Consider a function $f:\mathbb{R}^n \to \mathbb{R}$ and an arbitrary vector $u \in \mathbb{R}^n$. Is $$f(x) = \log(1+e^{u^Tx})$$ Lipschitz? In other words, can we prove - $$\exists L,\ ||y-x|| \leq L |f(y)-f(x)| \ \forall x,y$$
I have been staring at this problem for an hour but do not know how to solve it! In particular, I don't know what to do after writing $\log(\frac{1+e^{u^Ty}}{1+e^{u^Tx}})$. Any help/hint is appreciated.
One way to do this is to check that the derivative $Df$ is bounded in $\mathbb{R}^n$. Notice that $Df(x)=\dfrac{e^{u^Tx}u}{1+e^{u^Tx}}$, so that $\| Df(x)\|\leq \| u\|$, where $\| \cdot\|$ denotes the Euclidean norm in $\mathbb{R}^n$.