Difference between revisions of "Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/2 implies 3"

Latest revision as of 00:37, 28 February 2016

Statement

Given two normed spaces $(X,\Vert\cdot\Vert_X)$ and $(Y,\Vert\cdot\Vert_Y)$ and also a linear map $L:X\rightarrow Y$ then we have:

If $L$ is continuous at a point (say $p\in X$ ) then
L is a bounded linear map, that is to say:
- $\exists A\ge 0\ \forall x\in X[\Vert L(x)\Vert_Y \le A\Vert x\Vert_X]$

Proof

The key to this proof is exploiting the linearity of $L$ . As will be explained in the blue box.

Suppose that L:X→Y is continuous at p∈X. Then:
- $\forall\epsilon>0\ \exists\delta>0[\Vert x-p\Vert_X<\delta\implies\Vert L(x-p)\Vert_Y<\epsilon]$ (note that $L(x-p)=L(x)-L(p)$ due to linearity of $L$ )
Define u:=x−p, then:
- $\forall\epsilon>0\ \exists\delta>0[\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon]$ (writing $Lu:=L(u)$ as is common for linear maps)

[Expand]

We now know we may assume that $\forall\epsilon>0\ \exists\delta>0[\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon]$ - see this box for a guide on how to use this information and how I constructed the rest of the proof. The remainder of the proof follows this box.

The key to this proof is in the norm structure and the linearity of $L$ .

Pick (fix) some ϵ>0 we now know:
- $\exists\delta>0$ such that if we have $\Vert x\Vert_X<\delta$ then we have $\Vert Lx\Vert<\epsilon$

In the proof we will have to show at some point that: $\forall x\in X[\Vert Lx\Vert_Y\le A\Vert x\Vert_X]$ this means:

At some point we'll be given an arbitrary $x\in X$

However we have a $\delta$ such that $\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon$

All we have to do is make $\Vert x\Vert_X$ so small that it is less than $\delta$ - we can do this using scalar multiplication.
Note that if $x=0$ the result is trivial, so assume that arbitrary $x$ is $\ne 0$
With this in mind the task is clear: we need to multiply x by something such that the actual vector part has ∥⋅∥X<δ
- Then we can say (supposing x=αp for some positive α and ∥p∥X<δ) that ∥L(αp)∥Y=α∥Lp∥Y
  - But $\Vert p\Vert_X<\delta\implies\Vert Lp\Vert_Y<\epsilon$
- So $\Vert L(\alpha p)\Vert_Y=\alpha\Vert Lp\Vert_Y<\alpha\epsilon$ , if we can get a $\Vert x\Vert_X$ involved in $\alpha$ the result will follow.

Getting an arbitrary $\Vert x\Vert_X$ to have magnitude $<\delta$

Lets normalise x and then multiply it by the magnitude again (so as to do nothing)
- So notice $x=\overbrace{\Vert x\Vert_X}^\text{scalar part}\cdot\overbrace{\frac{1}{\Vert x\Vert_X}x}^\text{vector part}$ ,
Now the vector part has magnitude 1 we can get it to within δ by multiplying it by δ2
- So $x=\overbrace{\Vert x\Vert_X\cdot\frac{2}{\delta} }^\text{scalar part}\cdot\overbrace{\frac{\delta}{2\Vert x\Vert_X}x}^\text{vector part}$ .

Now we have $\left\Vert\frac{\delta}{2\Vert x\Vert_X}x\right\Vert_Y<\delta$ which $\implies \left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\epsilon$

Thus $\Vert Lx\Vert_Y=\frac{2\Vert x\Vert_X}{\delta}\cdot\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\frac{2\Vert x\Vert_X}{\delta}\cdot\epsilon=\frac{2\epsilon}{\delta}\Vert x\Vert_X$

But $\epsilon$ was fixed, and so was the $\delta$ we know to exist based off of this, so set $A=\frac{2\epsilon}{\delta}$ after fixing them and the result will follow!

