# Derivative problem in R^n space

I have been trying to solve the following problem:

Let $$f,g: \mathbb{R}^n \rightarrow \mathbb{R}$$ be differentiable functions and supose $$f(x_0)=g(x_0)$$, and $$f(x)\leq g(x)$$ in a certain region with center in $$x_0$$. Show that the Jacob Matrixes of $$f$$ and $$g$$ are equal at $$x_0$$.

Here's what I have so far:

$$\cdot$$First I started by defining the function $$h(x)=f(x)-g(x)$$. Therefore I have that $$h(x)\leq 0$$ and $$h(x_0)=0$$

$$\cdot$$ I write down the definition of the derivative at $$x_0$$ $$\lim_{t\to\ 0} \frac{h(x_0 + tv)-h(x_0)}{t}$$ $$\iff \lim_{t\to\ 0} \frac{h(x_0 + tv)-0}{t}$$

Suposing that $$t$$ approaches from the right (doesn't make diference), It's easy to conclude that if the limit is positive then $$h(x_0+tv) > 0$$, which is not possible base on the fact that $$h(x)<0$$.

What's left to conclude is that $$h(x_0+tv)$$ can't be negative, but I don't see where to go from here.

• Do you perhaps mean $f, g: \mathbb R^n \longrightarrow \mathbb R$? If not, what does $f(x) \leq g(x)$ mean? Mar 20, 2021 at 23:07
• What does $f(x)\leqslant g(x)$ mean when $f(x),g(x)\in\Bbb R^n$? Mar 20, 2021 at 23:07
Let $$h=g-f$$. Then $$h \geq 0$$ and $$h(x_0)=0$$ Hence, $$h$$ has a local miminimum at $$x_0$$. This implies that all the partial derivatives of $$h$$ at $$x_0$$ are $$0$$. So $$f$$ and $$g$$ have the same partial derivatives at $$x_0$$.