Does the Zygmund class on a closed interval include all everywhere differentiable functions? Let $\mathcal{C}$ be the class of continuous functions in the Zygmund class on the closed interval $[0, 1]$.
In this question, a continuous function $f$ is in the Zygmund class if there is a constant $c>0$ such that— $$|f(x) + f(y) - 2f((x+y)/2)| \le c\epsilon, $$ for every $\epsilon>0$, whenever $x$ and $y$ are points in $f$'s domain such that $|x-y|\le\epsilon$.
Then, does $\mathcal{C}$ include every function that is differentiable everywhere on $[0, 1]$? (I know the answer is yes for functions that are continuously differentiable everywhere on $[0, 1]$.) If not, what is an example of an everywhere differentiable function on a closed interval that is not in the Zygmund class?
To be clear, this question is not homework or a self-study assignment.
 A: I think we could find counterexamples with a bounded derivative (edit: wrong, see remark at the end), but here's one with an unbounded derivative, as a variation of a well-known differentiable function with non-continuous derivative does the trick: define $$f(x) = \begin{cases} x\sqrt{x} \sin(1/x) & \mbox{ if } x>0 \\ 0 & \mbox{ if } x=0\end{cases}$$
You can check that the derivative of $f$ at $0$ is zero (bounded in absolute value by $x^{3/2}$), and for $x>0$, the derivative is $f'(x)=\frac{3}{2}\sqrt{x}\sin(1/x) - \frac{1}{\sqrt{x}}\cos(1/x)$. The idea is that the derivative oscillates faster and faster as you approach zero.
Now for an integer $k > 0$, take $x = \frac{1}{2k\pi}$ and $y = \frac{1}{(2k+1)\pi}$: then $f(x)=f(y)=0$, but $\frac{x+y}{2} = \frac{2k+1/2}{2k(2k+1)\pi}$, so \begin{align*} \Big(\frac{x+y}{2}\Big)^{3/2} 
 & \underset{k\to \infty}{\sim} \frac{1}{(2k\pi)^{3/2}}, \mbox{ and} \\
\frac{2}{x+y} & = (2k+1)\pi \frac{1}{1+1/(4k)} \\
& = (2k+1)\pi \Big(1 - 1/(4k) + O\big(1/k^2\big)\Big) \\
& = (2k+1)\pi - \pi/2 + O\Big(\frac{1}{k}\Big)\end{align*}
Hence $\sin(2/(x+y)) = -1 + O(1/k)$, and we conclude that $f\Big(\frac{x+y}{2}\Big) = \frac{-1}{(2k\pi)^{3/2}} + O\Big(\frac{1}{k^{5/2}}\Big)$. Finally, $\big|f(x)+f(y)-2f\big((x+y)/2\big)\big| \underset{k \to \infty}{\sim} \frac{2}{(2k\pi)^{3/2}}$, but at the same time $|x-y| = \frac{1}{2k(2k+1)\pi} \sim \frac{1}{4\pi k^2}$.
Thus there cannot exist a finite constant $c$ as required, since $\frac{1}{4\pi k^2}$ does not dominate $\frac{2}{(2k\pi)^{3/2}}$. The function $f$ has a derivative everywhere, but is not in $\mathcal{C}$.

I thought some more about the bounded derivative question, and I was mistaken: if $f$ is differentiable and $f'$ is bounded, then $f \in \mathcal{C}$.
It simply follows from Taylor's theorem (with a Lagrange remainder): with $x < y$, there exists $\zeta_1 \in \big[x, \frac{x+y}{2}\big]$ such that $f\big(\frac{x+y}{2}\big)-f(x) = f'(\zeta_1) \frac{y-x}{2}$, and $\zeta_2 \in \big[\frac{x+y}{2}, y\big]$ such that $f(y)-f\big(\frac{x+y}{2}\big) = f'(\zeta_2) \frac{y-x}{2}$. Hence $$\Big|f(x)+f(y)-2f\big(\frac{x+y}{2}\big)\Big| = \Big|\big(f'(\zeta_2)-f'(\zeta_1)\big)\frac{y-x}{2}\Big| \le \frac{|f'(\zeta_1)|+|f'(\zeta_2)|}{2}(y-x) \le ||f'||_{\infty} (y-x)$$
So we can simply take $c = ||f'||_{\infty}$.
