Let $X$ be a vector space, and $\|\cdot\|_1$ and $\|\cdot\|_2$ two different (non-equivalent) Norms on $X.$ Let $(x_n)\subset X$ be a sequence and $x\in X$ such that $\lim_{n\to\infty}\|x_n-x\|_1=0.$ The question is: If $\lim_{n\to\infty}\|x_n-x\|_2>0$ can you say that the sequence $(x_n)$ does not converge (to any other limit) with respect to $\|\cdot\|_2$? Proof?, example? Thanks in advance!
Does the limit of a convergent sequence depend on the norm?
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analysis
convergence
norm
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0Is $X$ a finite dimensional space or infinite dimensional space? – 2012-12-05
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0@Mhenni: As you rightly pointed out, the hypothesis implies that $X$ must be infinite dimensional. – 2012-12-05
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0Albert: Note that $\lim\limits_{n\to\infty}\|x_n-x\|_2>0$ is not the negation of $\lim\limits_{n\to\infty}\|x_n-x\|_2=0$. You just want to say that $\|x_n-x\|_2$ does not converge to $0$, i.e., that $(x_n)$ does not converge to $x$ in $\|\cdot\|_2$, which does not imply that $\lim\limits_{n\to\infty}\|x_n-x\|_2$ exists. (However, it is equivalent to the limsup being positive.) – 2012-12-05
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0You are right, thanks for pointing that out! – 2012-12-06
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0In finite dimensions space the norms are equivalent [here](http://www.math.colostate.edu/~yzhou/course/math560_fall2011/norm_equiv.pdf) . However, that is not true in infinite dimensions spaces. See [here](http://www.physicsforums.com/showthread.php?t=451562). – 2013-04-30