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I'm reading the paper by Dijkstra and van Mill called Topological equivalence of discontinuous norms, and in its introduction there is this:

Consider the case of $\ell^p$ and $\ell^q$, where $p<q$. Then as vector spaces, $\ell^p$ and $\ell^q$ are isomorphic (they have Hamel basis of the same cardinality) and so under this equivalence the norm on $\ell^q$ defines a norm on $\ell^p$, which is badly discontinuous of course.

Now, i get how the norm is constructed, by I absolutely don't understand, why it is "badly discontinuous of course". I tried driving the continuousnes to contradiction by applying Pitt's compactness theorem on the isomorphism and then using the compactnes on sequence of vectors with one coordinate being 1 (changing places) and the others being 0, but I can't get to prove boundedness of the isomorphism (taken as going from $\ell^q$ to $\ell^p$, I can prove boundedness in the other direction quite easily, considering continuity of the norm we want to contradict).

What can I do?

Rado Švarc
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1 Answers1

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Suppose you have two Banach spaces $(X,\|\cdot\|_X), (Y,\|\cdot\|_Y)$ that are isomorphic as vector spaces, that is there is a linear bijective map $F:X\to Y$. What you are doing is using this isomorphism to pull the norm of $Y$ back to $X$, that is defining a norm $\|\cdot \|_{F^*(Y)}: X\to\Bbb R$ by $\|x\|_{F^*(Y)} = \|F(x)\|_Y$.

Now suppose the map $\|\cdot\|_{F^*(Y)}$ is continuous and let $x_n\to x$, which means that $x_n-x\to 0$ and then by continuity of $\|\cdot\|_{F^*(Y)}$ one gets $\|x_n-x\|_{F^*(Y)} =\|F(x_n-x)\|_Y = \|F(x_n)-F(x)\|_Y \to \|0\|_{F^*(Y)}=0$, ie $F(x_n)\to F(x)$. Since this holds for any convergent sequence $x_n\to x$, this just means that $F$ is continuous.

So $F:X\to Y$ is a continuous bijective linear map between two Banach spaces. By the open mapping theorem $F$ must also have a continuous inverse and $X$ and $Y$ must be isomorphic as topological vector spaces.

Now in your special scenario you are looking at $\ell^p$ and $\ell^q$ with their usual norms. If the pull-back of one of the norms to the other space is continuous you will find that the two spaces must be isomorphic (in the sense that there is a linear homeomorphism between them). However that cannot be true unless $p=q$, this classical fact is discussed in more detail in this answer.

s.harp
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