Hölder regularity for $f^2$

Question

Let $f$ be Hölder-continuous, which means in that case $$\sup_{|x-y|\leq \tau}|f(x)-f(y)|\leq C\tau^\alpha,$$ for a $\alpha\in(1/2,1]$ and $\tau < 1$ as well as $x,y\in[0,1]$. I'm trying to show, that $$\sup_{|x-y|\leq \tau}|f^2(x)-f^2(y)|\leq C'\tau^\alpha.$$

My approach: $$\sup_{|x-y|\leq \tau}|f^2(x)-f^2(y)|=| \sup_{|x-y|\leq \tau}|f(x)-f(y)|\big(f(x)+f(y)\big),$$ but don't know how to go further.

To be clear, it's about to show that also the quadratic functions have the same Hölder regularity $\alpha$. It doesn't matter that you need to have another constant to fulfill the inequality in the second line.

Answer 1

The easiest way, as Kosh commented, is just using that $C^\alpha[0,1]\subset C^0[0,1]$ so you can bound your right-hand side by $2\lvert f \rvert_{L^\infty[0,1]} C \tau^\alpha$.

In general we can prove that the composition $g(f(x))$ where $f\in C^\alpha([0,1])$ and $g\in C^\beta([0,1])$ is a $\alpha \beta$-Holder continuous function.

In your case $g=x^2$ is Lipschitz (that means $1$-Holder) continuous on $[0,1]$ (since $\lvert g'(x) \rvert_{L^\infty[0,1]}<\infty$), hence $f^2$ is still $\alpha$-Holder.

Answer 2

Counter-example, $x\mapsto x$ is $1-$ Hölder continuous but $x\mapsto x^2$ is not Holder continuous.

Indeed, assume $x\mapsto x^2$ isHolder continuous which implies the uniform continuity. Hence for $\varepsilon=1/2$ there is $r>0$ such that $$|x-y|<r\implies |x^2-y^2|<1/2$$

However, for $x=2r+\frac1{2r}$ and $y=2r$ then
$$x^2= y^2+2+ \frac{1}{4r^2}$$

Thus we have $$2+\frac{1}{4r^2}=|x^2-y^2|<1/2.$$ which is impossible