Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

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

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

share|improve this answer
    
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.