description/proof of Fermat's theorem for partial derivatives at local maximum or minimum point
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of Fermat's theorem for partial derivatives at local maximum or minimum point: for any differentiable map from any open subset of any Euclidean \(C^\infty\) manifold into the \(1\)-dimensional Euclidean \(C^\infty\) manifold, if the map has the local maximum or the local minimum at any point, the partial derivatives at the point are \(0\).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{R}^{d_1}\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\)
\(\mathbb{R}\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\)
\(U\): \(\in \{\text{ the open subsets of } \mathbb{R}^{d_1}\}\)
\(f\): \(: U \to \mathbb{R}\), \(\in \{\text{ the differentiable maps }\}\)
\(u\): \(\in U\)
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Statements:
\(f \text{ has the local maximum or the local minimum at } u\)
\(\implies\)
\(\forall j \in \{1, ..., d_1\} ({\partial_j f}_u = 0)\)
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2: Note
The reverse does not necessarily hold, because \(p\) may be an inflection point: for example, let \(U = \mathbb{R}\), \(f = r^3\), and \(u = 0\), then, \({\partial_1 f}_u = 0\), but \(f\) does not have any local maximum or any local minimum at \(u\), because for each \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\), \(f (- \delta) \lt 0 \land 0 \lt f (\delta)\).
3: Proof
Whole Strategy: Step 1: take an open neighborhood of \(u\), \(U_u\), over which \(f\) has the maximum or the minimum at \(u\); Step 2: see that \((f (u^1, ..., u^j + \delta, ..., u^{d_1}) - f (u)) / \delta \le 0\) or \(0 \le (f (u^1, ..., u^j + \delta, ..., u^{d_1}) - f (u)) / \delta\) and \(0 \le (f (u) - f (u^1, ..., u^j - \delta, ..., u^{d_1})) / \delta\) or \((f (u) - f (u^1, ..., u^j - \delta, ..., u^{d_1})) / \delta \le 0\); Step 3: see that \(lim_{\delta \to 0} (f (u^1, ..., u^j + \delta, ..., u^{d_1}) - f (u)) / \delta \le 0\) or \(0 \le lim_{\delta \to 0} (f (u^1, ..., u^j + \delta, ..., u^{d_1}) - f (u)) / \delta\) and \(0 \le lim_{\delta \to 0} (f (u) - f (u^1, ..., u^j - \delta, ..., u^{d_1})) / \delta\) or \(lim_{\delta \to 0} (f (u) - f (u^1, ..., u^j - \delta, ..., u^{d_1})) / \delta \le 0\).
Step 1:
There is an open neighborhood of \(u\), \(U_u \subseteq U\), over which \(f\) has the maximum or the minimum at \(u\), by the definition of local maximum or local minimum.
Step 2:
There is a \(2 \delta\)-'open cube' around \(u\), \(C_{u, 2 \delta} \subseteq U_u\): refer to Note for the definition of topology induced by metric.
Let \(j \in \{1, ..., d_1\}\) be any.
When \(f\) has the local maximum at \(u\), \((f (u^1, ..., u^j + \delta, ..., u^{d_1}) - f (u)) / \delta \le 0\) and \(0 \le (f (u) - f (u^1, ..., u^j - \delta, ..., u^{d_1})) / \delta\), because \(f (u)\) is the maximum over \(C_{u, 2 \delta}\).
When \(f\) has the local minimum at \(u\), \(0 \le (f (u^1, ..., u^j + \delta, ..., u^{d_1}) - f (u)) / \delta\) and \((f (u) - f (u^1, ..., u^j - \delta, ..., u^{d_1})) / \delta \le 0\), because \(f (u)\) is the minimum over \(C_{u, 2 \delta}\).
Step 3:
Let us suppose that \(f\) has the local maximum at \(u\).
\({\partial_j f}_u = lim_{\delta \to 0} (f (u^1, ..., u^j + \delta, ..., u^{d_1}) - f (u)) / \delta \le 0\) and \(0 \le lim_{\delta \to 0} (f (u) - f (u^1, ..., u^j - \delta, ..., u^{d_1})) / \delta = {\partial_j f}_u\), which implies that \({\partial_j f}_u = 0\).
Let us suppose that \(f\) has the local minimum at \(u\).
\(0 \le lim_{\delta \to 0} (f (u^1, ..., u^j + \delta, ..., u^{d_1}) - f (u)) / \delta = {\partial_j f}_u\) and \({\partial_j f}_u = lim_{\delta \to 0} (f (u) - f (u^1, ..., u^j - \delta, ..., u^{d_1})) / \delta \le 0\), which implies that \({\partial_j f}_u = 0\).
