Example of sequences with different limits for two norms

Question

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.

Answer 1

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.

Answer 2

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 $\ell_2$.

Answer 3

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 $y \neq x$.

Answer 4

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 $Q\neq0$.

Answer 5

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.