Equivalent conditions for a linear map between two normed spaces to be continuous everywhere
Contents
Statement
Given two normed spaces [ilmath](X,\Vert\cdot\Vert_X)[/ilmath] and [ilmath](Y,\Vert\cdot\Vert_Y)[/ilmath] and a linear map [ilmath]L:X\rightarrow Y[/ilmath] between them, then the following are equivalent (meaning if you have 1 you have all the others):
- If we have a sequence [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] with [ilmath]x_n\rightarrow 0[/ilmath] then [ilmath](\Vert L(x_n)\Vert_Y)_{n=1}^\infty[/ilmath] is a bounded sequence[1][Note 1]
- [ilmath]L[/ilmath] is continuous at [ilmath]0\in X[/ilmath][1][2][Note 2]
- [ilmath]L[/ilmath] is a bounded linear map, that is [ilmath]\exists A>0\ \forall x\in X[\Vert L(x)\Vert_Y\le A\Vert x\Vert_X][/ilmath][1][2]
- [ilmath]L[/ilmath] is continuous everywhere[1][2]
Proof
[ilmath]1)\implies 2)[/ilmath]: That [ilmath]L[/ilmath] maps all null sequences to bounded sequences [ilmath]\implies[/ilmath] [ilmath]L[/ilmath] is continuous at [ilmath]0\in X[/ilmath]
This is a proof by contrapositive. That is we will show that if [ilmath]L[/ilmath] is not continuous at [ilmath]p[/ilmath] [ilmath]\implies[/ilmath] [ilmath]L[/ilmath] takes a null sequence to one that isn't bounded (an unbounded one).
- Let the normed spaces [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] be given, as well as a linear map [ilmath]L:X\rightarrow Y[/ilmath]
- Suppose that [ilmath]L[/ilmath] is not continuous at [ilmath]p[/ilmath], this means:
- [ilmath]\exists (x_n)_{n=1}^\infty\rightarrow p[/ilmath] such that [math]\lim_{n\rightarrow\infty}\left(L(x_n)\right)\ne L(p)[/math]
Recall that continuity states that:
- [ilmath]L[/ilmath] is continuous at [ilmath]p[/ilmath] [ilmath]\iff[/ilmath] [math]\forall(x_n)_{n=1}^\infty\subseteq X \left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\implies\left(\lim_{n\rightarrow\infty}\left(L(x_n)\right)=L\left(\lim_{n\rightarrow\infty}(x_n)\right)=L(p)\right)\right][/math]
So it follows that to not be continuous at [ilmath]p[/ilmath]:
- [math]\exists (x_n)_{n=1}^\infty\subseteq X\left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\wedge\neg\left(\lim_{n\rightarrow\infty}(L(x_n))=L(p)\right)\right][/math], by negation of implies. Additionally we may negate the [ilmath]=[/ilmath] and thus we see this is the same as:
- [math]\exists (x_n)_{n=1}^\infty\subseteq X\left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\wedge\left(\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)\right)\right][/math]
Which is exactly "there exists a sequence in [ilmath]X[/ilmath] whose limit is [ilmath]p[/ilmath] and where [ilmath]\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)[/ilmath]"
- Let us now take [ilmath]L(x_n)\not\rightarrow L(p)[/ilmath] and subtract [ilmath]L(p)[/ilmath] from both sides. We see:
- [ilmath]L(x_n)-L(p)\not\rightarrow L(p)-L(p)[/ilmath], using the fact that [ilmath]L[/ilmath] is linear we see that:
- [ilmath]L(x_n-p)\not\rightarrow L(0)[/ilmath] and [ilmath]L(0)=0\in Y[/ilmath] so:
- [ilmath]L(x_n-p)\not\rightarrow 0[/ilmath]
- [ilmath]L(x_n)-L(p)\not\rightarrow L(p)-L(p)[/ilmath], using the fact that [ilmath]L[/ilmath] is linear we see that:
- Thus [ilmath]\Vert L(x_n-p)\Vert_Y\not\rightarrow 0[/ilmath] (as [ilmath]\Vert0\Vert_Y=0[/ilmath] by definition)
- Let us now take [ilmath]L(x_n)\not\rightarrow L(p)[/ilmath] and subtract [ilmath]L(p)[/ilmath] from both sides. We see:
- So [math]\exists C>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge\Vert L(x_n-p)\Vert_Y>\epsilon][/math]
If a sequence converges to [ilmath]0[/ilmath] then we have:
- [ilmath]\forall\epsilon>0\ \exists N\in\mathbb{N}\ \forall n\in\mathbb{N}[n>N\implies d(x_n,x)<\epsilon][/ilmath] we know we don't have this, so we negate it:
- [ilmath]\exists\epsilon>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge d(x_n,x)>\epsilon][/ilmath].
