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

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

Proof

The key to this proof is exploiting the linearity of $L$. As will be explained in the blue box.

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

[Expand]

We now know we may assume that $\forall\epsilon>0\ \exists\delta>0[\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon]$ - 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 $L$.

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

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

• At some point we'll be given an arbitrary $x\in X$

However we have a $\delta$ such that $\Vert u\Vert_X<\delta\implies\Vert Lu\Vert_Y<\epsilon$

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

Getting an arbitrary $\Vert x\Vert_X$ to have magnitude $<\delta$

1. Lets normalise $x$ 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 $1$ we can get it to within $\delta$ 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 $\epsilon$ was fixed, and so was the $\delta$ we know to exist based off of this, so set $A=\frac{2\epsilon}{\delta}$ after fixing them and the result will follow!

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