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Diffstat (limited to 'pages/functional-analysis-basics/the-fundamental-four')
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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 f8b8254..f6a9783 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 @@ -3,29 +3,40 @@ title: Closed Graph Theorem parent: The Fundamental Four grand_parent: Functional Analysis Basics nav_order: 4 -# cspell:words --- # {{ page.title }} -{: .theorem-title } -> {{ page.title }} -> {: #{{ page.title | slugify }} } -> -> An (everywhere-defined) linear operator between Banach spaces is bounded -> iff its graph is closed. +{% theorem * Closed Graph Theorem %} +An (everywhere-defined) linear operator between Banach spaces is bounded +iff its graph is closed. +{% endtheorem %} We prove a slightly more general version: -{: .theorem-title } -> {{ page.title }} -> {: #{{ page.title | slugify }}-variant } -> -> Let $X$ and $Y$ be Banach spaces -> and $T : \dom{T} \to Y$ a linear operator -> with domain $\dom{T}$ closed in $X$. -> Then $T$ is bounded if and only if -> its graph $\graph{T}$ is closed. +{% theorem * Closed Graph Theorem (Variant) %} +Let $X$ and $Y$ be Banach spaces +and $T : \dom{T} \to Y$ a linear operator +with domain $\dom{T}$ closed in $X$. +Then $T$ is bounded if and only if +its graph $\graph{T}$ is closed. +{% endtheorem %} {% proof %} +Let us assume first that $T$ is bounded. +Let $(x_n,Tx_n)_n$ be a sequence in $\graph{T}$ that converges to some element $(x,y) \in X \times Y$. +This means that $x_n \to x$ and $Tx_n \to y$ for $n \to \infty$. +The continuity of $T$ implies $Tx_n \to Tx$. +Since a convergent series in a Hausdorff space has a unique limit, +it follows that $Tx = y$; hence $(x,y)$ lies in $\graph{T}$. +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 $\norm{(x,Tx)} = \norm{x} + \norm{Tx}$. +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 +[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. +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 9d21d41..18cf64a 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 @@ -6,6 +6,7 @@ nav_order: 1 --- # {{ page.title }} +{: .no_toc } In fact, there are multiple theorems and corollaries which bear the name Hahn–Banach. @@ -13,111 +14,112 @@ All have in common that they guarantee the existence of linear functionals with various additional properties. -{: .definition-title } -> Definition (Sublinear Functional) -> -> A functional $p$ on a real vector space $X$ -> is called *sublinear* if it is -> {: .mb-0 } -> -> {: .mt-0 .mb-0 } -> - *positive-homogenous*, that is -> {: .mt-0 .mb-0 } -> -> $$ -> p(\alpha x) = \alpha \, p(x) \qquad \forall \alpha \ge 0, \ \forall x \in X, -> $$ -> -> - and satisfies the *triangle inequality* -> {: .mt-0 .mb-0 } -> -> $$ -> p(x+y) \le p(x) + p(y) \qquad \forall x,y \in X. -> $$ -> {: .katex-display .mb-0 } +<details open markdown="block"> + <summary> + Table of contents + </summary> + {: .text-delta } +- TOC +{:toc} +</details> + +{% definition Sublinear Functional %} +A functional $p$ on a real vector space $X$ +is called *sublinear* if it is +{: .mb-0 } + +{: .mt-0 .mb-0 } +- *positively homogenous*, that is + {: .mt-0 .mb-0 } + + $$ + p(\alpha x) = \alpha \, p(x) \qquad \forall \alpha \ge 0, \ \forall x \in X, + $$ + +- and *subadditive*, that is + {: .mt-0 .mb-0 } + +$$ +p(x+y) \le p(x) + p(y) \qquad \forall x,y \in X. +$$ +{% enddefinition %} If $p$ is a sublinear functional, then $p(0)=0$ and $p(-x) \ge -p(x)$ for all $x$. Every norm on a