Consider the picture here:
The function taken into account is $f(x) = e^x$ and also the halfplane $x\ge 0$ is highlighted in blue. The point $(0, 1)$ is a minimum for the function in the halfplane $x\ge 0$. During calc I class, it was said that this condition can be expressed in this case availing of first derivative, i.e. saying that $$\frac{d}{ds} f(s)\Big|_{s=0}\ge 0. $$
More or less I agree with that condition which means to me, roughly speaking, that the function is increasing on the RHS of the point. However, I am trying to understand how this condition should be redrafted in the case $f$ is a function in several variables (e.g. $f=f(x, y)$) for which the minimum is achieved on the boundary.
Intuitively I think it should be something related to its first partial derivatives, but it sounds a bit weird to me that we need to ask something that $$\frac{\partial f}{\partial x} (x, y)\Big|_{s=x_0}\ge 0,\qquad \frac{\partial f}{\partial y} (x, y)\Big|_{s=x_0}\ge 0, $$
where $(x_0, y_0)$ is the point where the minimum is achieved.
Do someone have an idea about the conditions to require in this more challenging case? I am very intrigued by this more challenging situation.
Thank you in advance.
If $f: \Bbb R^2 \to \Bbb R$ is differentiable and attains its minimum on the right halfplane at some point $(0, y_0)$: $$ f(0, y_0) = \min \{ f(x, y) \mid x \ge 0, y \in \Bbb R\} $$ then
$x \mapsto g(x) := f(x, y_0)$ satisfies $g(x) \ge g(0)$ for $x \ge 0$, so that $$ g'(0) = \frac{\partial f}{\partial x} (x, y)\Big|_{(0, y_0)} \ge 0 $$ as you observed for the one-dimensional case, and
$y \mapsto h(y) := f(0, y)$ satisfies $h(y) \ge h(y_0)$ for all $y \in \Bbb R$, so that $$ h'(y_0) = \frac{\partial f}{\partial x} (x, y)\Big|_{(0, y_0)} = 0 $$ because $h$ has a minimum at $y = y_0$.
As in the one-dimensional case, this conditions are necessary, but not sufficient for a minimum on the boundary.