# Does the limit of a convergent sequence depend on the norm?

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!

-
Is $X$ a finite dimensional space or infinite dimensional space? – Mhenni Benghorbal Dec 5 '12 at 8:39
@Mhenni: As you rightly pointed out, the hypothesis implies that $X$ must be infinite dimensional. – Jonas Meyer Dec 5 '12 at 8:40
Albert: 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.) – Jonas Meyer Dec 5 '12 at 8:56
You are right, thanks for pointing that out! – Albert Dec 6 '12 at 10:46
In finite dimensions space the norms are equivalent here . However, that is not true in infinite dimensions spaces. See here. – Mhenni Benghorbal Apr 30 '13 at 5:55

This was asked and answered in a MathOverflow question, Example of sequences with different limits for two norms. The answer is that there are examples where $(x_n)$ converges in both norms, to different limits.
@Albert: That is incorrect. The answers give distinct limits that exist within a single set, with the two different norms. Take the accepted answer by Denis Serre, for example. The limit in $\|\cdot\|_1$ is the zero function, and the limit in $\|\cdot\|_2$ is the constant function always equal to $1$. Both of these are trigonometric polynomials functions, hence in the space, and they are the (unique) limits for the respective norms of the same sequence. The fact that the sequence is constructed from a function not in the space is not a problem; that function is not either of the limits. – Jonas Meyer Dec 6 '12 at 15:13