# Example of sequences with different limits for two norms

I was explaining to my students that if there is an inequality between two norms, then there is an inclusion between their spaces of convergent sequences, with matching limits. I then proceeded to show examples of such inequalities on the normed spaces they knew, and counterexamples of sequences which converge for a norm and not for another, stating the equivalence of norms in finite dimension, etc.

It is then that I wondered about the following : does there exist a vector space, two norms on that vector space and a single sequence which converges for both norms, but with different limits?

The first remark is that such a counter-example cannot exist in finite dimension ; and one first has to find "really inequivalent norms", which do exist : consider the space of polynomials in one variable, and define norms on it by summing the absolute values of the coefficients :

• first with a weight $1$ for every coefficient ;
• second with $2^n$ or $2^{-n}$ depending on the parity of the degree $n$.

It's now easy to find a sequence going to zero for the first and not for the second, and a sequence going to zero for the second and not for the first - so there can't be an inequality between those.

Notice this is all over the real or complex numbers, though the question could be amusing in a more general setting.

• In Koblitz's book on p-adic analysis, in the chapter on power series, he gives an example of an infinite series of rational numbers which converges in both R and some Q_p (maybe p = 2?) and the limits are different rational numbers. Using Zorn's lemma, Q_p can be embedded into C and the p-adic abs. value on Q_p can be extended to an absolute value on C. Therefore C, as a vector space over Q, equipped with its usual absolute value and a (non-constructive) extension of the p-adic absolute value, admits a sequence which converges for both norms, but with different limits. – KConrad Feb 2 '11 at 4:27
• That's another interesting example ; even higher level than Bill Johnson's, but good. – Julien Puydt Feb 2 '11 at 11:30
• Dear Julien Puydt, As you said:" consider the space of polynomials in one variable, and define norms on it by summing the absolute values of the coefficients : first with a weight 1 for every coefficient ; second with 2n or 2−n depending on the parity of the degree n." Could define the sequence so that the two limit in these two different norms are different ? Thanks a lot! – zhongjie Sep 11 '15 at 8:53
• @zhongjie: defining a single sequence with two different limits for two different norms is precisely what the question is about! – Julien Puydt Sep 11 '15 at 14:55
• Two $L^1$ norms on $[0,1]$ w.rto two mutually singular measures. – Pietro Majer Mar 2 at 0:08

Consider the space $X$ of trigonometric polynomials (with period $1$, say). Choose the norms $$\|f\|_1=\sup\{|f(x)|;\frac16\le x\le\frac13\},\qquad \|f\|_2=\sup\{|f(x)|;\frac23\le x\le\frac56\}.$$ Now consider the partial sums $f_N$ of the Fourier series of the periodic function $F$ defined by $F(x)=0$ if $x\in(0,1/2)$ and $F(x)=1$ if $x\in(1/2,1)$. In the first norm, $f_N$ converges to $g\equiv0$, whereas in the second one, $f_N$ converges to $h\equiv1$.

Remark that $F$ does not belong to $X$, but this has no importance at all. Perhaps it is even natural in order to construct practical examples.

• A remark: I do get the feeling (with your last paragraph in mind), this example is showing more that for a fixed vector space $X$ and two different norms w.r.t. which $X$ is not complete, the completion of $X$ by the two norms are not necessarily the same. – Willie Wong Feb 1 '11 at 13:55
• @Willie. Of course they aren't. If a sequence has two distinct limits for two norms, then these norms are not comparable. Not only the completions are distinct, but there is not natural embedding from one to the other. Here the completions are $C([1/6,1/3])$ and $C[2/3,5/6])$ thanks to Stone-Weierstrass. – Denis Serre Feb 1 '11 at 14:05
• @Denis: I don't think I phrased my remark quite the way I meant to. Let me try again (hope you don't mind). I guess what I am trying to say is similar to the point you just made in the comments. What I think I want to say is that I think there is a conceptual difference (which may just be something that is in my head only) between a set $E$ and two topologies $T,S$, and a sequence such that $x_k \to x$ in $(E,S)$ and $x_k \to y$ in $(E,T)$ with $x\neq y$, $x,y\in E$; and the case of two sets $X,Y$ such that $X\cap Y = E$, a topology $S$ on $X$, a topology $T$ on $Y$, and a sequence $(x_k)$ – Willie Wong Feb 1 '11 at 14:28
• Instead of $X$ (in the answer), it seems that you could easily take a much more general space---for example, the space of bounded real functions. Then you don't need the remark in the last paragraph. – John Bentin Feb 3 '11 at 22:25
• In a late response to John Bentin's comment, if $X$ were replaced by the space of bounded real functions, then the norms would only be seminorms, and would not satisfy $\|x\|=0\implies x=0$. Limits would not be unique. – Jonas Meyer Dec 5 '12 at 14:50

Note first that your example spaces cannot give what you want because in both spaces the coordinate evaluation functionals are continuous and separate points.

