diff options
Diffstat (limited to 'pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md')
| -rw-r--r-- | pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md | 44 |
1 files changed, 28 insertions, 16 deletions
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 |