Let the function $f:\mathbb R^n \rightarrow \mathbb R^n$ be Lipschitz with constant $L$. In a paper, they say the condition that $f:\mathbb R^n \rightarrow \mathbb R^n$ is Lipschitz is equivalent to the condition that there exists a $B < \infty$ such that
\begin{align} (i)&: \quad \Vert f(x) \Vert \leq B(1+\Vert x\Vert) \\ (ii)&: \quad (f(x)-f(y))^\top(x-y) \leq B\Vert x-y\Vert^2 \end{align}
holds. Condition (i) is clear for me: Lipschitz implies that the function grows at most linearly with a finite $L$, which implies (i). However, condition (ii) I do not see how it is fulfilled only requiring that $f$ is Lipschitz. In fact, condition (ii) has for me some similarity with the (reversed) condition for strongly convex functions, implying in this case that the primitive function of $f$ is less than $B \Vert x \Vert^2$, i.e. less than a quadratic function. This kind of gives the idea that $f$ has grow at most linearly. However, that is not really a proof.
Does someone know why condition (ii) is fulfilled when $f$ is Lipschitz?
Thanks in advance!
I would say that, thanks to Cauchy-Schwarz inequality, \begin{align} (f(x)-f(y))^{T}(x-y) &\leq \|f(x)-f(y)\|\,\|x-y\|\\ &\leq L\|x-y\|\,\|x-y\|=L\|x-y\|^2 \end{align}