Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function such that $|f|$ is Hölder continuous with exponent $0<\alpha\leq1$. Does it follow that $f$ is also Hölder continuous with the same exponent?
I thought of this statement a few days ago and had trouble proving it or generating a counterexample. If we don't stipulate that $f$ is itself continuous, then the result is false (take $f(0)=1$ and $f(x)=-1$, otherwise). Additionally, the converse follows from the reverse of the triangle inequality, for if $f$ is Hölder continuous, then $$ \big||f(x)|-|f(y)|\big| \leq |f(x)-f(y)| \leq C|x-y|^\alpha. $$
Yes, this is corrrect. Let $x,y\in\mathbb R$. If $f(x)$ and $f(y)$ have the same sign, then $$|f(x)-f(y)|=\big||f(x)|-|f(y)|\big|\leq C|x-y|^{\alpha}.$$ Now, suppose that $f(x)$ and $f(y)$ have different signs. Then, from continuity of $f$, there exists $z\in(x,y)$ such that $f(z)=0$. Then, $$|f(x)-f(y)|\leq|f(x)|+|f(y)|=\big||f(x)|-|f(z)|\big|+\big||f(z)|-|f(y)|\big|\leq$$$$C|x-z|^{\alpha}+C|y-z|^{\alpha}\leq C|x-y|^{\alpha}+C|x-y|^{\alpha}=2C|x-y|^{\alpha}.$$ So, if $x,y\in\mathbb R$, you have that $|f(x)-f(y)|\leq 2C|x-y|^{\alpha}$, so $f$ is Hölder continuous with exponent $\alpha$.