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The lecture notes "The open mapping theorem and related theorems" by Anton R Schep begins with a lemma that asks that the identity map $I : (X, ||\cdot||_1) \to (X, ||\cdot||_2)$ is continuous.

I want an example where the identity map is (a) everywhere defined, and (b) discontinuous. All the examples that I can think of involve imposing different norms on the same space, (say , imposing $L_p$ versus $L_q$ on $C[0, 1]$). But these kinds of impositions mean that not all elements $f \in C[0, 1]$ with finite $L_p$ norm has finite $L_q$ norm for $q > p$, and thus the identity mapping fails to be well defined everywhere in the domain.

A promising line of attack appears to be to consider a Sobolev space of functions in $C[0, 1]$ with bounded first derivative. I was then hoping to show that the differentiation operator is discontinuous. But I do not actually know a concrete example where this is the case.

I would prefer a discrete example in one of the sequence spaces, if at all possible.

Siddharth Bhat
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  • Continuous functions on a compact interval are in every $L^p$ though... and yes all the examples involve different norms on the same space because that's what you require? – Bruno B Aug 07 '23 at 08:17
  • If the identity map $I:X\to X$ should be discontinuous you need two inequivalet norms on $X.$ It is possible to find two inequivalent norms $|\cdot|_1$ and $|\cdot|_2$ such that both spaces $(X,|\cdot|_1)$ and $(X,|\cdot|_2)$ are complete. – Ryszard Szwarc Aug 07 '23 at 08:28
  • @BrunoB I want a concrete example of such norms. Could you provide me one? – Siddharth Bhat Aug 07 '23 at 08:42
  • Well, $L^p$ norms on $C[0,1]$ as mentioned. Because unlike what you said with your sentence beginning by "but these kinds of impositions ...", continuous functions on a compact measurable set (and $[0,1]$ is one such set) are $L^p$ for every $p \in [1, +\infty]$. – Bruno B Aug 07 '23 at 09:10
  • @RyszardSzwarc Technically that's not true if $X$ is of countably infinite dimension for example, since $X$ then cannot be Banach ever, but I'm guessing you meant that it is possible if one of the norms is already supposed complete, in which case I won't doubt you on that. – Bruno B Aug 07 '23 at 09:18
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    Related: https://mathoverflow.net/questions/184464/are-banach-space-norms-up-to-equivalence-unique – Gerd Aug 07 '23 at 10:55
  • Thanks to Gerd's link, I'm realising that $L^p$ norms on $C[0,1]$ won't do the trick for what you're asking since $C[0,1]$ is not complete for those norms and you want a complete space. (I had misread your post slightly and thought those were examples you saw elsewhere, not yours, my bad). You won't have explicit examples it seems due to their existence using some form of the axiom of choice according to the MO thread. – Bruno B Aug 07 '23 at 11:16
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    @BrunoB A more striking example can be given. There is a space $X$ complete with respect to two norms $|\cdot |_1$ and $|\cdot |_2$ and a sequence $x_n$ convergent with respect to either norm, but having different limits. Clearly the norms cannot be equivalent. One of these spaces could be $\ell^1$ with the standard norm. – Ryszard Szwarc Aug 07 '23 at 15:39
  • @RyszardSzwarc could you spell out the example for me? It would be quite enlightening to see. – Siddharth Bhat Aug 07 '23 at 15:43

1 Answers1

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I will use the fact that the Hamel basis of every infinite dimensional separable Banach space has the same cardinality continuum see. If we assume the continuum hypothesis the conclusion is easy to obtain. Indeed it is well known that the Hamel basis is uncountable, hence its cardinality is at least continuum. On the other hand the space contains a dense countable subset $A,$ hence every element is a limit of a sequence with terms in $S.$ There are at most continuum of such sequences, hence we are done. It turns out the continuum hypothesis is actually not required, the result is due to H. E. Lacey.

Consider the space $X=\mathcal{F}(\mathbb{N})$ consisting of sequences with finitely many nonzero terms. We introduce two norms on $X$ $$\|x\|_1=\sum_{n=1}^\infty |x_n|,\qquad \|x\|_2= \left |x_1-\sum_{n=2}^\infty nx_n\right |+\sum_{n=2}^\infty |x_n|$$ For the sequence $v_k=e_1-{1\over k}e_k$ we have $$\|v_k-e_1\|_1\to 0,\qquad \|v_k\|_2\to 0$$ We complete $X$ with respect to either norm and obtain two separable Banach spaces $X_1=\ell^1$ and $X_2.$ We have $$X_1=\ell^1=X\oplus Y_1,\qquad X_2=X\oplus Y_2,$$ where $Y_1,Y_2$ are linear subspaces and $\oplus$ denotes the algebraic direct sum. The Hamel bases of $Y_1$ and $Y_2$ have the same cardinality. Indeed the Hamel bases of $X_1$ and $X_2$ have the same cardinality continuum, while the Hamel basis of $X$ is countable. Therefore there exists an algebraic linear isomorphism $\varphi:Y_1\to Y_2.$ Hence the mapping $$\Phi(x\oplus y)=x\oplus\varphi(y)$$ is an algebraic isomorphism between $X_1$ and $X_2.$ We introduce the norm $\|\cdot\|_3$ on $X_1$ by $\|x\|_3=\|\Phi(x)\|_2.$ Then the space $(X_1,\|\cdot\|_3)$ is isometrically isomorphic to $(X_2,\|x\|_2),$ therefore it is complete. Moreover $$\|v_k-e_1\|_1\to 0,\qquad \|v_k\|_3=\|\Phi(v_k)\|_2=\|v_k\|_2\to 0$$ In particular the norms $\|\cdot\|_1$ and $\|\cdot\|_3$ are not equivalent. Hence the identity map $I:(X_1,\|\cdot\|_1)\to (X_1,\|\cdot\|_3)$ is not continuous.

Ryszard Szwarc
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