# Let $U$ be a convex open set in $\mathbb{R}^n$ and $f:U\longrightarrow \mathbb{R}$ such that $\left| \frac{\partial f}{\partial x_i}(x)\right| \le M$

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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Note that the points $(y_1,...,y_j,x_{j+1},...,x_n),(y_1,...,y_{j-1},x_j,...,x_n)$ not necessarily belong to $U$. – felipeuni Jul 7 '12 at 3:09
@felipeuni They belong to $U$ because $U$ is open and convex so $U$ is open and connected and then $U$ path connected. For example in $\Bbb R^2$ if you have $(x_1,x_2)$ and $(y_1,y_2)$ in $U$ then $(x_1,x_2)--(x_1,y_2) --(y_1,y_2)$ is path between $(x_1,x_2)$ and $(y_1,y_2)$ so $(x_1,y_2)$ belongs to $U$ – leo Jul 7 '12 at 6:16
@leo But $(x_1,y_2)$ not necessarily belong to $U$.If the set $U$ is a ball(maximum-norm) then $(x_1,y_2)$ belong to $U$. – felipeuni Jul 7 '12 at 7:24
@felipeuni see this. That show the possible path in $\Bbb R^3$. – leo Jul 7 '12 at 16:27
@felipeuni If $U$ is open and connected the points involved in the path always belongs to $U$. See here and here for further explanation. – leo Jul 7 '12 at 16:32