# Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/2 implies 3

## 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 continuous at a point (say [ilmath]p\in X[/ilmath]) then
• [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]

## Proof

The key to this proof is exploiting the linearity of [ilmath]L[/ilmath]. As will be explained in the blue box.

• Suppose that [ilmath]L:X\rightarrow Y[/ilmath] is continuous at [ilmath]p\in X[/ilmath]. Then:
• [ilmath]\forall\epsilon>0\ \exists\delta>0[\Vert x-p\Vert_X<\delta\implies\Vert L(x-p)\Vert_Y<\epsilon][/ilmath] (note that [ilmath]L(x-p)=L(x)-L(p)[/ilmath] due to linearity of [ilmath]L[/ilmath])
• Define [ilmath]u:=x-p[/ilmath], then:
• [ilmath]\forall\epsilon>0\ \exists\delta>0[\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon][/ilmath] (writing [ilmath]Lu:=L(u)[/ilmath] as is common for linear maps)

We now know we may assume that [ilmath]\forall\epsilon>0\ \exists\delta>0[\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon][/ilmath] - see this box for a guide on how to use this information and how I constructed the rest of the proof. The remainder of the proof follows this box.

The key to this proof is in the norm structure and the linearity of [ilmath]L[/ilmath].

• Pick (fix) some [ilmath]\epsilon>0[/ilmath] we now know:
• [ilmath]\exists\delta>0[/ilmath] such that if we have [ilmath]\Vert x\Vert_X<\delta[/ilmath] then we have [ilmath]\Vert Lx\Vert<\epsilon[/ilmath]

In the proof we will have to show at some point that: [ilmath]\forall x\in X[\Vert Lx\Vert_Y\le A\Vert x\Vert_X][/ilmath] this means:

• At some point we'll be given an arbitrary [ilmath]x\in X[/ilmath]

However we have a [ilmath]\delta[/ilmath] such that [ilmath]\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon[/ilmath]

• All we have to do is make [ilmath]\Vert x\Vert_X[/ilmath] so small that it is less than [ilmath]\delta[/ilmath] - we can do this using scalar multiplication.
• Note that if [ilmath]x=0[/ilmath] the result is trivial, so assume that arbitrary [ilmath]x[/ilmath] is [ilmath]\ne 0[/ilmath]
• With this in mind the task is clear: we need to multiply [ilmath]x[/ilmath] by something such that the actual vector part has [ilmath]\Vert\cdot\Vert_X<\delta[/ilmath]
• Then we can say (supposing [ilmath]x=\alpha p[/ilmath] for some positive [ilmath]\alpha[/ilmath] and [ilmath]\Vert p\Vert_X<\delta[/ilmath]) that [ilmath]\Vert L(\alpha p)\Vert_Y=\alpha\Vert Lp\Vert_Y[/ilmath]
• But [ilmath]\Vert p\Vert_X<\delta\implies\Vert Lp\Vert_Y<\epsilon[/ilmath]
• So [ilmath]\Vert L(\alpha p)\Vert_Y=\alpha\Vert Lp\Vert_Y<\alpha\epsilon[/ilmath], if we can get a [ilmath]\Vert x\Vert_X[/ilmath] involved in [ilmath]\alpha[/ilmath] the result will follow.

Getting an arbitrary [ilmath]\Vert x\Vert_X[/ilmath] to have magnitude [ilmath]<\delta[/ilmath]

1. Lets normalise [ilmath]x[/ilmath] and then multiply it by the magnitude again (so as to do nothing)
• So notice $x=\overbrace{\Vert x\Vert_X}^\text{scalar part}\cdot\overbrace{\frac{1}{\Vert x\Vert_X}x}^\text{vector part}$,
2. Now the vector part has magnitude [ilmath]1[/ilmath] we can get it to within [ilmath]\delta[/ilmath] by multiplying it by $\frac{\delta}{2}$
• So $x=\overbrace{\Vert x\Vert_X\cdot\frac{2}{\delta} }^\text{scalar part}\cdot\overbrace{\frac{\delta}{2\Vert x\Vert_X}x}^\text{vector part}$.

Now we have $\left\Vert\frac{\delta}{2\Vert x\Vert_X}x\right\Vert_Y<\delta$ which $\implies \left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\epsilon$

• Thus $\Vert Lx\Vert_Y=\frac{2\Vert x\Vert_X}{\delta}\cdot\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\frac{2\Vert x\Vert_X}{\delta}\cdot\epsilon=\frac{2\epsilon}{\delta}\Vert x\Vert_X$

But [ilmath]\epsilon[/ilmath] was fixed, and so was the [ilmath]\delta[/ilmath] we know to exist based off of this, so set [ilmath]A=\frac{2\epsilon}{\delta}[/ilmath] after fixing them and the result will follow!

• Fix some arbitrary [ilmath]\epsilon>0[/ilmath] (it doesn't matter what)
• We know there [ilmath]\exists\delta>0[\Vert x\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon][/ilmath] - take such a [ilmath]\delta[/ilmath] (which we know to exist by hypothesis) and fix it also.
• Define [ilmath]A:=\frac{2\epsilon}{\delta}[/ilmath] (if you are unsure of where this came from, see the blue box)
• Let [ilmath]x\in A[/ilmath] be given (this is the [ilmath]\forall x\in X[/ilmath] part of our proof, we have just claimed an [ilmath]A[/ilmath] exists on the above line)
• If [ilmath]x=0[/ilmath] then
• Trivially the result is true, as [ilmath]L(0_X)=0_Y[/ilmath], and [ilmath]\Vert 0_Y\Vert_Y=0[/ilmath] by definition, [ilmath]A\Vert x\Vert_X=0[/ilmath] as [ilmath]\Vert 0_X\Vert_X=0[/ilmath] so we have [ilmath]0\le 0[/ilmath] which is true. (This is more workings than the line is worth)
• Otherwise ([ilmath]x\ne 0[/ilmath])
• Notice that $x=\frac{2\Vert x\Vert_X}{\delta}\cdot\frac{\delta}{2\Vert x\Vert_X}x$ and $\left\Vert \frac{\delta}{2\Vert x\Vert_X}x\right\Vert_X<\delta$
• and $\left\Vert \frac{\delta}{2\Vert x\Vert_X}x\right\Vert_X<\delta\implies\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y<\epsilon$
• Thus we see $\Vert Lx\Vert_Y=\left\Vert L\left(\frac{2\Vert x\Vert_X}{\delta}\cdot\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y=\frac{2\Vert x\Vert_X}{\delta}\cdot\overbrace{\left\Vert L\left(\frac{\delta}{2\Vert x\Vert_X}x\right)\right\Vert_Y}^{\text{remember this is }<\epsilon}$ $<\frac{2\Vert x\Vert_X}{\delta}\cdot\epsilon=\frac{2\epsilon}{\delta}\Vert x\Vert_X=A\Vert x\Vert_X$
• Shortening the workings this states that: $\Vert Lx\Vert_Y< A\Vert x\Vert_X$
• So if [ilmath]x=0[/ilmath] we have equality, otherwise $\Vert Lx\Vert_Y< A\Vert x\Vert_X$

In either case, it is true that $\Vert Lx\Vert_Y\le A\Vert x\Vert_X$
This completes the proof