real vector space is a sublinear functional. -{: .theorem-title } -> {{ page.title }} (Basic Version) -> -> Let $p$ be a sublinear functional on a real vector space $X$. -> Then there exists a linear functional $f$ on $X$ satisfying -> $f(x) \le p(x)$ for all $x \in X$. +{% theorem * Hahn–Banach Theorem (Basic Version) %} +Let $p$ be a sublinear functional on a real vector space $X$. +Then there exists a linear functional $f$ on $X$ satisfying +$f(x) \le p(x)$ for all $x \in X$. +{% endtheorem %} ## Extension Theorems -{: .theorem-title } -> {{ page.title }} (Extension, Real Vector Spaces) -> -> Let $p$ be a sublinear functional on a real vector space $X$. -> Let $f$ be a linear functional -> which is defined on a linear subspace $Z$ of $X$ -> and satisfies -> -> $$ -> f(x) \le p(x) \qquad \forall x \in Z. -> $$ -> -> Then $f$ has a linear extension $\tilde{f}$ to $X$ such that -> -> $$ -> \tilde{f}(x) \le p(x) \qquad \forall x \in X. -> $$ +{% theorem * Hahn–Banach Theorem (Extension, Real Vector Spaces) %} +Let $p$ be a sublinear functional on a real vector space $X$. +Let $f$ be a linear functional +which is defined on a linear subspace $Z$ of $X$ +and satisfies + +$$ +f(x) \le p(x) \qquad \forall x \in Z. +$$ + +Then $f$ has a linear extension $\tilde{f}$ to $X$ such that + +$$ +\tilde{f}(x) \le p(x) \qquad \forall x \in X. +$$ +{% endtheorem %} {% proof %} {% endproof %} -{: .definition-title } -> Definition (Semi-Norm) -> -> We call a real-valued functional $p$ on a real or complex vector space $X$ -> a *semi-norm* if it is -> {: .mb-0 } -> -> {: .mt-0 .mb-0 } -> - *absolutely homogenous*, that is -> {: .mt-0 .mb-0 } -> -> $$ -> p(\alpha x) = \abs{\alpha} \, p(x) \qquad \forall \alpha \in \KK \ \forall x \in X, -> $$ -> - and satisfies the *triangle inequality* -> {: .mt-0 .mb-0 } -> -> $$ -> p(x+y) \le p(x) + p(y) \qquad \forall x,y \in X. -> $$ -> {: .katex-display .mb-0 } - -{: .theorem-title } -> {{ page.title }} (Extension, Real and Complex Vector Spaces) -> -> Let $p$ be a semi-norm on a real or complex vector space $X$. -> Let $f$ be a linear functional -> which is defined on a linear subspace $Z$ of $X$ -> and satisfies -> -> $$ -> \abs{f(x)} \le p(x) \qquad \forall x \in Z. -> $$ -> -> Then $f$ has a linear extension $\tilde{f}$ to $X$ such that -> -> $$ -> \abs{\tilde{f}(x)} \le p(x) \qquad \forall x \in X. -> $$ - -{: .theorem-title } -> {{ page.title }} (Extension, Normed Spaces) -> -> Let $X$ be a real or complex normed space -> and let $f$ be a bounded linear functional -> defined on a linear subspace $Z$ of $X$. -> Then $f$ has a bounded linear extension $\tilde{f}$ to $X$ such that $\norm{\tilde{f}} = \norm{f}$. +{% definition Semi-Norm %} +We call a real-valued functional $p$ on a real or complex vector space $X$ +a *semi-norm* if it is +{: .mb-0 } + +{: .mt-0 .mb-0 } +- *absolutely homogenous*, that is + {: .mt-0 .mb-0 } + + $$ + p(\alpha x) = \abs{\alpha} \, p(x) \qquad \forall \alpha \in \KK \ \forall x \in X, + $$ +- and satisfies the *triangle inequality* + {: .mt-0 .mb-0 } + + $$ + p(x+y) \le p(x) + p(y) \qquad \forall x,y \in X. + $$ +{% enddefinition %} + +{% theorem * Hahn–Banach Theorem (Extension, Real and Complex Vector Spaces) %} +Let $p$ be a semi-norm on a real or complex vector space $X$. +Let $f$ be a linear functional +which is defined on a linear subspace $Z$ of $X$ +and satisfies + +$$ +\abs{f(x)} \le p(x) \qquad \forall x \in Z. +$$ + +Then $f$ has a linear extension $\tilde{f}$ to $X$ such that + +$$ +\abs{\tilde{f}(x)} \le p(x) \qquad \forall x \in X. +$$ +{% endtheorem %} + +{% theorem * Hahn–Banach Theorem (Extension, Normed Spaces) %} +Let $X$ be a real or complex normed space +and let $f$ be a bounded linear functional +defined on a linear subspace $Z$ of $X$. +Then $f$ has a bounded linear extension $\tilde{f}$ to $X$ such that $\norm{\tilde{f}} = \norm{f}$. +{% endtheorem %} {% proof %} We apply the preceding theorem with $p(x) = \norm{f} \norm{x}$ @@ -131,17 +133,66 @@ Corollaries Important consequence: canonical embedding into bidual +{% theorem * Hahn–Banach Theorem (Existence of Functionals) %} +Let $X$ be a real or complex normed space +and let $x$ be a nonzero element of $X$. +Then there exists a bounded linear functional $f$ on $X$ +with $f(x) = \norm{x}$ and $\norm{f} = 1$. +{% endtheorem %} + +{% proof %} +On the linear subspace $\KK x \subset X$ spanned by $x$ +we define a functional $f_0$ by $f_0(\alpha x) = \alpha \norm{x}$ for $\alpha \in \KK$. +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 $\norm{f} = \norm{f_0} = 1$. +{% endproof %} + +Recall that for a normed space $X$ we denote its (topological) dual space by $X'$. + +{% corollary %} +For every element $x$ of a real or complex normed space $X$ one has + +$$ +\norm{x} = \sup_{f \in X' \setminus \braces{0}} \frac{\abs{f(x)}}{\norm{f}} +$$ + +and the supremum is attained. +{% endcorollary %} + +{% corollary %} +The elements of a real or complex normed space $X$ +are separated by the elements of its dual $X'$. +{% endcorollary %} + ## Separation Theorems -{: .theorem-title } -> {{ page.title }} (Separation, Point and Closed Subspace) -> -> Suppose $Z$ is a closed subspace -> of a normed space $X$ and $x$ lies in $X \setminus Z$. -> Then there exists a bounded linear functional on $X$ -> which vanishes on $Z$ but has a nonzero value at $x$. - -{: .theorem-title } -> {{ page.title }} (Separation, Convex Sets) -> -> TODO +{% theorem * Hahn–Banach Theorem (Separation, Point and Closed Subspace) %} +Suppose $Z$ is a closed subspace of a normed space $X$ +and $x$ lies in $X \setminus Z$. +Then there exists a bounded linear functional $f$ on $X$ +vanishing on $Z$ and with nonzero value $f(x) = \dist{x,Z}$. +{% endtheorem %} + +{% proof %} +Since $Z$ is a closed subspace of $X$, +the quotient vector space $X/Z$ becomes a normed space +with the quotient norm given by + +$$ +\norm{y + Z} = \dist{y,Z} = \inf_{z \in Z} \norm{y-z} \quad \forall y \in X. +$$ + +Moreover, the canonical mapping $\pi : X \to X/Z$, $y \mapsto y+Z$, is bounded. +Given a $x \in X$ that does not lie in $Z$, the null space of $\pi$, +we see that $\pi(x)$ is a nonzero element of $X/Z$. +By Hahn–Banach, there exists a bounded linear functional $g$ on $X/Z$ +with $g(\pi(x)) = \norm{x} = \dist{x,Z} \ne 0$. +Now the composition $f = g \circ \pi$ is a bounded functional on $X$ +with the desired properties. +{% endproof %} + +{% theorem * Hahn–Banach Theorem (Separation, Convex Sets) %} +TODO +{% endtheorem %} 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 53da008..b191bb2 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 @@ -3,7 +3,6 @@ title: Open Mapping Theorem parent: The Fundamental Four grand_parent: Functional Analysis Basics nav_order: 3 -# cspell:words surjective bijective --- # {{ page.title }} @@ -13,12 +12,10 @@ where $X$ and $Y$ are topological spaces, is called *open* if the image under $T$ of each open set of $X$ is open in $Y$. -{: .theorem-title } -> {{ page.title }} -> {: #{{ page.title | slugify }} } -> -> A bounded linear operator between Banach spaces is open -> if and only if it