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diff --git a/pages/functional-analysis-basics/banach-alaoglu-theorem.md b/pages/functional-analysis-basics/banach-alaoglu-theorem.md new file mode 100644 index 0000000..59e4a92 --- /dev/null +++ b/pages/functional-analysis-basics/banach-alaoglu-theorem.md @@ -0,0 +1,19 @@ +--- +title: Banach–Alaoglu Theorem +parent: Functional Analysis Basics +nav_order: 3 +# cspell:words +--- + +# {{ page.title }} + +{: .theorem-title } +> {{ page.title }} +> {: #{{ page.title | slugify }} } +> +> The closed unit ball in the dual of a normed space is weak\* compact. + +{% proof %} +{% endproof %} + +## Generalization: Alaoglu–Bourbaki diff --git a/pages/functional-analysis-basics/compact-operators.md b/pages/functional-analysis-basics/compact-operators.md new file mode 100644 index 0000000..b114c24 --- /dev/null +++ b/pages/functional-analysis-basics/compact-operators.md @@ -0,0 +1,44 @@ +--- +title: Compact Operators +parent: Functional Analysis Basics +nav_order: 4 +published: false +# cspell:words +--- + +# {{ page.title }} + +{: .definition-title } +> Definition (Compact Linear Operator) +> {: #compact-operator } +> +> A linear operator $T : X \to Y$, +> where $X$ and $Y$ are normed spaces, +> is said to be a *compact linear operator*, +> if for every bounded subset $M \subset X$ +> the image $TM$ is relatively compact in $Y$. + +{: .proposition-title } +> Proposition (Characterisation of Compactness) +> +> Let $X$ and $Y$ be normed spaces. +> A linear operator $T : X \to Y$ is compact if and only if +> for every bounded sequence $(x_n)$ in $X$ +> the image sequence $(Tx_n)$ in $Y$ has a convergent subseqence. + +{: .proposition-title } +> Every compact linear operator is bounded. + +{: .proposition-title } +> Proposition (Compactness of Zero and Identity) +> +> The zero operator on any normed space is compact. +> The indentity operator on a normed space $X$ is compact if and only if $X$ has finite dimension. + +{: .proposition-title } +> Proposition (The Space of Compact Linear Operators) +> +> The set $C(X,Y)$ of compact linear operator from a normed space $X$ into a normed space $Y$ +> form a linear subspace of the space $B(X,Y)$ of bounded linear operators from $X$ into $Y$. +> If $Y$ is a Banach space, then $C(X,Y)$ is a closed linear subspace of the Banach space +> $B(X,Y)$ and hence itself a Banach space. diff --git a/pages/functional-analysis-basics/index.md b/pages/functional-analysis-basics/index.md new file mode 100644 index 0000000..1b7fd69 --- /dev/null +++ b/pages/functional-analysis-basics/index.md @@ -0,0 +1,11 @@ +--- +title: Functional Analysis Basics +nav_order: 3 +has_children: true +--- + +# {{ page.title }} + +## Recommended Textbooks + +{% bibliography --file functional-analysis-basics %} diff --git a/pages/functional-analysis-basics/reflexive-spaces.md b/pages/functional-analysis-basics/reflexive-spaces.md new file mode 100644 index 0000000..dee0e55 --- /dev/null +++ b/pages/functional-analysis-basics/reflexive-spaces.md @@ -0,0 +1,123 @@ +--- +title: Reflexive Spaces +parent: Functional Analysis Basics +nav_order: 2 +# cspell:words +--- + +# {{ page.title }} + +{: .definition-title } +> Definition (Canonical Embedding) +> +> Let $X$ be a normed space. +> The mapping +> +> $$ +> C : X \longrightarrow X'', \quad x \mapsto g_x, +> $$ +> +> where the functional $g_x$ on $X'$ is defined by +> +> $$ +> g_x(f) = f(x) \quad \text{for $f \in X'$,} +> $$ +> +> is called the *canonical embedding* of $X$ into its bidual $X''$. + +{: .lemma } +> The canonical embedding $C : X \to X''$ of a normed space into its bidual +> is well-defined and an embedding of normed spaces. + +{% proof %} +{% endproof %} + +In particular, $C$ is isometric, hence injective. + +{: .definition-title } +> Definition (Reflexivity) +> +> A normed space is said to be *reflexive* +> if the canonical embedding into its bidual +> is surjective. + +If a normed space $X$ is reflexive, +then $X$ is isomorphic with $X''$, its bidual. +James gives a counterexample for the converse statement. + +{: .theorem } +> If a normed space is reflexive, +> then it is complete; hence a Banach space. + +{% proof %} +{% endproof %} + +{: .theorem } +> If a normed space $X$ is reflexive, +> then