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The theorem says "If the function $f:\Bbb{R}^n→\Bbb{R}$ has a local extremum at $α∈\Bbb{R}^n$, then $α$ is a critical point".
For $n=1$, its quite simple by using the definition that WLOG assuming $f$ has local maximum at $x=\alpha$, $f(x)\le f(\alpha)$ whenever $|x-\alpha|<\delta$. So we can say that
$\dfrac{f(\alpha+h)-f(\alpha)}{h}\le 0$ for all $0<h<\delta$ and from this, it can be shown that $f'(\alpha)\le0$
and similarly considering $-\delta<h<0$, we get $f'(\alpha)\ge0$ and from this we can conclude that $f'(\alpha)=0$ which completes the theorem.

But how do I choose the neighbourhood when it comes to $n>1$? I can anticipate that $|x-\alpha|<\delta$ will be changed to $||x-\alpha||<\delta$ but it is also true that $\dfrac{df(\alpha)}{dx}$ will not suit in that case to the definition of critical point.

Ted Shifrin
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Manjoy Das
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3 Answers3

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If we suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is differentiable and has a local extremum in $\mathbf{v} = (v_1, \ldots, v_n)$, then for each $1 \leq i \leq n$, you can define $g(t) = f(v_1, \ldots, v_{i-1}, t, v_{i+1}, \ldots, v_n)$ and use the same argument as functions of one variable to achieve that $g'(v_i) = f_{x_i}(\mathbf{v}) = 0$. Because all partial derivatives vanish, so $\mathbf{v}$ is a critical point for $f$.

on1921379
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  • can you please elaborate "then for each $1≤i≤n$, you can define $g(t)=f(v_1,…,v_{i−1},t,v_{i+1},…,v_n)$"? perhaps with an example. – Manjoy Das Aug 09 '22 at 20:32
  • @ManjoyDas Let $f(x, y) = x^2 + 2y^2$. $f$ has a minimum at $(0, 0)$, so we define $g_1(t) = f(t, 0) = t^2$ and $g_2(t) = f(0, t) = 2t^2$. Because $g_1$ and $g_2$ have a minimum at $0$, so $g_1'(0) = g_2'(0) = 0$, but these values equal the partial derivatives of $f$ respect to $x$ and $y$. Therefore $f_x(0, 0) = f_y(0, 0) = 0$. – on1921379 Aug 09 '22 at 21:11
  • the trick is really sweet – Manjoy Das Aug 09 '22 at 21:51
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Note: We need to assume that $f: \mathbb{R}^n \to \mathbb{R}$ is differentiable at a local extremum $\mathbf{a} \in \mathbb{R}^n$ in order to have $$\lim_{h \to 0} \frac{f(\mathbf{a} + h\mathbf{u} ) - f(\mathbf{a} )}{h}=\lim_{h \to 0^+} \frac{f(\mathbf{a} + h\mathbf{u} ) - f(\mathbf{a} )}{h} \, ,$$ and $$\lim_{h \to 0} \frac{f(\mathbf{a} + h\mathbf{u} ) - f(\mathbf{a} )}{h}=\lim_{h \to 0^-} \frac{f(\mathbf{a} + h\mathbf{u} ) - f(\mathbf{a} )}{h}\, .$$

Let $\{\mathbf{e}_1, \ldots, \mathbf{e}_n\}$ be the standard basis of $\mathbb{R}^n$.

Suppose $f: \mathbb{R}^n \to \mathbb{R}$ has a local extremum and is differentiable at $\mathbf{a} \in \mathbb{R}^n$.

WLOG, assume $f$ has a local maximum at $\mathbf{a}$. This means there is $\delta>0$ so that $f(\mathbf{x})\leq f(\mathbf{a})$ whenever $\|\mathbf{x}-\mathbf{a}\|<\delta$. So for $j=1, \ldots, n$ we have

$$f(\mathbf{a}+h\mathbf{e}_j)-f(\mathbf{a})\leq 0 \text{ whenever } \|(\mathbf{a}+h\mathbf{e}_j)-\mathbf{a}\|=|h|<\delta.$$

Thus if $h \in (0, \delta)$, then for $j=1, \ldots, n$ we have $$\frac{f(\mathbf{a}+h\mathbf{e}_j)-f(\mathbf{a})}{h}\leq 0$$ which means $\lim_{h \to 0^+} \frac{f(\mathbf{a} + h\mathbf{e}_j ) - f(\mathbf{a} )}{h} \leq 0. $ Since $\frac{\partial }{\partial x_j}f(\mathbf{a})$ exists $(j=1, \ldots, n)$, we have $$\frac{\partial }{\partial x_j}f(\mathbf{a})=\lim_{h \to 0^+} \frac{f(\mathbf{a} + h\mathbf{e}_j ) - f(\mathbf{a} )}{h} \leq 0.$$

On the other hand, if $h \in (-\delta,0)$, then for $j=1, \ldots, n$ we have $$\frac{f(\mathbf{a}+h\mathbf{e}_j)-f(\mathbf{a})}{h}\geq 0$$ which means $\lim_{h \to 0^-} \frac{f(\mathbf{a} + h\mathbf{e}_j ) - f(\mathbf{a} )}{h} \geq 0. $

Since $\frac{\partial }{\partial x_j}f(\mathbf{a})$ exists $(j=1, \ldots, n)$, we have $$\frac{\partial }{\partial x_j}f(\mathbf{a})=\lim_{h \to 0^-} \frac{f(\mathbf{a} + h\mathbf{e}_j ) - f(\mathbf{a} )}{h} \geq 0.$$

We combine to conclude that $\frac{\partial }{\partial x_j}f(\mathbf{a})=0$, for each of $j=1, \ldots, n$.

M A Pelto
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    I am just wondering why I need $a+he_j$, precisely $e_j$ to denote the point close to $a$. – Manjoy Das Aug 09 '22 at 21:54
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    $\mathbf{a}+h \mathbf{e}j=(a_1, \ldots, a{j-1}, a_j+h, a_{j+1}, \ldots, a_n)$, and we define the partial derivative of $f$ at $\mathbf{a}$ with respect to the $j^{\text{th}}$ variable $x_j$ by $\frac{\partial }{\partial x_j}f(\mathbf{a})=\lim_{h \to 0} \frac{f(\mathbf{a} + h\mathbf{e}_j ) - f(\mathbf{a} )}{h}$ – M A Pelto Aug 09 '22 at 22:15
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Here is how I would have proceeded with the proof.

Let $\delta>0$ so that $f(\alpha)\geq f(v)$ for all $v\in \mathbb{R}^n$ with $\|v-\alpha\|<\delta$.

Choose $x\in \{v\in \mathbb{R}^n:\|v-\alpha\|<\delta\}$ arbitrarily.

Define the function $l:(-1,1)\mapsto\mathbb{R}$ by $l(t)=f\left(\alpha+(x-\alpha)t\right)$

Notice $l(0)$ is a global max of $l(t)$. Since $0$ belongs to the interior of $(-1,1)$ we must have $l'(0)=0$.

However, $l'(0)=(\vec{\nabla} f)(\alpha)\cdot (x-a)$, and because $x$ is chosen arbitraily, we must have $(\vec{\nabla} f)(\alpha)=0$ i.e. $\alpha$ is a critical point of $f$.

Matthew H.
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