Difference between revisions of "Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/1 implies 2"

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

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

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.

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

TODO: Finish this proof