the weak and weak$^*$ topologies on $X'$ agree. + +{% proof %} +By definition, the weak and weak$^*$ topologies on $X'$ +are the initial topologies induced by the sets of functionals +$X''$ and $C(X)$, respectively. +Since $X$ is reflexive, those sets are equal. +{% endproof %} + +The converse is true as well. Proof: TODO + +{: .theorem } +> If a normed space $X$ is reflexive, +> then its dual $X'$ is reflexive. + +{% proof %} +Since $X$ is reflexive, +the canonical embedding + +$$ +C : X \longrightarrow X'', \quad C(x)(f) = f(x), \quad x \in X, f \in X', +$$ + +is an isomorphism. +Therefore, the the dual map + +$$ +C' : X''' \longrightarrow X', \quad C'(h)(x) = h(C(x)), \quad x \in X, h \in X''', +$$ + +is an isomorphism as well. +A priori, it is not clear how $C'$ is related to +the canonical embedding + +$$ +D : X' \longrightarrow X''', \quad D(f)(g) = g(f), \quad f \in X', g \in X''. +$$ + +To show that $D$ is surjective, +consider any element $h$ in $X'''$. +We claim that $h=D(f)$ with $f=C'(h)$. +Let $g$ be any element of $X''$. +It is of the form $g=C(x)$ with $x \in X$ unique, because $X$ is reflexive. +We have + +$$ +h(g) = h(C(x)) = C'(h)(x) = f(x) +$$ + +by the definition of $C'$. +On the other hand, + +$$ +D(f)(g) = g(f) = C(x)(f) = f(x) +$$ + +by the definitions of $D$ and $C$. +This shows that $D$ is surjective, hence $X'$ is reflexive. +In fact, we have shown more: $D = (C')^{-1}$. +{% endproof %} + +{: .theorem } +> Every finite-dimensional normed space is reflexive. +> + +{: .theorem } +> Every Hilbert space is reflexive. 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 new file mode 100644 index 0000000..f8b8254 --- /dev/null +++ b/pages/functional-analysis-basics/the-fundamental-four/closed-graph-theorem.md @@ -0,0 +1,31 @@ +--- +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. + +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. + +{% proof %} +{% 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 new file mode 100644 index 0000000..9d21d41 --- /dev/null +++ b/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md @@ -0,0 +1,147 @@ +--- +title: Hahn–Banach Theorem +parent: The Fundamental Four +grand_parent: Functional Analysis Basics +nav_order: 1 +--- + +# {{ page.title }} + +In fact, there are multiple theorems and corollaries +which bear the name Hahn–Banach. +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 } + +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 $x \in X$. + +## 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 +> +> $$ +> \tilde{f}(x) \le p(x) \qquad \forall x \in X. +> $$ + +{% 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 $\norm{\tilde{f}} = \norm{f}$. + +{% proof %} +We apply the preceding theorem with $p(x) = \norm{f} \norm{x}$ +and obtain a linear extension $\tilde{f}$ of $f$ to $X$ +satisfying $\abs{\tilde{f}(x)} \le \norm{f} \norm{x}$ for all $x \in X$. +This implies that $\tilde{f}$ is bounded and $\norm{\tilde{f}} \le \norm{f}$. +We have $\norm{\tilde{f}} \ge \norm{f}$, because $\tilde{f}$ extends $f$. +{% endproof %} + +Corollaries + +Important consequence: canonical embedding into bidual + +## 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 diff --git a/pages/functional-analysis-basics/the-fundamental-four/index.md b/pages/functional-analysis-basics/the-fundamental-four/index.md new file mode 100644 index 0000000..e814571 --- /dev/null +++ b/pages/functional-analysis-basics/the-fundamental-four/index.md @@ -0,0 +1,8 @@ +--- +title: The Fundamental Four +parent: Functional Analysis Basics +nav_order: 2 +has_children: true +--- + +# {{ page.title }} 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 new file mode 100644 index 0000000..53da008 --- /dev/null +++ b/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.md @@ -0,0 +1,109 @@ +--- +title: Open Mapping Theorem +parent: The Fundamental Four +grand_parent: Functional Analysis Basics +nav_order: 3 +# cspell:words surjective bijective +--- + +# {{ page.title }} + +Recall that a mapping $T : X \to Y$, +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. + +{% proof %} +Let $X$ and $Y$ be Banach spaces +and let $T : X \to Y$ be a bounded linear operator. +Let $B_X$ and $B_Y$ denote the open unit balls in $X$ and $Y$, respectively. + +First, suppose