Proving $\log(\log(4+y^2)$ is Lipschitz with $L=\frac{1}{2\log(4)}$

Question

I need to prove $f(y)=\log(\log(4+y^2)$ is Lipschitz with $L=\frac{1}{2\log(4)}$. Calculating $\max f'(y)$ should do the trick but is a lot of work.

So I tried looking up other methods and stumbled upon this question: Prove that $\sqrt {f(x)}$ is Lipschitz with an answer I tried to understand, but failed untill now. Can I use the taylor series for my problem?

I tried expanding around zero, but I got nothing usefull I think: $ f(y) = \log(\log(4)) + \frac{y^2}{\log(256)} + \mathcal{O}^4.$ This looks nothing like the desired $L$. looking further, I found the Taylor expansion around zero of $f''$ is exacty $\frac{1}{2\log(4)}$, but that is the global maximum of $f''$ and not of $f'$. What is going on here?

My other option is of course finding the zeros of $f''$. Is there a smart way to do that?

Answer 1

You can first show that $\log(4+y^2)$ is Lipschitz using the derivative: the maximum value of $\dfrac{2y}{4+y^2}$ is $\dfrac 12$ and the minimum value is $-\dfrac 12$ so that $$|\log(4+y^2) - \log(4+z^2)| \le \frac 12 |y-z|.$$

Note that the range of $\log(4+y^2)$ is $[\log 4,\infty)$. The maximum value of $(\log x)' = \dfrac 1x$ on this set is $\dfrac 1{\log 4}$, so that $$x,w \in [\log 4,\infty) \implies |\log x - \log w| \le \frac{1}{\log 4}|x-w|.$$

Thus $$|\log(\log(4+y^2)) - \log(\log(4+z^2))| \le \frac{1}{\log 4} |\log(4+y^2) - \log(4+z^2)| \le \frac{1}{2 \log 4} |y-z|.$$

Answer 2

I would write $$\frac{\log(4+y_2^2))-\log(\log(4+y_1^2))}{y_2-y_1}=\frac{2\xi}{(\xi^2+4)\log(\xi^2+4)}$$ Now use that $$\xi^2+4\geq 4\xi$$ and $$\log(\xi^2+4)\geq \log(4)$$