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Let $U$ be a convex open set in $\mathbb{R}^n$ and $f:U\longrightarrow \mathbb{R}$ such that $ \left| \large \frac{\partial f}{\partial x_i}(x)\right| \le M (\text{constant}) \; ,\forall x\in U$ and $\forall i=1\,,\cdots ,n.$ Prove that $|f(x)-f(y)|\le M||x-y||_1$ (1-norm) $\forall x,y \in U$

$\large{\frac{\partial f}{\partial x_i}}:$ Partial derivatives

$f:$ not necessarily differentiable

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For any $x,y \in U$,

$f(y) - f(x) = \sum_{j=1}^{n}f(y_1,...,y_j,x_{j+1},...,x_n) - f(y_1,...,y_{j-1},x_j,...,x_n)$

We apply the mean value theorem to get

$f(y_1,...,y_j,x_{j+1},...,x_n) - f(y_1,...,y_{j-1},x_j,...,x_n) = (y_j - x_j) \frac{\partial f}{\partial x_j}(c^{j})$

for some $c^{j}$. But we have $\bigg |\dfrac{\partial f}{\partial x_j}(c^{j})(y_j - x_j)\bigg| \le M|y_j - x_j|$. So

$|f(y) - f(x) | \le \sum_{j=1}^{n}M|y_j - x_j| = M\|y- x\|_1$

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    @leo Any idea as it is shown that: Let $[a,b]\subset U$(open set in $\mathbb R^n$) there is a path $\lambda:[0,1]\to U$ such that $\lambda(0)=a \, , \lambda(1)=b$ (polygonal path) whose linear segments are parallel to the coordinate axes.2012-07-07