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A sequence of complex-valued functions converges uniformly if and only if it is uniformly Cauchy
Statement
Let be a set and . Then converges uniformly on if and only if it is uniformly Cauchy (Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary). This includes .
Facts & Assumptions
Given: A set and functions .
The complex plane is complete (The complex plane is complete, and convergence is equivalent to convergence of real and imaginary parts).
For complex numbers, and if and only if (Conjugation is an involutive real-field automorphism, , and modulus is definite, multiplicative, and subadditive).
Proof
If uniformly, then for choose with for every and ; for , [L2] gives , so is uniformly Cauchy.
Conversely, suppose is uniformly Cauchy. For each , the sequence is Cauchy and hence has a limit by [L1]; this defines , including the unique empty function when .
Given , choose such that for all and . Fixing and passing in the continuous modulus gives for every , so uniformly.
Depends on
- Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary
- The complex plane is complete, and convergence is equivalent to convergence of real and imaginary parts
- Conjugation is an involutive real-field automorphism, $z\overline z=|z|^2$, and modulus is definite, multiplicative, and subadditive
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 54 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- L. Ahlfors, Complex Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)