16 October, 2013

Erdős' Anti-Calculus Proposition


Proposition

Erdös contends that in Complex Analysis, versus Single Variable Calculus, a function will not have a derivative of zero when it reaches a maximum value.

In Single Variable Calculus:

A function \(f(x)\), which is smooth (continuous and differentiable) on a open domain \(D\), will have a derivative of zero when it attains a maximum on \(D\).

However, for a complex function \(g(z)\), on a closed domain \(U\) with border \(B\), the maximum (Guaranteed by the Extreme Value Theorem), may not have a derivative of zero.

Proof

Let there exist a complex \(\epsilon\) such that \(|\epsilon| > 0\) and \(\delta\) such that \(|\delta| > 0\) where \(\epsilon\) and \(\delta\) are arbitrarily small.

Furthermore, let the complex function \(f(z)\) be analytic and non-constant on an open domain \(D\), bounded by a contour \(B\) along which \(f(z)\) is continuous and differentiable. Thus, \(f\) is continuous and differentiable on the closed region \(U = D \cup B\).

By the Maximum Modulus Theorem, the maximum of \(f\) occurs on \(B\) at \(z_0\).

Now, suppose that \(f'(z_0)\) is indeed zero. This would mean that for some \(\epsilon\), there is are points \(w = z_0 + \delta\) where \(\delta < \epsilon\) for which \(f(w)=f(z_0)\), since \(f(z)\) is continuous. \(z_0 + \delta\) is a maximum point for \(f(z)\) on \(D\). By the Maximum Modulus Theorem, this means that \(f\) must be constant on \(D\).

Since \(f(z)\) is non-constant on \(D\), this is a contradiction. \(#\)

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