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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.

Please help me ending this proof or show me a better way to solve this problem.

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    $\begingroup$ Do you perhaps mean $f, g: \mathbb R^n \longrightarrow \mathbb R$? If not, what does $f(x) \leq g(x)$ mean? $\endgroup$
    – paul blart math cop
    Mar 20, 2021 at 23:07
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    $\begingroup$ What does $f(x)\leqslant g(x)$ mean when $f(x),g(x)\in\Bbb R^n$? $\endgroup$
    – José Carlos Santos
    Mar 20, 2021 at 23:07
  • $\begingroup$ Yes, sry my bad, it's from R^n to R Corrected it $\endgroup$
    – José Pedro Ferreira
    Mar 20, 2021 at 23:08

1 Answer 1

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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 $x_0$.

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