# Prove $f_{\alpha,\beta}$ is Lipschitz.

Let $$\alpha\in\mathbb{R}$$ and $$\beta >0$$ consider the function:

$$f_{\alpha,\beta}(x)=f(x)= \left\{ \begin{array}{lcc} x^\alpha\sin(x^\beta) & if & 0 < x \leq 1 \\ \\ 0& if & x =0 \\ \end{array} \right.$$

Prove $$f$$ is Lipchitz in $$[0,1]$$

My attempt:

Let $$x,y\in [0,1]$$ and $$\alpha + \beta >0$$ then

Note that $$f(x)-f(y)=(x^\alpha-y^\alpha)\sin x^\beta+y^\alpha(\sin x^\beta-\sin y^\beta)$$

$$\dfrac{|f(x)-f(y)|}{|x-y|}=\dfrac{|(x^\alpha-y^\alpha)\sin x^\beta+y^\alpha(\sin x^\beta-\sin y^\beta)|}{|x-y|}\leq\dfrac{|(x^\alpha-y^\alpha)||\sin x^\beta|+|y^\alpha||\sin x^\beta-\sin y^\beta|}{|x-y|}$$

$$\leq \dfrac{|x^\alpha-y^\alpha|+2|y^\alpha|}{|x-y|}$$

Here, I'm stuck. Can someone help me?

• I think the question should say for what values of $\alpha$ and $\beta$ is the function Lipschitz? Apr 1, 2020 at 23:14
• If $\alpha + \beta < 0$ does it not make the function singular at origin?
– r9m
Apr 1, 2020 at 23:14
• You should write the condition on $\alpha,\beta$ at the beginning.
– zhw.
Apr 2, 2020 at 4:14

$$f'(x)=\alpha x^{\alpha-1}\sin(x^\beta)+\beta x^{\alpha+\beta-1}\cos(x^\beta)$$. Therefore, \begin{align} |f'(x)|&\leq |\alpha x^{\alpha-1}\sin(x^\beta)| + |\beta x^{\alpha+\beta-1}\cos(x^\beta)| \\ &\leq |\alpha|+|\beta| \end{align} under the right assumptions on $$\alpha$$ and $$\beta$$. From my inequality, sufficient conditions should be clear. Given that $$\beta>0$$, you can also show that they are necessary.
• If $\alpha + \beta>0$ how can you guarantee that $|\alpha x^{\alpha-1}\sin(x^\beta)| + |\beta x^{\alpha+\beta-1}\cos(x^\beta)| \\ \leq |\alpha|+|\beta|$ Thanks for answer. Apr 1, 2020 at 23:16
• You need more. Hint: $\lim_{x\to 0} \frac{\sin(x)}{x}=1$ How does changing the powers cause blow up? That's for the first term in the derivative. For the second, it's even easier since $\cos(0^\beta)=1$ so blow up is determined by the $x^{\alpha+\beta-1}$ term. Apr 1, 2020 at 23:21