Proof
(related to Proposition: Supremum Norm and Uniform Convergence)
In the following, $D$ is a set and $\mathbb F$ denotes either the field of real numbers or the field of complex numbers.
"$\Rightarrow$"
- Assume, the sequence of functions $f_n:D\to\mathbb F$ is uniformly convergent to a limit function $f:D\to\mathbb F.$
- This is equivalent to the statement: For any given $\epsilon > 0$ there is an index $N$ such that $|f_n(x)-f(x)| <\epsilon$ for all $n\ge N$ and all $x\in D.$
- By the definition of the supremum norm, this is equivalent to the statement: For any given $\epsilon > 0$ there is an index $N$ such that $||f_n-f||_\infty <\epsilon$ for all $n\ge N$ and all $x\in D.$
- By the definition of convergent sequence, this means $\lim_{n\to\infty}||f_n-f||_\infty=0.$
"$\Leftarrow$"
- Assume, $\lim_{n\to\infty}||f_n-f||_\infty=0.$
- Now, reverse the argumentation of the "$\Rightarrow$" block.
∎
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References
Bibliography
- Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
- Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition
