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Title: Uniform Convergence for Differentiable Functions
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Series: Real Analysis
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Chapter: Differentiable Functions
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YouTube-Title: Real Analysis 37 | Uniform Convergence for Differentiable Functions
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Bright video: https://youtu.be/PwkaPb_sCVQ
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Dark video: https://youtu.be/0gR7UWRn7m4
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Forum: Ask a question in Mattermost
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Quiz: Test your knowledge
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Dark-PDF: Download PDF version of the dark video
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Print-PDF: Download printable PDF version
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Exercise Download PDF sheets
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: ra37_sub_eng.srt missing
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Download bright video: Link on Vimeo
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Download dark video: Link on Vimeo
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Timestamps (n/a)
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Subtitle in English (n/a)
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Quiz Content
Q1: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of functions $f_n: [0,1] \rightarrow \mathbb{R}$. Which implication is correct?
A1: If it is pointwisely convergent, then also uniformly convergent.
A2: If it is uniformly convergent, then also pointwisely convergent.
Q2: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of continuous functions $f_n: [0,1] \rightarrow \mathbb{R}$. Which implication is correct?
A1: If it is pointwisely convergent to $f$, then $f$ is also continuous.
A2: If it is uniformly convergent to $f$, then $f$ is also continuous.
A3: If it is uniformly convergent to $f$, then $f$ is also differentiable.
Q3: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of differentiable functions $f_n: [0,1] \rightarrow \mathbb{R}$. If the sequence is uniformly convergent to $f: [0,1] \rightarrow \mathbb{R}$, is $f$ also differentiable?
A1: No, never!
A2: Yes, always!
A3: In general, it is not.
Q4: Let $(f_1, f_2, f_3, \ldots)$ be a sequence of differentiable functions $f_n: [0,1] \rightarrow \mathbb{R}$ that is pointwisely convergent to $f: [0,1] \rightarrow \mathbb{R}$. Which implication is correct?
A1: If $(f_1^\prime, f_2^\prime, f_3^\prime, \ldots)$ is uniformly convergent, then $f$ is also differentiable.
A2: If $(f_1^\prime, f_2^\prime, f_3^\prime, \ldots)$ is pointwisely convergent, then $f$ is also differentiable.
A3: If $(f_1^\prime, f_2^\prime, f_3^\prime, \ldots)$ is pointwisely convergent, then $f$ is not differentiable.
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Last update: 2025-01