that $T$ is surjective. +The balls $m B_X$, $m \in \NN$, cover $X$. +Since $T$ is surjective, +their images $mTB_X$ cover $Y$. +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 +[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 $\alpha > 0$ +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 +the translation by a vector and the multiplication with a nonzero scalar +are homeomorphisms and thus compatible with taking the closure. +We conclude $\alpha B_Y \subset \overline{T(mB_X-q)}$. +Since $mB_X-q$ is a bounded set, +it is contained in a ball $\beta B_X$ for some $\beta > 0$. +Thus, $\alpha B_Y \subset \overline{T \beta B_X} = \beta \overline{TB_X}$. +With $\gamma := \alpha / \beta > 0$ we obtain $\gamma B_Y \subset \overline{TB_X}$. + +Clearly, every $y \in \gamma B_Y$ is the limit of a sequence $(Tx_n)$, +where $x_n \in B_X$. +However, the sequence $(x_n)$ *may not converge*! +We show that it is possible to find a *convergent* sequence $(s_n)$ in $4B_X$ +such that $Ts_n \to y$. +To construct $(s_n)$, we recursively define a sequence $(y_k)$ +with $y_k \in 2^{-k} \gamma B_Y$ for $k \in \NN_0$. +The sequence starts with $y_0 := y \in 2^0 \gamma B_Y$. +Given $y_k \in 2^{-k} \gamma B_Y$, one has $y_k \in \overline{T 2^{-k} B_X}$. +By the definition of closure, there exists a $x_k \in 2^{-k} B_X$ +such that $Tx_k$ lies in the open $2^{-(k+1)} \gamma$-ball about $y_k$. +This means that $y_{k+1} := y_k - Tx_k \in 2^{-(k+1)}\gamma B_Y$. +Now define $s_n$ as the $n$-th partial sum of the series $\sum_{k=0}^{\infty} x_k$. +The series converges, +because it converges absolutely (Here we use the completeness of $X$). +The latter is true because $\sum \norm{x_k} \le \sum 2^{-k} = 3$. +This also shows that each $s_n$ and the limit $x := \lim s_n$ lie in $4B_X$. +The auxiliary sequence $(y_n)$ converges to $0$ by construction. +Therefore, in the limit $n \to \infty$ + +$$ +Ts_n = \sum_{k=0}^{n} Tx_k = \sum_{k=0}^{n} y_k - y_{k+1} += y_0 - y_{n+1} \to y_0 = y, +$$ + +as desired. +It follows from the continuity of $T$ that $Ts_n \to Tx$, thus $Tx = y$. + +In the preceding paragraph it was proven that $\gamma B_Y \subset 4TB_X$. +Hence, $\delta B_Y \subset TB_X$ where $\delta := \gamma/4$. +To show that $T$ is open, consider any open set $U \subset X$. +If $y$ lies in $TU$, there exists a $x \in U$ such that $Tx=y$. +Since $U$ is open, there is an $\epsilon > 0$ such that $x+\epsilon B_X \subset U$. +Applying $T$, we find $y + \epsilon TB_X \subset TU$. +Combine with $\delta B_Y \subset TB_X$ to see $y + \epsilon \delta B_X \subset TU$. +Hence, $TU$ is open. +This shows that $T$ is indeed an open mapping. + +Conversely, suppose that $T$ is open. TODO +{% endproof %} + +--- + +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 + +Also known as *Inverse Mapping Theorem*. 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 new file mode 100644 index 0000000..13460da --- /dev/null +++ b/pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md @@ -0,0 +1,76 @@ +--- +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. + +{% proof %} +For each $n \in \NN$ the set + +$$ +A_n = \bigcap_{T \in \mathcal{T}} \braces{x \in X : \norm{Tx} \le n} +$$ + +is closed, since it is the intersection +of the preimages of the closed interval $[0,n]$ +under the continuous maps $x \mapsto \norm{Tx}$. +Given any $x \in X$, +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. +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. +It follows that $A_m$ contains an open ball $B(y,\epsilon)$. + +To show that $\braces{\norm{T} : T \in \mathcal{T}}$ is bounded, +let $z \in X$ with $\norm{z} \le 1$. +Then $y+\epsilon z \in B(y,\epsilon)$. +Using the reverse triangle inequality and the linearity of $T$, we find + +$$ +\epsilon \norm{Tz} \le \norm{Ty} + \norm{T(y + \epsilon z)} \le 2m. +$$ + +This proves $\norm{T} \le 2m/\epsilon$ for all $T \in \mathcal{T}$. +{% endproof %} + +--- + +In particular, for a sequence of operators $(T_n)$, +if there are pointwise bounds $c_x$ such that + +$$ +\norm{T_n x} \le c_x \quad \forall n \in \NN, \forall x \in X, +$$ + +the theorem implies the existence of bound $c$ such that + +$$ +\norm{T_n} \le c \quad \forall n \in \NN. +$$ + +If $X$ is not complete, this may be false. |