# Local extrema in multiple variables

• Apr 20th 2009, 06:56 PM
Local extrema in multiple variables
Suppose that the function $\displaystyle f: \mathbb {R} ^n \rightarrow \mathbb {R}$ has first order partial derivatives and that the point $\displaystyle x \in \mathbb {R} ^n$ is a local minimizer for f.
Prove that $\displaystyle \nabla f(x) = 0$.

Proof so far.

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

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

So how should I incorporate them together? Thanks.
• Apr 21st 2009, 04:06 AM
CaptainBlack
Quote:

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

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

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

so:

$\displaystyle \nabla f(x)=0$

CB