I need to prove that the function $f(t,x):(a,b)\times \mathbb{R}\to \mathbb{R}$ $$ f(t,x)=(t^2-x) \frac{\log(1+x^2)}{1+x^2}$$
is locally lipschitz with respect to $x$.
I think that is globally Lipschitz, but I'm not sure if this is true because I'm not convinced with my proof. Of course, the key is that $\log(y)/y \leq 1$ when $y>1$. But I have
$$\|f(t,x)-f(t,y) \|=\left\| (t^2-x) \frac{\log(1+x^2)}{1+x^2} - (t^2-y) \frac{\log(1+y^2)}{1+y^2} \right\| $$
and I'm not sure if I can simplify this by removing the fraction. I 've tried
$$ \left\| (t^2-x) \frac{\log(1+x^2)}{1+x^2} - (t^2-y) \frac{\log(1+y^2)}{1+y^2} \right\|\leq \left\| (t^2-x)- (t^2-y) \frac{\log(1+y^2)}{1+y^2} \right\|= \left\| (t^2-y) \frac{\log(1+y^2)}{1+y^2} -(t^2-x)\right\|\leq \left\| (t^2-y) -(t^2-x)\right\|=\|x-y \|$$
But I think that this is not true if $t^2-x<0$, and It's suspicious that the Lipschitz constant is $L=1$
Any idea?
Thanks!
The partial derivative $\frac{\partial f}{\partial x}$ is continuous, hence, by the Mean value theorem, $f$ is locally Lipschitz.