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

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

Proof

This is a proof by contrapositive. That is we will show that if $L$ is not continuous at $p$ $\implies$ $L$ takes a null sequence to one that isn't bounded (an unbounded one).

• Let the normed spaces $X$ and $Y$ be given, as well as a linear map $L:X\rightarrow Y$
  • Suppose that $L$ is not continuous at $p$, this means:

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• $\exists (x_n)_{n=1}^\infty\rightarrow p$ such that $$\lim_{n\rightarrow\infty}\left(L(x_n)\right)\ne L(p)$$

Recall that continuity states that:

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

So it follows that to not be continuous at $p$:

• $$\exists (x_n)_{n=1}^\infty\subseteq X\left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\wedge\neg\left(\lim_{n\rightarrow\infty}(L(x_n))=L(p)\right)\right]$$ , by negation of implies. Additionally we may negate the $=$ 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 $X$ whose limit is $p$ and where $\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)$"

• Let us now take $L(x_n)\not\rightarrow L(p)$ and subtract $L(p)$ from both sides. We see:
  • $L(x_n)-L(p)\not\rightarrow L(p)-L(p)$, using the fact that $L$ is linear we see that:
    • $L(x_n-p)\not\rightarrow L(0)$ and $L(0)=0\in Y$ so:
  • $L(x_n-p)\not\rightarrow 0$
• Thus $\Vert L(x_n-p)\Vert_Y\not\rightarrow 0$ (as $\Vert0\Vert_Y=0$ by definition)

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• 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 $0$ then we have:

• $\forall\epsilon>0\ \exists N\in\mathbb{N}\ \forall n\in\mathbb{N}[n>N\implies d(x_n,x)<\epsilon]$ we know we don't have this, so we negate it:
• $\exists\epsilon>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge d(x_n,x)>\epsilon]$.

That is there exists an $\epsilon>0$ such that for all $N$ there exists a bigger $n$ such that $d(x_n,x)>\epsilon$, we shall later call such an $\epsilon$ $C$ and construct a subsequence out of the $n$s

• Thus it is possible to construct a subsequence, $(\Vert L(x_{n_k}-p)\Vert_Y)_{k=1}^\infty$ of the image $(x_n)$ where for every $k$ we have:

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• $\Vert L(x_{n_k}-p)\Vert_Y>C$

By the negation of convergent sequence we see that there exists a $C$ such that for all $N\in\mathbb{N}$ there exists another natural number, $n$ such that $\Vert L(x_n-p)\Vert_Y>C$, we construct a sequence, $(n_k)_{k=1}^\infty$ of such $n$-ns. That is we know there exists a $C$, so we pick $N=1$ and get the $n$ that works, this is our first term, we then set $N=2$ and the $n$ that works is our second term, and so forth.

Some of these $n$s may be the same, but that doesn't matter.

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

It would have been better if I used $n$ and $m$ as terms, to make which $n$ I am talking about clearer, but a reader able to attempt this proof should follow.

• We now have a sequence $(x_{n_k})$ such that $\Vert L(x_{n_k}-p)\Vert_Y>C$
• Define a new sequence $$b_k:=\frac{1}{\sqrt{\Vert x_{n_k}-p\Vert} }$$
  • It is easy to see that $b_k\rightarrow +\infty$ (as $(x_{n_k}-p)\rightarrow 0$)

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TODO: Prove that this tends to $+\infty$

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• Define a new sequence $$d_k:=b_k(x_{n_k}-p)$$

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• Clearly $$d_k\rightarrow 0$$

If $(d_k)_{k=1}^\infty\rightarrow 0$ then $\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies\Vert d_k-0\Vert_X<\epsilon]$.

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

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TODO: Finish this proof

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• 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 $L$ is not continuous at $p$ that the mapping of a null sequence is unbounded, the contrapositive of what we set out to claim

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TODO: At the bottom, explain how $\Vert L(d_k)\Vert_Y$ being unbounded related to $L(d_k)$

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