Let $\rho(t)$ be a function on the set $\mathbb{R}^+$ of nonnegative real numbers such that:
- $\rho$ is nondecreasing (and continuous - thanks for the correction)
- $\rho(t) = 0$ if and only if $t = 0$
Let $X$ be a metric space and let $f$ be a real valued function on $X$. Say that $f$ has modulus of continuity $\rho$ if $|f(x) - f(y)| \leq \rho(d(x,y))$ for every $x$ and $y$ in $X$. For example, a function is Lipschitz if and only if it has modulus of continuity $Ct$ for some positive real number $C$. Observe that a function with modulus of continuity $\rho$ is necessarily continuous.
Question: If $X$ is a compact metric space without isolated points, is it true that the set of all functions with modulus of continuity $\rho$ is nowhere dense (meaning its closure contains no open set) in $C(X)$ equipped with the supremum norm?
I am a TA in a class in which it was claimed that the answer is yes, but I don't completely believe the proof given and I can't seem to find a correct argument except in special cases. For example, one can show that the set of all Lipschitz functions on $[0,1]$ with Lipschitz constant $C$ is nowhere dense in $C[0,1]$ using the existence of piecewise linear functions of arbitrarily small norm whose linear pieces all have slope larger than $C$ (or smaller than -C). So the idea for general $X$ should be to construct continuous functions of arbitrarily small norm with arbitrarily rapid oscillation, but I don't see how to do this.
Thanks!