Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/3 implies 4

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 $L$ is a bounded linear map, that is to say;
  • $\exists A\ge 0\ \forall x\in X[\Vert L(x)\Vert_Y \le A\Vert x\Vert_X]$ then:
• $L$ is everywhere continuous

Proof

This is a straight forward implication proof. We suppose by hypothesis that $\exists A\ge 0\ \forall x\in X[\Vert L(x)\Vert_Y \le A\Vert x\Vert_X]$ is true.
We wish to show that $L$ is continuous everywhere, this means $\forall x\in X[L\ \text{is continuous at }x]$, I will use sequential continuity in this proof (that is: $\forall(x_n)_{n=1}^\infty\rightarrow x[\lim_{n\rightarrow\infty}(L(x_n))=L(x)]$, or the image of the sequence under $L$ converges to the image of $x$ under $L$)

• Let $x\in X$ be given (arbitrary point)

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• Let $(x_n)_{n=1}^\infty\rightarrow x$ be given (an arbitrary sequence that converges to $x$)

Remember that to converge means $\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(x_n,x)<\epsilon]$ and in our normed space:

• $d(x_1,x_2):=\Vert x_1-x_2\Vert_X$ (see metric induced by norm), so what we're really saying is:
• $\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies \Vert x_n-x\Vert_X<\epsilon]$

• Let $\epsilon >0$ be given (we will show that $$\left\Vert L(x_n)-L(x)\right\Vert_Y<\epsilon$$ )

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• Choose $N$ such that $$\forall n\in\mathbb{N}\left[n>N\implies\Vert x_n-x\Vert_X<\frac{\epsilon}{A}\right]$$ (Where $A$ is the $A$ in the hypothesis of $L$ being a bounded linear map)

We will want to show at some point that $$\lim_{n\rightarrow\infty}\left(L(x_n)\right)=L(x)$$ this means we'll need to do something along the lines of:

• $\Vert L(x_n)-L(x)\Vert_Y=\Vert L(x_n-x)\Vert_Y\le A\Vert x_n-x\Vert_X$

If we pick $N$ such that $$\Vert x_n-x\Vert_X<\frac{\epsilon}{A}$$ then:

• $\Vert L(x_n)-L(x)\Vert_Y=\Vert L(x_n-x)\Vert_Y\le A\Vert x_n-x\Vert_X$ $$<A\frac{\epsilon}{A}=\epsilon$$

and that'll complete the proof.
Note: we know such an $N$ exists as by definition $$\lim_{n\rightarrow\infty}(x_n)=x$$

• Let $n\in\mathbb{N}$ be given

• If $n\le N$ then we're done. By the truth table of implies if the RHS ($\Vert L(x_n)-L(x)\Vert_Y<\epsilon$) is true or false, the implication is true, so this literally doesn't matter.
• If $n>N$ then:

• $$\Vert L(x_n)-L(x)\Vert_Y=\overbrace{\Vert L(x_n-x)\Vert_Y}^\text{by linearity}$$ $$\le\overbrace{A\Vert x_n-x\Vert_X}^{\text{boundedness of }L}$$ $$<A\frac{\epsilon}{A}=\epsilon$$
• Thus we have shown $$n>N\implies \Vert L(x_n)-L(x)\Vert_Y<\epsilon$$ as required.

That completes the proof