Why? M Hall specifically wants to consider the ‘wildly oscillatory’ case. (I am supposing that $x \mapsto x\sin(1/x)$ is wildly oscillatory, anyway.) $\endgroup$
    – LSpice
    Commented May 9, 2011 at 2:13
  • $\begingroup$ Wildly oscilating refers to $\phi(x)/x$ being both close to zero and unbounded for very small $x$. The assumption WLOG refers to the fact that the limit would not exist if $\phi$ were not strictly positive or negative in a small $(0,\epsilon)$ (assuming continuity). $\endgroup$
    – Thomas Kragh
    Commented May 27, 2011 at 15:12

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