Suppose that $D\subset\mathbb{R}^n$ is a domain(connected open set),$f$ is differentiable everywhere in $D$,and there exists $L>0$ such that $$ \lVert Jf\left( x \right) -Jf\left( y \right) \rVert \le L\lVert x-y \rVert ,\forall x,y\in D. $$ Prove: $$ \lVert f\left( y \right) -f\left( x \right) -Jf\left( x \right) \left( y-x \right) \rVert \le \frac{1}{2}L\lVert x-y \rVert ^2,\forall x,y\in D. $$
Note:$Jf(x)$ represents the Jacobian determinant of $f$.$\lVert \cdot\rVert$ is the usual norm.
The title doesn't say what the domain of $f$ is.And I assume $f:D\to \mathbb{R}^m$.
I want to think about the one-dimensional case first,that is to say I want to prove the following problem:
$f$ is a differentiable function on $I=(a,b)$,and there exists $M>0$ s.t. $$ \left| f'\left( y \right) -f'\left( x \right) \right|\le M\left| y-x \right|. $$ Then $$ \left| f\left( y \right) -f\left( x \right) -f'\left( x \right) \left( y-x \right) \right|\le \frac{1}{2}M\left| y-x \right|^2. $$
It's easy to prove by Newton-Leibniz Formula.However,it seems hard to generalize it to higher dimensional condition.
We want $D$ to be convex, such that the following is well-defined: Set $g:[0, 1] \rightarrow \mathbb{R}^m$, $g(s) := f(sy+(1-s)x)$ for fixed $x, y \in D$. Note that $$ g'(s) = Jf(sy+(1-s)x)(y-x) $$ by the chain rule. Therefore, by the fundamental theorem of calculus (the integral is to be understood componentwise): $$ \lVert f(y) - f(x) - Jf(x)(y-x) \rVert = \left \lVert g(1) - g(0) - \int^1_0 Jf(x)(y-x)~\mathrm{d}s \right \rVert = \left \lVert \int^1_0 g'(s)~\mathrm{d}s - \int^1_0 Jf(x)(y-x)~\mathrm{d}s \right \rVert \underbrace{=}_{\text{Linearity of Integral and $Jf$}} \left \lVert \int^1_0 (Jf(sy + (1-s)x) - Jf(x))(y-x)~\mathrm{d}s \right \rVert $$ We now use the triangle-inequality and the fact that the operator norm is subordinate: $$ \left \lVert \int^1_0 (Jf(sy + (1-s)x) - Jf(x))(y-x)~\mathrm{d}s \right \rVert \leq \int^1_0 \lVert Jf(sy+(1-s)x) - Jf(x)\rVert \lVert y-x \rVert~\mathrm{d}s \underbrace{\leq}_{$L$-\text{continuity}}L\int^1_0 s\lVert x - y\rVert^2~\mathrm{d}s = \frac{L}{2} \lVert y - x\rVert^2 $$