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Title: Arzelà–Ascoli Theorem
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Series: Functional Analysis
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Chapter: Compactness
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YouTube-Title: Functional Analysis 17 | Arzelà–Ascoli Theorem
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Forum: Ask a question in Mattermost
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Quiz: Test your knowledge
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Subtitle on GitHub: fa17_sub_eng.srt missing
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Timestamps
00:00 Introduction
00:32 Examples
04:05 Continuous functions
06:07 Equicontinuity
07:43 Examples (Equicontinuity)
11:25 Arzelà–Ascoli theorem
12:59 Credits
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Subtitle in English (n/a)
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Quiz Content
Q1: Let $(X,| \cdot |)$ be a normed space. What is always correct for a set $A \subseteq X$.
A1: If $A$ is compact, then $A$ is also closed.
A2: If $A$ is closed, then $A$ is also bounded.
A3: If $A$ is closed and bounded, then $A$ is also compact.
A4: If $A$ is unbounded, then $A$ is also compact.
A5: If $A$ is not compact, then $A$ is unbounded.
Q2: Let $C([0,1])$ be given with the supremum norm. What is correct for a uniformly equicontinuous subset $A \subseteq C([0,1])$.
A1: $$ \sup_{f \in A} | f(t) - f(s) | \xrightarrow{|t-s| \rightarrow 0} 0$$
A2: $$ \sup_{f \in A} | f(t) - f(s) | \geq 0 $$
A3: $ \sup_{f \in A} | f(t) - f(s) | \leq 0 $ for every $t,s \in [0,1]$.
Q3: Let $C([0,1])$ be given with the supremum norm. Is the closed ball $A = { f \in C([0,1]) \mid | f | \leq 1 }$ uniformly equicontinuous?
A1: No, for every $s \neq t$ we have that $\sup_{f \in A} | f(t) - f(s) | \geq 1$ no matter how close $s$ and $t$ are.
A2: Yes, for every $s \neq t$ we have that $\sup_{f \in A} | f(t) - f(s) | \leq 1$.
A3: One needs more information.
Q4: Let $C([0,1])$ be given with the supremum norm. What is always correct for a subset $A \subseteq C([0,1])$?
A1: If $A$ is uniformly equicontinuous, closed, and bounded, then $A$ is compact.
A2: If $A$ is uniformly equicontinuous and closed, then $A$ is compact.
A3: If $A$ is uniformly bounded and closed, then $A$ is compact.
A4: If $A$ is compact, then it is not uniformly equicontinuous.
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Last update: 2025-09
