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This is a follow-up to my last question: Modulus of Continuity. I accidentally asked the wrong question there, so I'm going to start over and hopefully ask the right question.

I'll repeat the relevant definitions. Let $\rho: \mathbb{R}^+ \to \mathbb{R}^+$ be a continuous nondecreasing function such that $\rho(t) = 0$ if and only of $t = 0$. If you can answer my question in the special case $\rho(t) = Ct$ where $C$ is a constant then it will probably be possible to adapt your construction to the general case.

Say that a function $f: X \to \mathbb{R}$ on a metric space has modulus of continuity $\rho$ at a point $x_0 \in X$ if $|f(x) - f(x_0)| \leq \rho(d(x,x_0))$ for every $x \in X$. For example, a function has modulus of continuity $Ct$ at $x_0$ if and only if it is Lipschitz with Lipschitz constant $C$ at $x_0$.

Question If $X$ is a compact metric space without isolated points, is it true that the set of all continuous functions on $X$ which have modulus of continuity $\rho$ at some point of $X$ is nowhere dense in $C(X)$ equipped with the supremum norm?

To prove that the answer is affirmative for a given $X$ one must be able to construct functions of arbitrarily small norm which oscillate arbitrarily rapidly. For example, if $\rho(t) = Ct$ and $X = [0,1]$ then one can use a piecewise linear function such that the slope of each linear piece is larger than $C$ in absolute value. However, I don't see how to generalize this idea to an arbitrary compact metric space without isolated points.

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Posted even on Mathoverflow. Here: – Giuseppe Nov 21 '11 at 19:45

A detailed answer was posted by fedja on MathOverflow. His main points were:

  • the set of functions that are $\rho$-continuous somewhere is not nowhere dense. In fact it is dense, because every function can be flattened a bit: namely, $(f-a-\epsilon)^+-(f-a)^-+a$ is within $\epsilon$ of $f$ but is flat where $f$ took values between $a$ and $a+\epsilon$.

  • the set of functions that are $\rho$-continuous somewhere is of first category. The key steps are summarized below.

Fix a constant $A$ and consider the set $F_A$ of functions $f$ such that there exists a point $x\in X$ at which $f$ has global modulus of continuity $A\rho $. This set is closed: if $f_n\in F_A$ converge to $f$ uniformly and $x_n\in X$ are the corresponding bad points, then any accumulation point $x$ of the sequence $x_n$ is bad for $f$. Also, it contains no open set, because any function $f \in C(X)$ can be perturbed by small spikes placed along some $\delta$-net in $X$ so as to lose the membership in $F_A$.

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Try functions such as $f(x) = \min(\epsilon, 2 \rho(d(x,x_0)))$.

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When $\rho(t)=Ct^\alpha$ and $X=[0,1]$, you can try functions like $t^{\alpha/2}$. For arbitrary $\rho$, I imagine you can do something similar, maybe involve $\rho$ in the definition of the function. – Jeff Nov 17 '11 at 22:22
But for which $x_0$? To prove that the set of all functions which have modulus of continuity at some point in $X$ is nowhere dense we must show that for any such function there is a nearby continuous function which fails to have modulus of continuity at any point of $X$. Your function by itself is constant far enough away from $x_0$, and it's not clear how to fix that even by adding up a bunch of functions of the form you describe. – Paul Siegel Nov 17 '11 at 23:30

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