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I can prove the contrapositive:

$x_n$ does not tend to $0$ implies either:

  1. $x_n$ diverges (does not converge), in which case neither does $\|x_n\|$, or
  2. $x_n$ converges to $x \neq 0$ which implies $\|x_n\|$ converges to $\|x\|\neq 0$.

Okay, but is there a simpler way to do this?

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Write down what it means to have $x_n \to 0$! –  Dylan Moreland May 30 '12 at 18:27
In some works, $\|x_n\| \to 0$ would be the definition of $x_n \to 0$. What is your definition of $x_n \to 0$? –  Nate Eldredge May 30 '12 at 18:27
Also, your statement 1 is wrong. It is possible that $x_n$ diverges but $\|x_n\|$ converges. Consider, in $\mathbb{R}$, the sequence $x_n = (-1)^n$. –  Nate Eldredge May 30 '12 at 18:28
Or worse, $x_n=e^{in}$ as a sequence of complex numbers... –  Asaf Karagila May 30 '12 at 18:30
Statement 1 should be "In which case ||x_n|| does not tend to 0" obviously –  Adam Rubinson May 30 '12 at 18:49
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1 Answer

up vote 11 down vote accepted

You can prove this directly:

Recall that $x_n\to x$ if for every $\epsilon>0$ there is some $n_0$ such that for all $n>n_0$ we have $\|x_n-x\|<\epsilon$.

If $\|x_n\|\to 0$ this means that $\|x_n-0\|$ tends to zero and therefore $x_n$ converges to $0$.

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Yeah it was that last line I couldn't see. The reason this is not completely obvious to me is that you can only garuntee this works for 0. For all other points it isn't true... –  Adam Rubinson May 30 '12 at 18:52
@Adam: Usually writing down the definitions solves a lot of problems. –  Asaf Karagila May 30 '12 at 18:53
I might vote myself down for not seeing that trick - it is too easy and I should spot it even though I have only seen it the other way round before... edit: you're not allowed to vote on yourself! –  Adam Rubinson May 30 '12 at 18:54
Asaf: From my perspective, I have been prolongued to the continuity of ||.||, i.e. the ||x_n - x||--> 0 implies | ||x_n|| - ||x|| | --> 0 implies ||x_n|| --> ||x||. It is funny that although the other argument only works for 0 and so I would expect it to be some lengthy argument involving the Kernel etc, it is really simple... but yeah...what can I say lol –  Adam Rubinson May 30 '12 at 18:57
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