View source for Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/1 implies 2

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{{Requires proof|Needs some finishing off with the conclusion}}
==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}} maps a sequence, {{M|1=(x_n)_{n=1}^\infty\rightarrow 0}} (a [[null sequence]]) to a [[bounded sequence]] '''then'''
* {{M|L}} is continuous at some {{M|p\in X}}
==Proof==
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This is a proof by [[contrapositive]]. That is we will show that if {{M|L}} is not continuous at {{M|p}} {{M|\implies}} {{M|L}} takes a null sequence to one that isn't bounded (an unbounded one).
* Let the normed spaces {{M|X}} and {{M|Y}} be given, as well as a [[linear map]] {{M|L:X\rightarrow Y}}
** Suppose that {{M|L}} is not [[continuous]] at {{M|p}}, this means:
{{Begin Notebox}}
:* {{M|1=\exists (x_n)_{n=1}^\infty\rightarrow p}} such that {{MM|1=\lim_{n\rightarrow\infty}\left(L(x_n)\right)\ne L(p)}}
{{Begin Notebox Content}}
Recall that [[continuity]] states that:
* {{M|L}} is continuous at {{M|p}} {{M|\iff}} {{MM|1=\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]}}
So it follows that to ''not'' be continuous at {{M|p}}:
* {{MM|1=\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]}}, by [[negation of implies]]. Additionally we may negate the {{M|1= =}} and thus we see this is the same as:
** {{MM|1=\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]}}
Which is exactly "there exists a [[sequence]] in {{M|X}} whose limit is {{M|p}} and where {{M|1=\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)}}"
{{End Notebox Content}}{{End Notebox}}
:* Let us now take {{M|L(x_n)\not\rightarrow L(p)}} and subtract {{M|L(p)}} from both sides. We see:
:** {{M|L(x_n)-L(p)\not\rightarrow L(p)-L(p)}}, using the fact that {{M|L}} is linear we see that:
:*** {{M|L(x_n-p)\not\rightarrow L(0)}} and {{M|1=L(0)=0\in Y}} so:
:** {{M|1=L(x_n-p)\not\rightarrow 0}}
:* Thus {{M|1=\Vert L(x_n-p)\Vert_Y\not\rightarrow 0}} (as {{M|1=\Vert0\Vert_Y=0}} by definition)
{{Begin Notebox}}
:* So {{MM|1=\exists C>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge\Vert L(x_n-p)\Vert_Y>\epsilon]}}
{{Begin Notebox Content}}
If a sequence converges to {{M|0}} then we have:
* {{M|\forall\epsilon>0\ \exists N\in\mathbb{N}\ \forall n\in\mathbb{N}[n>N\implies d(x_n,x)<\epsilon]}} we know we ''don't'' have this, so we negate it:
* {{M|\exists\epsilon>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge d(x_n,x)>\epsilon]}}.
That is there ''exists an {{M|\epsilon>0}}'' such that for all {{M|N}} there exists a bigger {{M|n}} such that {{M|d(x_n,x)>\epsilon}}, we shall later call such an {{M|\epsilon}} {{M|C}} and construct a subsequence out of the {{M|n}}s
{{End Notebox Content}}{{End Notebox}}
:* Thus it is possible to construct a subsequence, {{M|1=(\Vert L(x_{n_k}-p)\Vert_Y)_{k=1}^\infty}} of the image {{M|(x_n)}} where for every {{M|k}} we have:
{{Begin Notebox}}
::* {{M|\Vert L(x_{n_k}-p)\Vert_Y>C}}
{{Begin Notebox Content}}
By the negation of convergent sequence we see that there exists a {{M|C}} such that for all {{M|N\in\mathbb{N} }} there exists another natural number, {{M|n}} such that {{M|\Vert L(x_n-p)\Vert_Y>C}}, we construct a sequence, {{M|1=(n_k)_{k=1}^\infty}} of such {{N|n}}s. That is we know there exists a {{M|C}}, so we pick {{M|1=N=1}} and get the {{M|n}} that works, this is our first term, we then set {{M|1=N=2}} and the {{M|n}} that works is our second term, and so forth. 

Some of these {{M|n}}s may be the same, but that doesn't matter.

{{M|(x_{n_k})}} is the [[subsequence]] of {{M|(x_n)}} which contains only the terms that satisfy {{M|\Vert L(x_n-p)\Vert_Y> C}}

It would have been better if I used {{M|n}} and {{M|m}} as terms, to make which {{M|n}} I am talking about clearer, but a reader able to attempt this proof should follow.
{{End Notebox Content}}{{End Notebox}}
:* We now have a [[sequence]] {{M|(x_{n_k})}} such that {{M|\Vert L(x_{n_k}-p)\Vert_Y>C}}
:* Define a new sequence {{MM|1=b_k:=\frac{1}{\sqrt{\Vert x_{n_k}-p\Vert} } }}
:** It is easy to see that {{M|b_k\rightarrow +\infty}} (as {{M|(x_{n_k}-p)\rightarrow 0}}) 
:* Define a new sequence {{MM|1=d_k:=b_k(x_{n_k}-p) }}
{{Todo|How does {{M|d_k\rightarrow 0\in X}}? Use the metric induced by the {{M|\Vert\cdot\Vert_Y}}?}}
:* Clearly {{MM|1=\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} } }} {{MM|1= =\sqrt{\Vert x_{n_k}-p\Vert_X} \rightarrow 0}}
:* But {{MM|1=\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}}
* Thus we have shown if {{M|L}} is not continuous at {{M|p}} that the mapping of a null sequence is unbounded, the [[contrapositive]] of what we set out to claim
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{{Theorem Of|Linear Algebra|Functional Analysis}}
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