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[parent] proof of calculus theorem used in the Lagrange method (Proof)

Let $ f(\mathbf{x})$ and $ g_i(\mathbf{x}), i=0,{\ldots},m$ be differentiable scalar functions; $ \mathbf{x} \in R^n$.

We will find local extremes of the function $ f(\mathbf{x})$ where $ \nabla f=0$. This can be proved by contradiction:

$\displaystyle \nabla f \neq 0 $
$\displaystyle \Leftrightarrow \exists \epsilon_0 > 0, \forall \epsilon; 0<\epsi... ...mathbf{x}-\epsilon \nabla f) < f(\mathbf{x}) < f(\mathbf{x+\epsilon \nabla f}) $
but then $ f(\mathbf{x})$ is not a local extreme.

Now we put up some conditions, such that we should find the $ \mathbf{x} \in S \subset R^n$ that gives a local extreme of $ f$. Let $ S=\bigcap_{i=1}^m S_i$, and let $ S_i$ be defined so that $ g_i(\mathbf{x})=0 \forall \mathbf{x} \in S_i$.

Any vector $ \mathbf{x} \in R^n$ can have one component perpendicular to the subset $ S_i$ (for visualization, think $ n=3$ and let $ S_i$ be a flat surface). $ \nabla g_i$ will be perpendicular to $ S_i$, because:

$\displaystyle \exists \epsilon_0>0, \forall \epsilon; 0<\epsilon<\epsilon_0: g_... ...f{x}-\epsilon \nabla g_i)<g_i(\mathbf{x})< g_i(\mathbf{x}+\epsilon \nabla g_i) $
But $ g_i(\mathbf{x})=0$, so any vector $ \mathbf{x}+\epsilon \nabla g_i$ must be outside $ S_i$, and also outside $ S$. (todo: I have proved that there might exist a component perpendicular to each subset $ S_i$, but not that there exists only one; this should be done)

By the argument above, $ \nabla f$ must be zero - but now we can ignore all components of $ \nabla f$ perpendicular to $ S$. (todo: this should be expressed more formally and proved)

So we will have a local extreme within $ S_i$ if there exists a $ \lambda_i$ such that

$\displaystyle \nabla f = \lambda_i \nabla g_i $

We will have local extreme(s) within $ S$ where there exists a set $ \lambda_i, i=1,{\ldots},m$ such that

$\displaystyle \nabla f = \sum \lambda_i \nabla g_i $



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Cross-references: argument, surface, flat, subset, perpendicular, component, vector, contradiction, functions, scalar, differentiable

This is version 3 of proof of calculus theorem used in the Lagrange method, born on 2003-03-04, modified 2004-03-23.
Object id is 4073, canonical name is ProofOfCalculusTheoremUsedInTheLagrangeMethod.
Accessed 4193 times total.

Classification:
AMS MSC: 15A18 (Linear and multilinear algebra; matrix theory :: Eigenvalues, singular values, and eigenvectors)
 15A42 (Linear and multilinear algebra; matrix theory :: Inequalities involving eigenvalues and eigenvectors)
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