From 7c66b227a494748e2a546fb85317accd00aebe53 Mon Sep 17 00:00:00 2001 From: Justin Gassner Date: Thu, 15 Feb 2024 05:11:07 +0100 Subject: Update --- .../the-fundamental-four/closed-graph-theorem.md | 7 ++++--- .../the-fundamental-four/hahn-banach-theorem.md | 14 ++------------ .../the-fundamental-four/open-mapping-theorem.md | 4 ++-- .../the-fundamental-four/uniform-boundedness-theorem.md | 6 ++++-- 4 files changed, 12 insertions(+), 19 deletions(-) (limited to 'pages/functional-analysis-basics/the-fundamental-four') diff --git a/pages/functional-analysis-basics/the-fundamental-four/closed-graph-theorem.md b/pages/functional-analysis-basics/the-fundamental-four/closed-graph-theorem.md index f6a9783..e0ec62b 100644 --- a/pages/functional-analysis-basics/the-fundamental-four/closed-graph-theorem.md +++ b/pages/functional-analysis-basics/the-fundamental-four/closed-graph-theorem.md @@ -33,10 +33,11 @@ This shows that $\graph{T}$ is closed. Conversely, suppose that $\graph{T}$ is a closed subspace of $X \times Y$. Note that $X \times Y$ is a Banach space with norm $\norm{(x,y)} = \norm{x} + \norm{y}$. -Therefore $\graph{T}$ is itself as Banach space in the restricted norm $\norm{(x,Tx)} = \norm{x} + \norm{Tx}$. +Therefore, $\graph{T}$ is itself as Banach space in the restricted norm $\norm{(x,Tx)} = \norm{x} + \norm{Tx}$. The canonical projections $\pi_X : \graph{T} \to X$ and $\pi_Y : \graph{T} \to Y$ are bounded. -Clearly, $\pi_X$ is bijective, so its inverse $\pi_X^{-1} : X \to \graph{T}$ is a bounded operator by the +Clearly, $\pi_X$ is bijective, +so its inverse $\pi_X^{-1} : X \to \graph{T}$ is a bounded operator by the [Bounded Inverse Theorem](/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.html#bounded-inverse-theorem). -Consequently the composition, $\pi_Y \circ \pi_X^{-1} : X \to Y$ is bounded. +Consequently, the composition, $\pi_Y \circ \pi_X^{-1} : X \to Y$ is bounded. To complete the proof, observe that $\pi_Y \circ \pi_X^{-1} = T$. {% endproof %} diff --git a/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md b/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md index 18cf64a..a2602ac 100644 --- a/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md +++ b/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md @@ -6,7 +6,6 @@ nav_order: 1 --- # {{ page.title }} -{: .no_toc } In fact, there are multiple theorems and corollaries which bear the name Hahn–Banach. @@ -14,15 +13,6 @@ All have in common that they guarantee the existence of linear functionals with various additional properties. -
- - Table of contents - - {: .text-delta } -- TOC -{:toc} -
- {% definition Sublinear Functional %} A functional $p$ on a real vector space $X$ is called *sublinear* if it is @@ -146,7 +136,7 @@ we define a functional $f_0$ by $f_0(\alpha x) = \alpha \norm{x}$ for $\alpha \i It is easy to check that $f_0$ is linear and bounded with norm $\norm{f_0} = 1$. By the Hahn–Banach Extension Theorem for Normed Spaces, there exists a bounded linear functional $f$ on $X$ extending $f_0$ with identical norm. -Hence we have $f(x) = f_0(x) = \norm{x}$ and $\norm{f} = \norm{f_0} = 1$. +Hence, we have $f(x) = f_0(x) = \norm{x}$ and $\norm{f} = \norm{f_0} = 1$. {% endproof %} Recall that for a normed space $X$ we denote its (topological) dual space by $X'$. @@ -155,7 +145,7 @@ Recall that for a normed space $X$ we denote its (topological) dual space by $X' For every element $x$ of a real or complex normed space $X$ one has $$ -\norm{x} = \sup_{f \in X' \setminus \braces{0}} \frac{\abs{f(x)}}{\norm{f}} +\norm{x} = \sup_{f \in X' \setminus \braces{0}} \frac{\abs{f(x)}}{\norm{f}} $$ and the supremum is attained. diff --git a/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.md b/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.md index b191bb2..e7f2b70 100644 --- a/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.md +++ b/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.md @@ -30,10 +30,10 @@ This remains true, if we take closures: $\bigcup \overline{mTB_X} = Y$. Hence, we have written the space $Y$, which is assumed to have a complete norm, -as the union of countably many closed sets. It follows form the +as the union of countably many closed sets. It follows from the [Baire Category Theorem]({% link pages/general-topology/baire-spaces.md %}) that $\overline{mTB_X}$ has nonempty interior for some $m$. -Thus there are $q \in Y$ and $\alpha > 0$ +Thus, there are $q \in Y$ and $\alpha > 0$ such that $q + \alpha B_Y \subset \overline{mTB_X}$. Choose a $p \in X$ with $Tp=q$. It is a well known fact, that in a normed space diff --git a/pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md b/pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md index 47ddd3f..1140e45 100644 --- a/pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md +++ b/pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md @@ -14,11 +14,13 @@ Let $X$, $Y$ be normed spaces. We say that a collection $\mathcal{T}$ of bounded linear operators from $X$ to $Y$ is {: .mb-0 } -- *pointwise bounded* if the set $\braces{\norm{Tx} : T \in \mathcal{T}}$ is bounded for every $x \in X$, +- *pointwise bounded* if the set $\braces{\norm{Tx} : T \in \mathcal{T}}$ is bounded + for every $x \in X$, - *uniformly bounded* if the set $\braces{\norm{T} : T \in \mathcal{T}}$ is bounded. {% enddefinition %} -Clearly, every uniformly bounded collection of operators is pointwise bounded since $\norm{Tx} \le \norm{T} \norm{x}$. +Clearly, every uniformly bounded collection of operators is pointwise bounded +since $\norm{Tx} \le \norm{T} \norm{x}$. The converse is true, if $X$ is complete: {% theorem * Uniform Boundedness Theorem %} -- cgit v1.2.3-54-g00ecf