Examples are easy. Take in $\ell_2$ a linearly independent sequence that converges to a non zero vector, such as $x_n := e_1 + n^{-1}e_n$, $n=2,3,...$. Map $x_n$ to $n^{-1}e_n$ in $\ell_2$ and extend to a linear isomorphism from $\ell_2$ onto $\ell_2$.

• The moral of this is "Any independent sequence can be made to converge to zero". – Ewan Delanoy Feb 1 '11 at 13:04
• The first note is nice -- my norms weren't smelling good and I didn't know why ; now I know! I should have thought about it. For your example, I'm not sure I got everything ; I assume $e_n$ is the sequence which takes the value $1$ in rank $n-1$ (where $n\geq1$) and zero elsewhere. The family $(e_n)_{n\geq1}$ is orthonormal and generates a dense subspace. The family $(x_n)_{n\geq1}$ generates the same subspace, and is free. Good. – Julien Puydt Feb 1 '11 at 20:44
• I can then define a linear operator on that subspace like you explain. If I can extend it to the whole space and keep it injective, I guess I'll be able to skew the usual norm with that. But it's not clear how I'll extend it... – Julien Puydt Feb 1 '11 at 20:45
• You just take $A$ and $B$ so that $A\cup (x_n)$ and $B \cup (e_n)$ are Hamel bases for $\ell_2$, map $A$ one to one onto $B$, and extend linearly. – Bill Johnson Feb 1 '11 at 21:14
• Ah, so you were thinking about an algebraic extension - I was wondering if you hadn't something like the density in mind, in which case it was more problematic. Good! – Julien Puydt Feb 1 '11 at 21:28

I was given IRL another beautiful answer to that question, and thought it would be nice to share.

Consider the space $\mathbb{K}[X]$ and a polynom $Q\neq0$ of degree $m$ ; define a new basis for the space by considering $\mathcal{B}_Q=1,X,\dots,X^m,X^{m+1}-Q,X^{m+2}-Q,\dots$, then a norm by $N_Q(P)=\sup_{n\in\mathbb N}\frac1{2^n}|a_n|$ where $(a_n)_{n\in\mathbb{N}}$ are the coefficients of $P$ in $\mathcal{B}_Q$.

The same sequence $(X^n)_{n\in\mathbb{N}}$ now converges to $Q$ for $N_Q$ for each $Q\neq0$.

Just thinking out loud - this could be totally irrelevant - you can find a sequence of integers $b_0,b_1,\dots$ such that $0\le b_i\le4$ and $(b_0+5b_1/7+25b_2/49+\dots+5^nb_n/7^n)^2\equiv-1\pmod{5^{n+1}}$. That makes the series $b_0+5b_1/7+25b_2/49+\dots$ converge, in the 5-adic norm, to a number whose square is minus one. In the usual norm on the rationals, the series converges to some real number, most assuredly not a square root of minus one.

• Well, it's not 100% irrelevant, but close. ;-) The original question is over the real or complex numbers. Even the generalization I had in mind in my last sentence didn't involve touching the valuation of the base field to find a (counter)example, but merely going over to other base fields. – Julien Puydt Feb 1 '11 at 11:50
• Similarly irrelevant examples can be found by considering ${\mathbb Q}(\sqrt{2})$, with two inequivalent topologies defined by the induced topology on either ${\mathbb R}$ or ${\mathbb R}^2$. – B R Feb 1 '11 at 18:17

I'm not sure if you would consider this an "example", but it seems this is not as widely known as I would expect.

Theorem. Let $$X$$ be a Banach space and let $$||\cdot||_1, ||\cdot||_2$$ be non-equivalent norms on $$X$$. Then there exists a sequence $$(x_n)$$ in $$X$$ and $$x\neq y \in X$$ such that $$x_n \to x$$ with respect to $$||\cdot||_1$$ and $$x_n \to y$$ with respect to $$||\cdot||_2$$.

Proof. By the bounded inverse theorem it must be that the identity map $$\iota: (X, ||\cdot||_1) \to (X, ||\cdot||_2)$$ is discontinuous. Thus by the closed graph theorem it must be that the graph of $$\iota$$ is not closed in $$(X, ||\cdot||_1) \times (X, ||\cdot||_2)$$. Since the graph is not closed, we may choose a sequence $$(x_n, \iota(x_n))$$ in the graph converging in $$(X, ||\cdot||_1) \times (X, ||\cdot||_2)$$ to some $$(x,y)$$ such that $$(x,y)$$ is not in the graph. Converging in $$(X, ||\cdot||_1) \times (X, ||\cdot||_2)$$ means that $$x_n \to x$$ in $$(X, ||\cdot||_1)$$ and $$\iota(x_n) = x_n \to y$$ in $$(X, ||\cdot||_2)$$. But $$(x,y)$$ not being in the graph means that $$y \neq x$$.