Let $n\ge 2$ be some positive integer. Let $f:\mathbf{R}^n\times\mathbf{R}^n\to\mathbf{R}$ be defined as $f(x,y)=x\cdot y$ where $x\cdot y$ denotes the dot product of two vectors. Is $f$ Lipschitz?
This question is a generalization of the 1D case I asked earlier. The dot product in the higher dimension case is now nontrivial, i.e., not the scalar multiplication.
After the 1D case was done, I tried to mimic the analysis there to show that $f$ is not Lipschitz. But here is an unsuccessful attempt.
Consider the vectors $x,u,v\in\mathbf{R}^n$. Then by the Cauchy-Schwarz inequality, $$ |f(x,u)-f(x,v)|=|x\cdot(u-v)|\le |x||u-v|_{\mathbf{R}^n}\tag{1} $$ On the other hand, $$ |(x,u)-(x,v)|_{\mathbf{R}^n\times \mathbf{R}^n} = |(0,u-v)|_{\mathbf{R}^n\times \mathbf{R}^n} = |u-v|_{\mathbf{R}^n}\tag{2} $$
The trouble to use the 1D case analysis is that one does not have the "equality" in (1) and thus cannot directly use (1) and (2) to reach a contradiction assuming a Lipschitz constant $K$.
I find that if I "reduce" the dimensions more, I can recover the equal sign in (1), and thus the 1D argument works again.
Let $x,u,v$ be three real numbers and $e_1=(1,0,\cdots,0)$ the unit vector in $\mathbf{R}^n$.
Now consider $$ |f(xe_1,ue_1)-f(xe_1,ve_1)|=|x||u-v|=|x|\cdot \|ue_1-ve_1\|_{\mathbf{R}^n}\tag{1} $$ and $$ \|(xe_1,ue_1)-(xe_1,ve_1)\|_{\mathbf{R}^n\times \mathbf{R}^n} =\|ue_1 -ve_1\|_{\mathbf{R}^n}\tag{2} $$
Comparing (1) and (2), one can see that $f$ cannot be Lipschitz, since otherwise, there exists $K>0$ such that $$ |x|\cdot \|ue_1-ve_1\|_{\mathbf{R}^n} =|f(xe_1,ue_1)-f(xe_1,ve_1)| \le K\|ue_1 -ve_1\|_{\mathbf{R}^n}\tag{3} $$ for all $x,u,v\in\mathbb{R}$, which is impossible.
If one looks at (3) further, one can fix $u=1$ and $v=0$ in the whole argument to simplify the contradiction to the estimate $$ |x|\le K,\quad x\in\mathbf{R}\;. $$