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.