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

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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:


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.

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

This completes the proof.