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Problem: Given a function $b:\mathbb{R}\rightarrow\mathbb{R}$ defined by $b(x)=3x^{\frac{1}{3}}$. Prove that this function is Hölder continuous but not Lipschitz continuous.

My attempt: In order to prove Hölder continuity we have to show that there exist some $C,\alpha>0$ such that the following inequality holds for any $x,y\in\mathbb{R}$ $$|3x^{\frac{1}{3}}-3y^{\frac{1}{3}}|\leq C|x-y|^\alpha$$ I tried to set $\alpha=\frac{1}{3}$, and prove that there exists some $C$ such that we have that $$|3x^{\frac{1}{3}}|\leq C|x-y|^{\frac{1}{3}}+|3y^{\frac{1}{3}}|$$ However, I keep getting stuck with the fact that $x,y$ do not necessarily have to be positive. This is kind of messing up my whole attempt because I'm not sure how to proceed from here.

My plan was to first attempt to prove Hölder continuity and then Lipschitz continuity. Since I wasn't able to prove the first thing, I didn't yet start on proving we do not have Lipschitz continuity.

Any help is appreciated!

Charlie Shuffler
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The factor $3$ is not relevant, so let us consider $f(x) = x^{1/3}$ instead. ($f$ is defined on $\Bbb R$ as the inverse function of $y \mapsto y^3$).

$f$ is indeed Hölder continuous with exponent $\alpha = \frac 13$, as you suspected. In order to prove an estimate $$ |f(x) - f(y) | \le C |x-y|^{1/3} $$ for all $x, y \in \Bbb R$ we distinguish two cases:

Case 1: $x, y$ have the same sign. Without loss of generality we can assume that $x, y \ge 0$. In this case, $$ |f(x) - f(y) | \le |x-y|^{1/3} $$ as shown in proving that $f(x) = x^s$ is holder continuous with holder exponent s.

Case 2: $x, y$ have opposite sign. Without loss of generality $x < 0 < y$. Then $$ |f(x) - f(y) | \le C |x-y|^{1/3} \\ \iff |x|^{1/3} + y^{1/3} \le C (|x| + y)^{1/3} \\ \iff \left( \frac{|x|}{|x| + y}\right)^{1/3} + \left( \frac{y}{|x| + y}\right)^{1/3} \le C $$ Now both fractions on the left-hand side are less than one, so that the estimate is satisfied with $C=2$: $$ |f(x) - f(y) | \le 2 |x-y|^{1/3} $$

This completes the proof of the Hölder continuity.

One can also determine the best possible constant $C$ by computing the maximum of $$ u^{1/3} + (1-u)^{1/3} \text{ for } 0 < u < 1 \, , $$ which turns out to be $C = 2^{2/3}$.

For the (non) Lipschitz continuity, consider $$ \frac{|f(x)-f(0)|}{|x-0|} $$ for $x \ne 0$, $x \to 0$.

Martin R
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  • Very concise and clear, thanks alot! – Charlie Shuffler May 04 '19 at 17:30
  • I have one more question though. How exactly can you be assume in case 1 that $x,y\geq 0$ without loss of generality? Isn't it a separate case when $x,y\leq 0$? Because in that case the solution on the link you sent does not work, right? – Charlie Shuffler May 05 '19 at 19:09
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    @S.Crim: We can assume that because $f(-x) = -f(x)$. If $x, y \le 0$ then $|f(x) -f(y)| = |f(-x) - f(-y)| \le |(-x) - (-y)| = |x-y|$. – Martin R May 05 '19 at 19:17