# could any one help with this inequality?

Suppose $f(x,y)$ is a continuous and differentiable function defined on the region $D$, $A$ and $B$ are two inner points in region $D$, line segment $AB$ lies in the region $D$,

denote $K=\max\limits_{t\in[0,1]}\left\{\sqrt{\left(\frac{\partial f(M)}{\partial x}\right)^2+\left(\frac{\partial f(M)}{\partial y}\right)^2},\text{where}\ M=A+t(B-A)\right\}$

How to prove that $|f(A)-f(B)| \leq K\cdot|AB|$,where $|AB|$ means the length of $AB$

• Did you mean $M=A+t(B-A)$?
– user491874
Commented Dec 11, 2017 at 14:18
• yes ,sorry for that Commented Dec 11, 2017 at 14:23

Let $A=(a_1, a_2)$, $B=(b_1, b_2)$ and let $g(t)=f(a_1+t(b_1-a_1), a_2+t(b_2-a_2))$. The idea is that we are turning the 2-variable function $f$ on the segment $AB$ into a function of one variable $g$ on $[0,1]$. Note $g(0)=f(A)$, $g(1)=f(B)$.
Now, by differentiating on $t$ we get: $$g'(t)=\frac{\partial f}{\partial x}(M(t))\cdot(b_1-a_1) + \frac{\partial f}{\partial y}(M(t))\cdot(b_2-a_2)$$ where $M(t)=(a_1+t(b_1-a_1), a_2+t(b_2-a_2))$. This means: $$|f(B)-f(A)|=|g(1)-g(0)|=|g'(\xi)|$$ for some $\xi\in(0,1)$ (mean value theorem), but then $$|g'(\xi)|\le\sqrt{\left(\frac{\partial f}{\partial x}(M(\xi))\right)^2 + \left(\frac{\partial f}{\partial y}(M(\xi))\right)^2}\sqrt{(b_1-a_1)^2+(b_2-a_2)^2}$$ (Cauchy-Schwarz inequality), so your conclusion follows directly.