is surjective. +{% theorem * Open Mapping Theorem %} +A bounded linear operator between Banach spaces is open +if and only if it is surjective. +{% endtheorem %} {% proof %} Let $X$ and $Y$ be Banach spaces @@ -91,19 +88,22 @@ Conversely, suppose that $T$ is open. TODO --- -XXX injective For a bijective mapping between topological spaces, to say that it is open, is equivalent to saying that its inverse is continuous. The inverse of a bijective linear map between normed spaces is automatically linear and thus continuous if and only if it is bounded. As a corollary to the {{ page.title }} we obtain the following: -{: .corollary-title } -> Bounded Inverse Theorem -> {: #bounded-inverse-theorem } -> -> If a bounded linear operator between Banach spaces is bijective, -> then its inverse is bounded. -XXX relax to injective +{% corollary * Bounded Inverse Theorem %} +If a bounded linear operator between Banach spaces is bijective, +then its inverse is bounded. +{% endcorollary %} Also known as *Inverse Mapping Theorem*. + +{% corollary %} +Let $T: X \to Y$ be a bounded linear operator between Banach spaces +and suppose that $T$ is injective, so that the inverse $T^{-1} : R(T) \to X$ +is defined on the range of $T$. +The linear operator $T^{-1}$ is bounded if and only if $R(T)$ is closed in $X$. +{% endcorollary %} 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 13460da..47ddd3f 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 @@ -3,27 +3,35 @@ title: Uniform Boundedness Theorem parent: The Fundamental Four grand_parent: Functional Analysis Basics nav_order: 2 -description: > - The -# spellchecker:words preimages pointwise --- # {{ page.title }} Also known as *Uniform Boundedness Principle* and *Banach–Steinhaus Theorem*. -{: .theorem-title } -> {{ page.title }} -> {: #{{ page.title | slugify }} } -> -> If $\mathcal{T}$ is a set of bounded linear operators -> from a Banach space $X$ into a normed space $Y$ such that -> $\braces{\norm{Tx} : T \in \mathcal{T}}$ -> is a bounded set for every $x \in X$, then -> $\braces{\norm{T} : T \in \mathcal{T}}$ -> is a bounded set. +{% definition Pointwise and Uniform Boundedness %} +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 $x \in X$, +- *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 $\norm{Tx} \le \norm{T} \norm{x}$. +The converse is true, if $X$ is complete: + +{% theorem * Uniform Boundedness Theorem %} +If a collection of bounded linear operators +from a Banach space into a normed space +is pointwise bounded, +then it is uniformly bounded. +{% endtheorem %} {% proof %} +Suppose $X$ is a Banach space, $Y$ is a normed space +and $\mathcal{T}$ is a pointwise bounded collection +of bounded linear operators from $X$ to $Y$. For each $n \in \NN$ the set $$ @@ -38,9 +46,8 @@ the set $\braces{\norm{Tx} : T \in \mathcal{T}}$ is bounded by assumption. This means that there exists a $n \in \NN$ such that $\norm{Tx} \le n$ for all $T \in \mathcal{T}$. In other words, $x \in A_n$. -Thus we have show that $\bigcup A_n = X$. -XXX Apart from the trivial case $X = \emptyset$, -the union $\bigcup A_n$ has nonempty interior. +Thus, we have shown that $\bigcup A_n = X$. +In particular, $\bigcup A_n$ has nonempty interior. Now, utilizing the completeness of $X$, the [Baire Category Theorem]({% link pages/general-topology/baire-spaces.md %}) implies that there exists a $m \in \NN$ such that $A_m$ has nonempty interior. @@ -74,3 +81,8 @@ $$ $$ If $X$ is not complete, this may be false. + +TODO: +- strong operator convergence +- Kreyszig 4.9-5 +- Haase 15.6 |