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For a normed space $X$, when does $$ \lim_{n \to \infty} \| x_n \|_X = \| x \|_X$$ imply $$ \lim_{n \to \infty} \| x_n - x \|_X = 0 \;\;?$$

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First of all, this would mean that if $||x||=||y||$ then $||x-y||=0$, and hence $x=y$. So your norm would have to be $1-1$ – Thomas Andrews Nov 13 '12 at 14:34
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Following on Thomas Andrews' comment, suppose $X$ has the property that $\lim_n\|x_n\|=\|x\|$ implies that $\lim_n\|x_n-x\|=0$. By considering constant sequences, this shows that $\|x\|=\|y\|$ implies $x=y$. Then $X$ is one-dimensional: indeed fix $x_0\in X$ with $x_0\ne0$. Write $c=\|x_0\|$. For any other nonzero $x\in X$, $\|x\|=\|\,\|x\|/c\,x_0\|$, so $x=\|x\|/c\,x_0$.

Now using Tao's example, we consider the sequence $\{(-1)^nx_0\}$. As $\|(-1)^nx_0\|=\|x_0\|$, we get $\lim_n\|(-1)^nx_0-x_0\|=0$. The only way this can converge is when $x_0=0$.

So the only possible space with this property is the trivial space $\{0\}$.

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maybe this is wrong counterexample: $x_n=-1,1,-1,1,-1,1----$, then we take the norm the absolute value. $X=R$.......

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the converse of the result is right duo to the simple inequality – Tao Nov 13 '12 at 14:23

Just take $X= \mathbb{R}$ with the Euclidean norm. Then $x_n= -1$ and $x=1$ satisfies the first but not the second requirement.

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