Difference between revisions of "Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/4 implies 1"
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(Created page with "<noinclude> ==Statement== Given two normed spaces {{M|(X,\Vert\cdot\Vert_X)}} and {{M|(Y,\Vert\cdot\Vert_Y)}} and also a linear map {{M|L:X\rightarrow Y}}...") |
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** So {{M|L}} maps {{M|1=(x_n)_{n=1}^\infty\rightarrow x}} to {{MM|1=\left(L(x_n)\right)_{n=1}^\infty\rightarrow L(x)}} | ** So {{M|L}} maps {{M|1=(x_n)_{n=1}^\infty\rightarrow x}} to {{MM|1=\left(L(x_n)\right)_{n=1}^\infty\rightarrow L(x)}} | ||
** Recall that [[if a sequence converges it is bounded]], as {{MM|1=\left(L(x_n)\right)_{n=1}^\infty}} converges, it must therefore be bounded. | ** Recall that [[if a sequence converges it is bounded]], as {{MM|1=\left(L(x_n)\right)_{n=1}^\infty}} converges, it must therefore be bounded. | ||
| + | <br/> | ||
| + | '''Corollary: ''' the image of every [[null sequence]] under {{M|L}} is bounded | ||
| + | * A [[null sequence]] is just a [[sequence]] that [[convergence (sequence)|converges]] to {{M|0}}. If | ||
| + | ** We are given a {{MM|1=(x_n)_{n=1}^\infty\rightarrow 0}} then: | ||
| + | *** We know that {{MM|1=\left(L(x_n)\right)_{n=1}^\infty\rightarrow L(0) = 0}} (as {{M|L}} is a [[linear map]]) | ||
| + | ** Again, [[if a sequence converges it is bounded]] | ||
| + | <br/> | ||
This completes the proof. | This completes the proof. | ||
<noinclude> | <noinclude> | ||
{{Theorem Of|Linear Algebra|Functional Analysis}} | {{Theorem Of|Linear Algebra|Functional Analysis}} | ||
</noinclude> | </noinclude> | ||
Latest revision as of 01:34, 28 February 2016
Statement
Given two normed spaces [ilmath](X,\Vert\cdot\Vert_X)[/ilmath] and [ilmath](Y,\Vert\cdot\Vert_Y)[/ilmath] and also a linear map [ilmath]L:X\rightarrow Y[/ilmath] then we have:
- If [ilmath]L[/ilmath] is everywhere continuous then
- [ilmath]L[/ilmath] maps every null sequence to a bounded sequence
Proof
There is actually a slightly stronger result to be had here, I shall prove that, to which the above statement is a corollary.
- Let [ilmath](x_n)_{n=1}^\infty\rightarrow x[/ilmath] be a sequence that converges - we must show that the image of this sequence under [ilmath]L[/ilmath] is bounded
- As [ilmath]L[/ilmath] is everywhere continuous we know that it is sequentially continuous at [ilmath]x[/ilmath]. This means that:
- [math]\forall\left((x_n)_{n=1}^\infty\rightarrow x\right)\left[\big(L(x_n)\big)_{n=1}^\infty\rightarrow L(x)\right][/math]
- So [ilmath]L[/ilmath] maps [ilmath](x_n)_{n=1}^\infty\rightarrow x[/ilmath] to [math]\left(L(x_n)\right)_{n=1}^\infty\rightarrow L(x)[/math]
- Recall that if a sequence converges it is bounded, as [math]\left(L(x_n)\right)_{n=1}^\infty[/math] converges, it must therefore be bounded.
- As [ilmath]L[/ilmath] is everywhere continuous we know that it is sequentially continuous at [ilmath]x[/ilmath]. This means that:
Corollary: the image of every null sequence under [ilmath]L[/ilmath] is bounded
- A null sequence is just a sequence that converges to [ilmath]0[/ilmath]. If
- We are given a [math](x_n)_{n=1}^\infty\rightarrow 0[/math] then:
- We know that [math]\left(L(x_n)\right)_{n=1}^\infty\rightarrow L(0) = 0[/math] (as [ilmath]L[/ilmath] is a linear map)
- Again, if a sequence converges it is bounded
- We are given a [math](x_n)_{n=1}^\infty\rightarrow 0[/math] then:
This completes the proof.
