Uniform continuity? Suppose that $f$ is continuous on the interval $I$, and $|f(x)|$ is uniformly continuous. Can we have the conclusion that $\sin^3f(x)$ is uniform continuous? 
My intuition is that it is not. But I could not construct a counterexample. Would you help me out? Thank you.
 A: We have the conclusion - assuming real values, and not complex ones. The composition of uniformly continuous functions is uniformly continuous - for $f\circ g$ and an arbitrary $\epsilon > 0$, there is a $\delta_1 > 0$ with $\lvert x-y\rvert \leqslant \delta_1 \Rightarrow \lvert f(x) - f(x)\rvert \leqslant \epsilon$ by the uniform continuity of $f$, and by the uniform continuity of $g$, there is a $\delta_2 > 0$ with $\lvert \xi-\eta\rvert \leqslant \delta_2 \Rightarrow \lvert g(\xi) - g(\eta)\rvert \leqslant \delta_1$; so for $\lvert\xi-\eta\rvert \leqslant \delta_2$ we have $\lvert f(g(\xi)) - f(g(\eta))\rvert \leqslant \epsilon$ then.
And for continuous (real-valued) $f$ (on an interval), the uniform continuity of $\lvert f\rvert$ implies uniform continuity of $f$. Let $\epsilon > 0$ be given. By the uniform continuity of $\lvert f\rvert$, there is a $\delta > 0$ such that $$\lvert x - y\rvert \leqslant \delta \Rightarrow \bigl\lvert \lvert f(x)\rvert - \lvert f(y)\rvert \bigr\rvert \leqslant \epsilon/2.$$
Then we have $\lvert x-y\rvert \leqslant\delta \Rightarrow \lvert f(x)-f(y)\rvert \leqslant \epsilon$. For if $f(x)$ and $f(y)$ have different sign - and for real valued functions, that is the only way to have close absolute modulus without being close - then by the intermediate value theorem, there must be a $z$ between $x$ and $y$ with $f(z) = 0$. But if $z$ is between $x$ and $y$, then $\lvert z - x\rvert \leqslant \lvert x-y\rvert \leqslant \delta \Rightarrow \lvert f(x)\rvert \leqslant \epsilon/2$; similarly, $\lvert f(y)\rvert \leqslant \epsilon/2$. So If $\lvert x -y \rvert \leqslant \delta$, either the two values $f(x)$ and $f(y)$ have the same sign (or at least one of them is $0$), in which case $\epsilon/2 \geqslant \bigl\lvert \lvert f(x)\rvert - \lvert f(y)\rvert\bigr\rvert = \lvert f(x)-f(y)\rvert$, or we have $\lvert f(x) - f(y)\rvert \leqslant \lvert f(x)\rvert + \lvert f(y)\rvert \leqslant \epsilon/2 + \epsilon/2 = \epsilon$.
Since $\varphi \mapsto \sin^3 \varphi$ is uniformly continuous (bounded derivative) on $\mathbb{R}$, the uniform continuity of $\sin^3 f(x)$ follows.
