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+---
+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.