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

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 everywhere continuous then
• $L$ 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 $(x_n)_{n=1}^\infty\rightarrow x$ be a sequence that converges - we must show that the image of this sequence under $L$ is bounded
  • As $L$ is everywhere continuous we know that it is sequentially continuous at $x$. This means that:
    • $$\forall\left((x_n)_{n=1}^\infty\rightarrow x\right)\left[\big(L(x_n)\big)_{n=1}^\infty\rightarrow L(x)\right]$$
  • So $L$ maps $(x_n)_{n=1}^\infty\rightarrow x$ to $$\left(L(x_n)\right)_{n=1}^\infty\rightarrow L(x)$$
  • Recall that if a sequence converges it is bounded, as $$\left(L(x_n)\right)_{n=1}^\infty$$ converges, it must therefore be bounded.

Corollary: the image of every null sequence under $L$ is bounded

• A null sequence is just a sequence that converges to $0$. If
  • We are given a $$(x_n)_{n=1}^\infty\rightarrow 0$$ then:
    • We know that $$\left(L(x_n)\right)_{n=1}^\infty\rightarrow L(0) = 0$$ (as $L$ is a linear map)
  • Again, if a sequence converges it is bounded

This completes the proof.