Fix some arbitrary ϵ>0 (it doesn't matter what)
- We know there $\exists\delta>0[\Vert x\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon]$ - take such a $\delta$ (which we know to exist by hypothesis) and fix it also.
Define $A:=\frac{2\epsilon}{\delta}$ (if you are unsure of where this came from, see the blue box)
Let x∈A be given (this is the ∀x∈X part of our proof, we have just claimed an A exists on the above line)
- If x=0 then
  - Trivially the result is true, as $L(0_X)=0_Y$ , and $\Vert 0_Y\Vert_Y=0$ by definition, $A\Vert x\Vert_X=0$ as $\Vert 0_X\Vert_X=0$ so we have $0\le 0$ which is true. (This is more workings than the line is worth)
- Otherwise (x≠0)
  - Notice that x=2∥x∥Xδ⋅δ2∥x∥Xx and ∥δ2∥x∥Xx∥X<δ
    - and $\left\Vert \frac{\delta}{2\Vert x\Vert_X}x\right\Vert_X<\delta\implies\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\epsilon$
- Thus we see ∥Lx∥Y=∥L(2∥x∥Xδ⋅δ2∥x∥Xx)∥Y=2∥x∥Xδ⋅remember this is <ϵ⏞∥L(δ2∥x∥Xx)∥Y <2∥x∥Xδ⋅ϵ=2ϵδ∥x∥X=A∥x∥X
  - Shortening the workings this states that: $\Vert Lx\Vert_Y< A\Vert x\Vert_X$
So if $x=0$ we have equality, otherwise $\Vert Lx\Vert_Y< A\Vert x\Vert_X$

In either case, it is true that $\Vert Lx\Vert_Y\le A\Vert x\Vert_X$
This completes the proof

