## Can a continuous, nowhere differentiable function have specified “shape” at every point?

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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Do you want the limit to be nonzero almost everywhere? – Douglas Zare Mar 15 2010 at 22:50
How about taking $\phi$ to be a discontinuous additive function and $f=\phi$? – Jonas Meyer Mar 15 2010 at 22:51
@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. – Mike Hall Mar 16 2010 at 0:32
@Mike, Any $C^{0,\alpha}$-function $f$ and $\phi(x)=|x|^\beta$ for $0<\beta<\alpha<1$ will do. – Anton Petrunin Mar 16 2010 at 1:12
But Anton, I don't think that meets the "wildly oscillating" criterion. – Jonas Meyer Mar 16 2010 at 1:18

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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 “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.) – L Spice May 9 2011 at 2:13 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). – Thomas Kragh May 27 2011 at 15:12

I'm new here, so I hope my answer is of any use and not too late.

I was wondering: Wouldn't it be perhaps more natural to consider limits of the form

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

If for example we take $f(x) = |x|$ for $x\in \bf{R}$ and $\phi = f$, the "derivative" would be equal to one everywhere, which makes sense, since $f$ and $\phi$ are really the same, therefore their shape should be the same, right?

By the previous definition which uses $\phi(y-x)$ in the denominator, we would get limits like

$\lim_{y\to x}\frac{|y|-|x|}{|y-x|}$

which exists only for $x = 0$, even though our intuition tells us the shape is supposed to be the same everywhere. Any comments?

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 Oh, I think I get it now. You're looking for a function that is locally of the same shape as $\phi$ is around zero. My answer is probably a bit off-topic then ... – Dejan Govc Jul 14 2011 at 20:31