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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!

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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
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1 Answer

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.

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Dear Jonas, thank you for the link. The statement of the problem is indeed the same, but I am not sure if the given answer corresponds to that question. In all the examples quoted there, they build a sequence such that in one norm it converges (to zero) and in the other norm it "converges" to an element not in the space... then, by definition, the sequence does not converge (the limit must be an element of the space), and it is no counterexample. Taking the compleition with respect to the second norm does not solve the problem, since this makes the first norm not well defined on that set. –  Albert Dec 6 '12 at 10:45
@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
You are right! I'm sorry. Also the example from Mar 23 2011 at 8:44 works in a very simple way. Thanks again! –  Albert Dec 6 '12 at 23:24
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