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Title: Uniform Boundedness Principle / Banach–Steinhaus Theorem
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Series: Functional Analysis
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Chapter: Core Results in Functional Analysis
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YouTube-Title: Functional Analysis 24 | Uniform Boundedness Principle / Banach–Steinhaus 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: fa24_sub_eng.srt missing
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Timestamps
00:00 Introduction
01:19 Theorem
03:20 Proposition
04:33 Proof
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Subtitle in English (n/a)
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Quiz Content
Q1: Let $X,Y$ be two Banach spaces and $\mathcal{B}(X,Y)$ the space of bounded linear operators from $X$ to $Y$. Is $\mathcal{B}(X,Y)$ also a Banach space?
A1: Yes, it is.
A2: No, there are finite-dimensional counterexamples.
A3: No, it’s only a Banach space for finite-dimensional spaces $X$ and $Y$.
A4: No, never.
Q2: Let $X,Y$ be two Banach spaces and $\mathcal{B}(X,Y)$ the space of bounded linear operators from $X$ to $Y$. What does it mean that a subset $\mathcal{M} \subseteq \mathcal{B}(X,Y)$ is pointwisely bounded?
A1: For all $x \in X$, there is a $C > 0$ such that for all $T \in \mathcal{M}$ we have $ | T x |_Y \leq C$.
A2: For all $C > 0$, there is a $x \in X$ such that for all $T \in \mathcal{M}$ we have $ | T x |_Y \leq C$.
A3: For all $x \in X$, there is a $T \in \mathcal{M}$ such that for all $C > 0$ we have $ | T x |_Y \leq C$.
A4: For all $T \in \mathcal{M}$, there is a $C > 0$ such that for all $ x \in X $ we have $ | T x |_Y \leq C$.
A5: For all $T \in \mathcal{M}$, there is a $C > 0$ such that $ | T | \leq C$.
A6: There is a $C>0$ such that for all $T \in \mathcal{M}$ we have $ | T | \leq C$.
Q3: Let $X,Y$ be two Banach spaces and $\mathcal{B}(X,Y)$ the space of bounded linear operators from $X$ to $Y$. What does it mean that a subset $\mathcal{M} \subseteq \mathcal{B}(X,Y)$ is uniformly bounded?
A1: There is a $C>0$ such that for all $T \in \mathcal{M}$ we have $ | T | \leq C$.
A2: For all $C > 0$, there is a $x \in X$ such that for all $T \in \mathcal{M}$ we have $ | T x |_Y \leq C$.
A3: For all $x \in X$, there is a $T \in \mathcal{M}$ such that for all $C > 0$ we have $ | T x |_Y \leq C$.
A4: For all $T \in \mathcal{M}$, there is a $C > 0$ such that for all $ x \in X $ we have $ | T x |_Y \leq C$.
A5: For all $T \in \mathcal{M}$, there is a $C > 0$ such that $ | T | \leq C$.
A6: There is a $C>0$ such that there is $T \in \mathcal{M}$ with $ | T | \leq C$.
Q4: Let $X,Y$ be two Banach spaces and $\mathcal{M} \subseteq \mathcal{B}(X,Y)$ a collection of bounded linear operators from $X$ to $Y$. What is the correct statement of the Banach-Steinhaus theorem?
A1: $\mathcal{M}$ is uniformly bounded if and only if it is pointwisely bounded.
A2: $\mathcal{M}$ is always uniformly bounded.
A3: $\mathcal{M}$ is never uniformly bounded.
A4: If $\mathcal{M}$ is uniformly bounded, then it’s also pointwisely bounded, but in general the converse is false.
A5: If $\mathcal{M}$ is pointwisely bounded, then it’s also uniformly bounded, but in general the converse is false.
A6: $\mathcal{M}$ is always pointwisely bounded.
Q5: Let $X,Y$ be two Banach spaces and $T_n$ a sequence of bounded linear operators from $X$ to $Y$. What is always correct?
A1: If $\lim_{n\rightarrow \infty} T_n(x)$ exists for all $x \in X$, then this defines a bounded linear operator again.
A2: $\lim_{n\rightarrow \infty} T_n$ always defines a bounded linear operator from $X$ to $Y$.
A3: $\lim_{n\rightarrow \infty} T_n$ converges with respect to the operator norm.
A4: ${ T_n \mid n \in \mathbb{N} }$ is pointwisely bounded.
A5: ${ T_n \mid n \in \mathbb{N} }$ is uniformly bounded.
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Last update: 2025-09
