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Proof: Local extremum implies partial derivatives = 0

  1. Apr 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f\colon\mathbb{R}^m\to\mathbb{R}[/itex]. All partial derivatives of [itex]f[/itex] are defined at point [itex]P_0\colon = (x_1, x_2, ... , x_m)[/itex].
    If [itex]f[/itex] has local extremum at [itex]P_0[/itex] prove that [itex]\frac{\partial f}{\partial x_j} (P_0) = 0, j\in \{1, 2, ..., m\}[/itex]

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

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

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

    From 2) and 3) - according to Fermat's theorem [itex]f_{x_j}'(P_0) = g'(x_j) = 0 _{\blacksquare}[/itex]
     
    Last edited: Apr 3, 2015
  2. jcsd
  3. Apr 3, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    St is Qt?

    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.
     
  4. Apr 4, 2015 #3
    Oops, sleight of hand, Qt is St, yes.
     
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