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.



Example of sequences with different limits for two norms
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.


