Editing Bounded operator (section)

===Characterizations of bounded linear operators===

Let <math>F : X \to Y</math> be a linear operator between topological vector spaces (not necessarily Hausdorff). 
The following are equivalent:
#<math>F</math> is (locally) bounded;{{sfn|Narici|Beckenstein|2011|pp=441-457}}
#(Definition): <math>F</math> maps bounded subsets of its domain to bounded subsets of its codomain;{{sfn|Narici|Beckenstein|2011|pp=441-457}}
#<math>F</math> maps bounded subsets of its domain to bounded subsets of its [[Image of a function|image]] <math>\operatorname{Im} F := F(X)</math>;{{sfn|Narici|Beckenstein|2011|pp=441-457}}
#<math>F</math> maps every null sequence to a bounded sequence;{{sfn|Narici |Beckenstein| 2011|pp=441-457}}
#* A ''null sequence'' is by definition a sequence that converges to the origin.
#* Thus any linear map that is sequentially continuous at the origin is necessarily a bounded linear map.
#<math>F</math> maps every Mackey convergent null sequence to a bounded subset of <math>Y.</math><ref group=note>Proof: Assume for the sake of contradiction that <math>x_{\bull} = \left(x_i\right)_{i=1}^{\infty}</math> converges to <math>0</math> but <math>F\left(x_{\bull}\right) = \left(F\left(x_i\right)\right)_{i=1}^{\infty}</math> is not bounded in <math>Y.</math> Pick an open [[Balanced set|balanced]] neighborhood <math>V</math> of the origin in <math>Y</math> such that <math>V</math> does not absorb the sequence <math>F\left(x_{\bull}\right).</math> Replacing <math>x_{\bull}</math> with a subsequence if necessary, it may be assumed without loss of generality that <math>F\left(x_i\right) \not\in i^2 V</math> for every positive integer <math>i.</math> The sequence <math>z_{\bull} := \left(x_i/i\right)_{i=1}^{\infty}</math> is Mackey convergent to the origin (since <math>\left(i z_i\right)_{i=1}^{\infty} = \left(x_i\right)_{i=1}^{\infty} \to 0</math> is bounded in <math>X</math>) so by assumption, <math>F\left(z_{\bull}\right) = \left(F\left(z_i\right)\right)_{i=1}^{\infty}</math> is bounded in <math>Y.</math> So pick a real <math>r > 1</math> such that <math>F\left(z_i\right) \in r V</math> for every integer <math>i.</math> If <math>i > r</math> is an integer then since <math>V</math> is balanced, <math>F\left(x_i\right) \in r i V \subseteq i^2 V,</math> which is a contradiction. Q.E.D. This proof readily generalizes to give even stronger characterizations of "<math>F</math> is bounded." For example, the word "such that <math>\left(r_i x_i\right)_{i=1}^{\infty}</math> is a bounded subset of <math>X.</math>" in the definition of "Mackey convergent to the origin" can be replaced with "such that <math>\left(r_i x_i\right)_{i=1}^{\infty} \to 0</math> in <math>X.</math>"</ref><!---Old proof:<ref group=note>Proof if <math>Y</math> is locally convex: Assume for the sake of contradiction that <math>x_{\bull} = \left(x_i\right)_{i=1}^{\infty}</math> converges to <math>0</math> but <math>F\left(x_{\bull}\right) = \left(F\left(x_i\right)\right)_{i=1}^{\infty}</math> is not bounded in <math>Y.</math> Then there exists a continuous seminorm <math>p</math> on <math>Y</math> such that <math>p\left(F\left(x_{\bull}\right)\right) = \left(p\left(F\left(x_i\right)\right)\right)_{i=1}^{\infty}</math> is unbounded. Let <math>r_i = \sqrt{p\left(x_i\right)}</math> for every <math>i</math> where by going to a subsequence, it can be assumed without loss of generality that <math>r_i > 0</math> for all <math>i</math> and also <math>r_i \to \infty.</math> Then <math>\left(x_i/r_i\right)_{i=1}^{\infty}</math> Mackey convergent to the origin but its image under <math>F</math> is not bounded in <math>Y.</math> <math>\blacksquare</math></ref> End: old proof --->
#* A sequence <math>x_{\bull} = \left(x_i\right)_{i=1}^{\infty}</math> is said to be ''[[Mackey convergence|Mackey convergent to the origin]] in <math>X</math>'' if there exists a divergent sequence <math>r_{\bull} = \left(r_i\right)_{i=1}^{\infty} \to \infty</math> of positive real number such that <math>r_{\bull} = \left(r_i x_i\right)_{i=1}^{\infty}</math> is a bounded subset of <math>X.</math>

if <math>X</math> and <math>Y</math> are [[Locally convex topological vector space|locally convex]] then the following may be add to this list:
<ol start=6>
<li><math>F</math> maps bounded [[Absolutely convex set|disks]] into bounded disks.{{sfn|Narici|Beckenstein|2011|p=444}}</li>
<li><math>F^{-1}</math> maps [[Bornivorous set|bornivorous]] disks in <math>Y</math> into bornivorous disks in <math>X.</math>{{sfn|Narici|Beckenstein|2011|p=444}}</li>
</ol>

if <math>X</math> is a [[bornological space]] and <math>Y</math> is locally convex then the following may be added to this list:
<ol start=8>
<li><math>F</math> is [[Sequential continuity at a point|sequentially continuous at some]] (or equivalently, at every) point of its domain.{{sfn|Narici|Beckenstein|2011|pp=451-457}}
* A [[Sequential continuity|sequentially continuous]] linear map between two TVSs is always bounded,{{sfn|Wilansky|2013|pp=47-50}} but the converse requires additional assumptions to hold (such as the domain being bornological and the codomain being locally convex).
* If the domain <math>X</math> is also a [[sequential space]], then <math>F</math> is [[Sequential continuity|sequentially continuous]] if and only if it is continuous.</li>
<li><math>F</math> is [[Sequential continuity at a point|sequentially continuous at the origin]].</li>
</ol>