@@ Line 1: / Line 1: @@
 <noinclude>
-{{Stub page|See page 154 in Maurin's Analysis, although the proof isn't hard}}
 ==Statement==
 Given two [[normed space|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}} then we have:
-* If {{M|L}] is continuous at a point (say {{M|p\in X}}) '''then'''
+* If {{M|L}} is continuous at a point (say {{M|p\in X}}) '''then'''
 * {{M|L}} is a [[bounded linear map]], that is to say:
 ** {{M|\exists A\ge 0\ \forall x\in X[\Vert L(x)\Vert_Y \le A\Vert x\Vert_X]}}
 ==Proof==
 </noinclude>
+The key to this proof is exploiting the linearity of {{M|L}}. As will be explained in the blue box.
+* Suppose that {{M|L:X\rightarrow Y}} is continuous at {{M|p\in X}}. Then:
+** {{M|1=\forall\epsilon>0\ \exists\delta>0[\Vert x-p\Vert_X<\delta\implies\Vert L(x-p)\Vert_Y<\epsilon]}} (note that {{M|1=L(x-p)=L(x)-L(p)}} due to [[linear map|linearity]] of {{M|L}})
+* Define {{M|1=u:=x-p}}, then:
+** {{M|1=\forall\epsilon>0\ \exists\delta>0[\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon]}} (writing {{M|1=Lu:=L(u)}} as is common for [[linear map|linear maps]])
+{{Begin Blue Notebox}}
+We now know we may assume that {{M|1=\forall\epsilon>0\ \exists\delta>0[\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon]}} - see this box for a guide on how to use this information and how I constructed the rest of the proof. The remainder of the proof follows this box.
+{{Begin Blue Notebox Content}}
+The key to this proof is in the norm structure and the linearity of {{M|L}}.
+* Pick (fix) some {{M|\epsilon>0}} we now know:
+** {{M|\exists\delta>0}} such that if we have {{M|\Vert x\Vert_X<\delta}} then ''we have'' {{M|\Vert Lx\Vert<\epsilon}}
+In the proof we will have to show at some point that: {{M|\forall x\in X[\Vert Lx\Vert_Y\le A\Vert x\Vert_X]}} this means:
+* At some point we'll be given an arbitrary {{M|x\in X}}
+However we have a {{M|\delta}} such that {{M|\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon}}
+* '''All we have to do is make {{M|\Vert x\Vert_X}} so small that it is less than {{M|\delta}}''' - we can do this using scalar multiplication.
+* ''Note that if {{M|1=x=0}} the result is trivial, so assume that arbitrary {{M|x}} is {{M|\ne 0}}''
+* With this in mind the task is clear: we need to multiply {{M|x}} by something such that the actual vector part has {{M|\Vert\cdot\Vert_X<\delta}}
+** Then we can say (supposing {{M|1=x=\alpha p}} for some positive {{M|\alpha}} and {{M|\Vert p\Vert_X<\delta}}) that {{M|1=\Vert L(\alpha p)\Vert_Y=\alpha\Vert Lp\Vert_Y}}
+*** But {{M|\Vert p\Vert_X<\delta\implies\Vert Lp\Vert_Y<\epsilon}}
+** So {{M|1=\Vert L(\alpha p)\Vert_Y=\alpha\Vert Lp\Vert_Y<\alpha\epsilon}}, if we can get a {{M|\Vert x\Vert_X}} involved in  {{M|\alpha}} the result will follow.
+'''Getting an arbitrary {{M|\Vert x\Vert_X}} to have magnitude {{M|<\delta}}'''
+# Lets [[normalise (vector)|normalise]] {{M|x}} and then multiply it by the magnitude again (so as to do nothing)
+#* So notice {{MM|1=x=\overbrace{\Vert x\Vert_X}^\text{scalar part}\cdot\overbrace{\frac{1}{\Vert x\Vert_X}x}^\text{vector part} }},
+# Now the vector part has magnitude {{M|1}} we can get it to within {{M|\delta}} by multiplying it by {{MM|1=\frac{\delta}{2} }}
+#* So {{MM|1=x=\overbrace{\Vert x\Vert_X\cdot\frac{2}{\delta} }^\text{scalar part}\cdot\overbrace{\frac{\delta}{2\Vert x\Vert_X}x}^\text{vector part} }}.
+Now we have {{MM|1=\left\Vert\frac{\delta}{2\Vert x\Vert_X}x\right\Vert_Y<\delta}} which {{MM|1=\implies \left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\epsilon}}
+* Thus {{MM|1=\Vert Lx\Vert_Y=\frac{2\Vert x\Vert_X}{\delta}\cdot\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\frac{2\Vert x\Vert_X}{\delta}\cdot\epsilon=\frac{2\epsilon}{\delta}\Vert x\Vert_X}}
+'' But {{M|\epsilon}} was fixed, and so was the {{M|\delta}} we know to exist based off of this, so set {{M|1=A=\frac{2\epsilon}{\delta} }} after fixing them and the result will follow!''
+{{End Blue Notebox Content}}{{End Blue Notebox}}
+* Fix some arbitrary {{M|\epsilon>0}} (it doesn't matter what)
+** We know there {{M|\exists\delta>0[\Vert x\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon]}} - take such a {{M|\delta}} (which we know to exist by hypothesis) and fix it also.
+* Define {{M|1=A:=\frac{2\epsilon}{\delta} }} (if you are unsure of where this came from, see the blue box)
+* Let {{M|x\in A}} be given (this is the {{M|\forall x\in X}} part of our proof, we have just claimed an {{M|A}} exists on the above line)
+** If {{M|1=x=0}} then
+*** Trivially the result is true, as {{M|1=L(0_X)=0_Y}}, and {{M|1=\Vert 0_Y\Vert_Y=0}} by definition, {{M|1=A\Vert x\Vert_X=0}} as {{M|1=\Vert 0_X\Vert_X=0}} so we have {{M|0\le 0}} which is true. (This is more workings than the line is worth)
+** Otherwise ({{M|x\ne 0}})
+*** Notice that {{MM|1=x=\frac{2\Vert x\Vert_X}{\delta}\cdot\frac{\delta}{2\Vert x\Vert_X}x}} and {{MM|\left\Vert \frac{\delta}{2\Vert x\Vert_X}x\right\Vert_X<\delta}}
+**** and {{MM|1=\left\Vert \frac{\delta}{2\Vert x\Vert_X}x\right\Vert_X<\delta\implies\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\epsilon}}
+** Thus we see {{MM|1=\Vert Lx\Vert_Y=\left\Vert L\left(\frac{2\Vert x\Vert_X}{\delta}\cdot\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y=\frac{2\Vert x\Vert_X}{\delta}\cdot\overbrace{\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y}^{\text{remember this is }<\epsilon} }} {{MM|1=<\frac{2\Vert x\Vert_X}{\delta}\cdot\epsilon=\frac{2\epsilon}{\delta}\Vert x\Vert_X=A\Vert x\Vert_X}}
+*** Shortening the workings this states that: {{MM|1=\Vert Lx\Vert_Y< A\Vert x\Vert_X}}
+* So if {{M|1=x=0}} we have equality, otherwise {{MM|1=\Vert Lx\Vert_Y< A\Vert x\Vert_X}}
+In either case, it is true that {{MM|1=\Vert Lx\Vert_Y\le A\Vert x\Vert_X}}
+<br/>
+''This completes the proof''
 <noinclude>
 {{Theorem Of|Linear Algebra|Functional Analysis}}
 </noinclude>

Difference between revisions of "Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/2 implies 3"

Latest revision as of 00:37, 28 February 2016

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