# Is a function with non-negative first partial derivatives increasing?

Suppose $f(x_1,x_2,\dots,x_n)$ is a multivariable function $f:\mathbb{R}^n\rightarrow \mathbb{R}$. Suppose that for all partial derivatives, $1\le i \le n$,

$$\frac{\partial f}{\partial x_i}(q_1,q_2,\ldots,q_n) \ge 0$$

for all $q_i \ge r_i$.

Is f a non-decreasing function in the set of all points $q_i \ge r_i$ for all $1\le i \le n$?

Edit: By "non-decreasing", I mean that $f(q_1,q_2,\ldots,q_n) \ge f(r_1,r_2,\ldots,r_n)$ if $q_i \ge r_i$ for all i.

Also, is the converse true? Does a function that is 'non-decreasing', by my definition, also have all first partial derivatives positive?

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You only can speak of an increasing function if it is between ordered sets. Your function is "increasing in each variable", i.e., when the other variables are kept fixed. – Christian Blatter Jan 25 '11 at 16:29
@Christian: what if you consider the partial order on $\mathbb{R}^n$ given by $q \geq r$ iff $q_i \geq r_i$ for all $i$? :) – Willie Wong Jan 25 '11 at 19:19

If all partial derivatives are non-negative, then yes: $f(q_1,q_2,\ldots,q_n) \ge f(r_1,r_2,\ldots,r_n)$ if $q_i \ge r_i$ for all i. And the converse is true as well.
For the first question, take a walk from $r$ to $q$, one coordinate at a time: $(r_1,r_2,\ldots,r_n)$ to $(q_1,r_2,\ldots,r_n)$ to $\ldots$ to $(q_1,q_2,\ldots,q_n)$. None of these steps can decrease the value of $f$.