Local extrema in multiple variables

**tttcomrader** · April 20th 2009, 06:56 PM

Suppose that the function $f: \mathbb {R} ^n \rightarrow \mathbb {R}$ has first order partial derivatives and that the point $x \in \mathbb {R} ^n$ is a local minimizer for f.
Prove that $\nabla f(x) = 0$ .

Proof so far.

Now, $\nabla f(x) = ( \frac { \partial f}{ \partial x_1 } , \frac { \partial f}{ \partial x_2 } , ... , \frac { \partial f}{ \partial x_n }$
$= ( \lim _{t \rightarrow 0 } \frac {f(x+te_1)-f(x)}{t}, \lim _{t \rightarrow 0 } \frac {f(x+te_2)-f(x)}{t}, ...,\lim _{t \rightarrow 0 } \frac {f(x+te_n)-f(x)}{t} )$

Now, x is a local min implies that if $dist(x,x+h)<r$ , we will have $f(x+h) \geq f(x)$

So how should I incorporate them together? Thanks.

**CaptainBlack** · April 21st 2009, 04:06 AM

Originally Posted by tttcomrader

Suppose that the function $f: \mathbb {R} ^n \rightarrow \mathbb {R}$ has first order partial derivatives and that the point $x \in \mathbb {R} ^n$ is a local minimizer for f.
Prove that $\nabla f(x) = 0$ .

As $\bf{x}$ is a local minimiser of $f$ , then $\bf{x}$ is a local minimum of the 1-D minimisation of $f({\bf{x}}+\lambda {\bf{e}}_i)$ where ${\bf{e}}_i$ is the unit vector in the direction of the $i$ -th axis. Hence:

$\left.\frac{d}{d\lambda}f({\bf{x}}+\lambda {\bf{e}}_i)\right|_{\lambda=0}=[\nabla f(x)]_i=0$

so:

$<br /> \nabla f(x)=0<br />$

CB

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