# Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/4 implies 1

From Maths

## 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 everywhere continuous then
- [ilmath]L[/ilmath] maps every null sequence to a bounded sequence

## Proof

There is actually a slightly stronger result to be had here, I shall prove that, to which the above statement is a corollary.

- Let [ilmath](x_n)_{n=1}^\infty\rightarrow x[/ilmath] be a sequence that converges - we must show that the image of this sequence under [ilmath]L[/ilmath] is bounded
- As [ilmath]L[/ilmath] is everywhere continuous we know that it is sequentially continuous at [ilmath]x[/ilmath]. This means that:
- [math]\forall\left((x_n)_{n=1}^\infty\rightarrow x\right)\left[\big(L(x_n)\big)_{n=1}^\infty\rightarrow L(x)\right][/math]

- So [ilmath]L[/ilmath] maps [ilmath](x_n)_{n=1}^\infty\rightarrow x[/ilmath] to [math]\left(L(x_n)\right)_{n=1}^\infty\rightarrow L(x)[/math]
- Recall that if a sequence converges it is bounded, as [math]\left(L(x_n)\right)_{n=1}^\infty[/math] converges, it must therefore be bounded.

- As [ilmath]L[/ilmath] is everywhere continuous we know that it is sequentially continuous at [ilmath]x[/ilmath]. This means that:

**Corollary: ** the image of every null sequence under [ilmath]L[/ilmath] is bounded

- A null sequence is just a sequence that converges to [ilmath]0[/ilmath]. If
- We are given a [math](x_n)_{n=1}^\infty\rightarrow 0[/math] then:
- We know that [math]\left(L(x_n)\right)_{n=1}^\infty\rightarrow L(0) = 0[/math] (as [ilmath]L[/ilmath] is a linear map)

- Again, if a sequence converges it is bounded

- We are given a [math](x_n)_{n=1}^\infty\rightarrow 0[/math] then:

This completes the proof.