Proof: Local extremum implies partial derivatives = 0

[nuuskur]

Homework Statement

Let $f\colon\mathbb{R}^m\to\mathbb{R}$. All partial derivatives of $f$ are defined at point $P_0\colon = (x_1, x_2, ... , x_m)$.
If $f$ has local extremum at $P_0$ prove that $\frac{\partial f}{\partial x_j} (P_0) = 0, j\in \{1, 2, ..., m\}$

Homework Equations

Fermat's theorem:
Let $f\colon\mathbb{R}^m\to\mathbb{R}$. If $f$ is differentiable and has local extremum at point $P_0$ then $\nabla f(P_0) = \overrightarrow{0}$

The Attempt at a Solution

Assume $f$ has local minimum at point $P_0$, then there exists $\varepsilon > 0$ such that $f(P)\geq f(P_0)$ for every $P\in B(P_0,\varepsilon)$.
Let $j\in \{1,2, ..., m\}$. Observe the function:
$$g(t) = f(x_1, x_2, ... x_{j-1}, t, x_{j+1}, ... , x_m)$$
Note that:
1) $g$ is defined within $t\in (x_j -\varepsilon, x_j +\varepsilon)$
2) the function $g$ has local minimum at the point $t = x_j$
3) $g$ is differentiable at $t = x_j$, also $g'(x_j) = f_{x_j}'(P_0)$

Define for every $t\in\mathbb{R}\ \ \ S_t\colon = (x_1, ...,x_{j-1}, t, x_{j+1}, ..., x_m)$ then for every $t\in (x_j -\varepsilon, x_j +\varepsilon)$ it follows that $d(Q_t, P_0) = |t - x_j| < \varepsilon$, therefore $Q_t\in D\subset\mathbb{R}^m$ i.e $g$ is defined at point $t$ and for every $t$:
$$g(t) = f(Q_t)\geq f(P_0) = g(x_j)$$ which means $g$ has local minimum at the point $x_j$.

From 2) and 3) - according to Fermat's theorem $f_{x_j}'(P_0) = g'(x_j) = 0 _{\blacksquare}$

 

[mfb]

S_t is Q_t?

Fermat's theorem, as you write it here, covers $\mathbb{R}^m\to\mathbb{R}$ already, if you can use ít where is the point of reducing it to the 1-dimensional case? But then the whole problem looks trivial.

 

[nuuskur]

Oops, sleight of hand, Qt is St, yes.