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

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

## Proof

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

• Let [ilmath]x\in X[/ilmath] be given (arbitrary point)
• Let [ilmath](x_n)_{n=1}^\infty\rightarrow x[/ilmath] be given (an arbitrary sequence that converges to [ilmath]x[/ilmath])

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

• [ilmath]d(x_1,x_2):=\Vert x_1-x_2\Vert_X[/ilmath] (see metric induced by norm), so what we're really saying is:
• [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies \Vert x_n-x\Vert_X<\epsilon][/ilmath]
• Let [ilmath]\epsilon >0[/ilmath] be given (we will show that $\left\Vert L(x_n)-L(x)\right\Vert_Y<\epsilon$)
• Choose [ilmath]N[/ilmath] such that $\forall n\in\mathbb{N}\left[n>N\implies\Vert x_n-x\Vert_X<\frac{\epsilon}{A}\right]$ (Where [ilmath]A[/ilmath] is the [ilmath]A[/ilmath] in the hypothesis of [ilmath]L[/ilmath] 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:

• [ilmath]\Vert L(x_n)-L(x)\Vert_Y=\Vert L(x_n-x)\Vert_Y\le A\Vert x_n-x\Vert_X[/ilmath]

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

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

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

• Let [ilmath]n\in\mathbb{N} [/ilmath] be given
• If [ilmath]n\le N[/ilmath] then we're done. By the truth table of implies if the RHS ([ilmath]\Vert L(x_n)-L(x)\Vert_Y<\epsilon[/ilmath]) is true or false, the implication is true, so this literally doesn't matter.
• If [ilmath]n>N[/ilmath] 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