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

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Needs some finishing off with the conclusion

## 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] maps a sequence, [ilmath](x_n)_{n=1}^\infty\rightarrow 0[/ilmath] (a null sequence) to a bounded sequence then
• [ilmath]L[/ilmath] is continuous at some [ilmath]p\in X[/ilmath]

## Proof

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 $\lim_{n\rightarrow\infty}\left(L(x_n)\right)\ne L(p)$

Recall that continuity states that:

• [ilmath]L[/ilmath] is continuous at [ilmath]p[/ilmath] [ilmath]\iff[/ilmath] $\forall(x_n)_{n=1}^\infty\subseteq X \left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\implies\left(\lim_{n\rightarrow\infty}\left(L(x_n)\right)=L\left(\lim_{n\rightarrow\infty}(x_n)\right)=L(p)\right)\right]$

So it follows that to not be continuous at [ilmath]p[/ilmath]:

• $\exists (x_n)_{n=1}^\infty\subseteq X\left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\wedge\neg\left(\lim_{n\rightarrow\infty}(L(x_n))=L(p)\right)\right]$, by negation of implies. Additionally we may negate the [ilmath]=[/ilmath] and thus we see this is the same as:
• $\exists (x_n)_{n=1}^\infty\subseteq X\left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\wedge\left(\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)\right)\right]$

Which is exactly "there exists a sequence in [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]
• Thus [ilmath]\Vert L(x_n-p)\Vert_Y\not\rightarrow 0[/ilmath] (as [ilmath]\Vert0\Vert_Y=0[/ilmath] by definition)
• So $\exists C>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge\Vert L(x_n-p)\Vert_Y>\epsilon]$

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 $b_k:=\frac{1}{\sqrt{\Vert x_{n_k}-p\Vert} }$
• 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 $d_k:=b_k(x_{n_k}-p)$
• Clearly $d_k\rightarrow 0$

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 $\Vert d_k\Vert_X=b_k\Vert x_{n_k}-p\Vert_X=\frac{\Vert x_{n_k}-p\Vert_X}{\sqrt{\Vert x_{n_k}-p\Vert_X} }$ $=\sqrt{\Vert x_{n_k}-p\Vert_X}$

TODO: Finish this proof

• But $\Vert L(d_k)\Vert_Y=\Vert L(b_k(x_{n_k}-p))\Vert_Y=b_k\Vert L(x_{n_k}-p)\Vert_Y\ge Cb_k\rightarrow +\infty$
• Thus we have shown if [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

TODO: At the bottom, explain how [ilmath]\Vert L(d_k)\Vert_Y[/ilmath] being unbounded related to [ilmath]L(d_k)[/ilmath]