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 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80  --- 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](pages/general-topology/baire-spaces.html#baire-category-theorem) 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 $T \in \mathcal{T}$. {{ site.qed }} --- 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. --- {% bibliography %}