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Let $E$ be a vector space (over $\mathbb{R}$) with a positive definite hermitian form and let $\{x_{n}\}$ ($x_{n} \not= 0$) be a sequence converging to $x$ in the $L^{2}$ norm $\lvert\cdot\rvert=\sqrt{\langle\cdot,\cdot\rangle}$. Why does $\left\{x_{n}/\lvert x_{n}\rvert\right\}$ converge to $x/\lvert x \rvert$?

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    $\begingroup$ Short answer: The norm is continuous, scalar multiplication is continuous. Caveat: You must assume that $x\neq 0$, or things don't make sense. $\endgroup$ – Daniel Fischer Jun 13 '14 at 20:24
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What you are really asking is why $|x_n| \rightarrow |x|$ the rest is just the limit theorem of elementary analysis.

Now we can use the inequality, which is an application of the triangle inequality,

$$||x_n|-|x||\leq |x_n-x|$$ to prove the limit.

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  • $\begingroup$ @jaebond it seems so but its kind of circular, since convergence is defined in terms of that norm. $\endgroup$ – Rene Schipperus Jul 7 '14 at 19:46
  • $\begingroup$ Since this equality holds for any norm, is this how one would show that every norm is continuous? $\endgroup$ – Jacob Bond Jul 7 '14 at 19:47

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