Problem: If function $f$ defined on $\mathbb R$ is differentiable, its derivative is Lipschitz continuous and $f(0)=0$.
Show that exists constant $C$ such that $$ |f(a)-f(b)-f(c)+f(d)| \le C|a-b-c+d|+C|a-b|(|a-c|+|b-d|).$$ I considered this problem for several days, but I could not prove it. I hope you can give me some suggestions.
My question comes from an analysis conclusion used in the proof of the uniqueness of the solution in this article "Stochastic heat equation with rough dependence in space" written by Yaozhong Hu. The author says that the conclusion holds on the rectangular increments of $f$. At present, I only proved that the situation $a,b,c=a+h,d=b+h$, where I posted as an answer.
I only proved that the situation $a, b, c=a+h, d=b+h$.
Proof. If $c=a+h, d=b+h, a,b,h\in \mathbb R$. Then $$|f(a)-f(b)-f(c)+f(d)|=|f(a)-f(b)-f(a+h)+f(b+h)|,$$ Let $$g(t)=f(b+th)-f(a+th),$$ then $$|f(a)-f(b)-f(a+h)+f(b+h)|=|g(1)-g(0)|=|f'(b+\theta h)h-f'(a+\theta h)h|,$$ by using the Lipschitz continuous of $f'$, then $$|f(a)-f(b)-f(a+h)+f(b+h)|\le C|a-b||h|.$$