That is there exists an [ilmath]\epsilon>0[/ilmath] such that for all [ilmath]N[/ilmath] there exists a bigger [ilmath]n[/ilmath] such that [ilmath]d(x_n,x)>\epsilon[/ilmath], we shall later call such an [ilmath]\epsilon[/ilmath] [ilmath]C[/ilmath] and construct a subsequence out of the [ilmath]n[/ilmath]s
- Thus it is possible to construct a subsequence, [ilmath](\Vert L(x_{n_k}-p)\Vert_Y)_{k=1}^\infty[/ilmath] of the image [ilmath](x_n)[/ilmath] where for every [ilmath]k[/ilmath] we have:
- [ilmath]\Vert L(x_{n_k}-p)\Vert_Y>C[/ilmath]
By the negation of convergent sequence we see that there exists a [ilmath]C[/ilmath] such that for all [ilmath]N\in\mathbb{N} [/ilmath] there exists another natural number, [ilmath]n[/ilmath] such that [ilmath]\Vert L(x_n-p)\Vert_Y>C[/ilmath], we construct a sequence, [ilmath](n_k)_{k=1}^\infty[/ilmath] of such [ilmath]n[/ilmath]-ns. That is we know there exists a [ilmath]C[/ilmath], so we pick [ilmath]N=1[/ilmath] and get the [ilmath]n[/ilmath] that works, this is our first term, we then set [ilmath]N=2[/ilmath] and the [ilmath]n[/ilmath] that works is our second term, and so forth.
Some of these [ilmath]n[/ilmath]s may be the same, but that doesn't matter.
[ilmath](x_{n_k})[/ilmath] is the subsequence of [ilmath](x_n)[/ilmath] which contains only the terms that satisfy [ilmath]\Vert L(x_n-p)\Vert_Y> C[/ilmath]
It would have been better if I used [ilmath]n[/ilmath] and [ilmath]m[/ilmath] as terms, to make which [ilmath]n[/ilmath] I am talking about clearer, but a reader able to attempt this proof should follow.
- We now have a sequence [ilmath](x_{n_k})[/ilmath] such that [ilmath]\Vert L(x_{n_k}-p)\Vert_Y>C[/ilmath]
- Define a new sequence [math]b_k:=\frac{1}{\sqrt{\Vert x_{n_k}-p\Vert} }[/math]
- It is easy to see that [ilmath]b_k\rightarrow +\infty[/ilmath] (as [ilmath](x_{n_k}-p)\rightarrow 0[/ilmath])
TODO: Prove that this tends to [ilmath]+\infty[/ilmath]
- Define a new sequence [math]d_k:=b_k(x_{n_k}-p)[/math]
- Clearly [math]d_k\rightarrow 0[/math]
If [ilmath](d_k)_{k=1}^\infty\rightarrow 0[/ilmath] then [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies\Vert d_k-0\Vert_X<\epsilon][/ilmath].
- Notice [math]\Vert d_k\Vert_X=b_k\Vert x_{n_k}-p\Vert_X=\frac{\Vert x_{n_k}-p\Vert_X}{\sqrt{\Vert x_{n_k}-p\Vert_X} }[/math] [math]=\sqrt{\Vert x_{n_k}-p\Vert_X}[/math]
TODO: Finish this proof
- But [math]\Vert L(d_k)\Vert_Y=\Vert L(b_k(x_{n_k}-p))\Vert_Y=b_k\Vert L(x_{n_k}-p)\Vert_Y\ge Cb_k\rightarrow +\infty[/math]
- Thus we have shown if [ilmath]L[/ilmath] is not continuous at [ilmath]p[/ilmath] that the mapping of a null sequence is unbounded, the contrapositive of what we set out to claim
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Notes
- ↑ this may be better said (perhaps) as "[ilmath]L[/ilmath] maps all sequences convergent to [ilmath]0[/ilmath] to bounded sequences"
- ↑ I believe there is an error in Maurin's Analysis book, as it suggests that it is continuous at a particular point, and lists 0 as an example, Introduction to Topology (Gamelin and Greene) suggests that only at zero is the case, also the proof in Maurin's book only works for continuity at 0, not an arbitrary point
References
- ↑ 1.0 1.1 1.2 1.3 1.4 Analysis - Part 1: Elements - Krzysztof Maurin
- ↑ 2.0 2.1 2.2 Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene
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