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Suppose we have a vector space $X$. Let $\|\cdot\|_1$ and $\|\cdot\|_2$ be two different complete norms on $X$ s.th. $X$ equipped with $\|\cdot\|_j, \ j\in\{1,2\}$ is a Hilbert space.

Are there simple examples of such norms, which are inequivalent ?

This question arises from this discussion.

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  • $\begingroup$ Can you clarify what you mean by "inequivalent"? That is, for example, they should give the same topology but not be mere scalar multiples of one another? Or would you even ask whether they can give different topologies (but still have both be complete...)? $\endgroup$ – paul garrett Dec 28 '15 at 21:10
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    $\begingroup$ @paulgarrett, they should be complete norms, but the statement : "there exist $c_1,c_2$ s.th. $c_1\|x\|_1\le \|x\|_2\le c_2 \|x\|_1$" should be false. $\endgroup$ – mikis Dec 28 '15 at 21:20
  • $\begingroup$ Here's a good discussion on MathOverflow : mathoverflow.net/questions/108550/… $\endgroup$ – DisintegratingByParts Dec 29 '15 at 10:55
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I doubt that there are any simple examples. Note that if the two norms are comparable they must be equivalent, by, say, the open mapping theorem. (If a bounded linear map from one Banach space to another is invertible qua linear map then the inverse is also bounded.) This makes it hard to give a simple explicit example; when you start with a Hilbert space and write down another norm it tends to be dominated by the first norm, and is hence equivalent to the first norm. I don't know any set theory, but I suspect the existence of two inequivalent complete norms requires the axiom of choice. (Edit: Nate Eldredge says yes it does require AC; see his comment below for details.)

(Regarding my claim that if you write down an explicit new complete norm on a Banach space it tends to be dominated by the original norm, there's the closed graph theorem, which does not actually say this: "Suppose $X$ and $Y$ are Banach spaces and $T:X\to Y$ is linear. If $T$ was given by an explicit formula then $T$ is bounded." The CGT doesn't actually say that, but that's what it comes down to in practice, in my experience. There's a reason that those unbounded operators people study are only defined on a dense subspace...)

There do exist non-simple examples. Let $X$ and $Y$ be any two separable Hilbert spaces. Let $A$ be a Hamel basis for $X$ and let $B$ be a Hamel basis for $Y$. Multiply the basis elements by constants so every element of $A$ has norm $1$ but the norm of the elements of $B$ are unbounded.

Any bijection from $A$ onto $B$ extends to a linear isomorphism of $X$ and $Y$, hence induces a new norm on $X$ with respect to which $X$ is a Hilbert space. The two norms on $X$ are not equivalent, and hence they are incomparable.

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  • $\begingroup$ @David why would the new norm on $X$ be complete? $\endgroup$ – Justpassingby Dec 28 '15 at 22:10
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    $\begingroup$ @Justpassingby We have a (wildly discontinuous) linear isomorphism $T:X\to Y$. By definition the new norm on $X$ is $||x||_2=||Tx||_Y$. So completeness of the new norm on $X$ is simply the same thing as completeness of $Y$. $\endgroup$ – David C. Ullrich Dec 29 '15 at 0:05
  • $\begingroup$ @Justpassingby Or in staggering detail: If $||x_n-x_m||_2\to0$ then $||Tx_n-Tx_m||_Y=||T(x_n-x_m)||_Y\to0$. So $Tx_n\to y=Tx$ in $Y$. So $||x_n-x||_2=||T(x_n-x)||_Y=||Tx_n-y||_Y\to0.$ $\endgroup$ – David C. Ullrich Dec 29 '15 at 0:07
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    $\begingroup$ No. Take a Hamel basis $B$ where every element has norm one. Consider the linear map $T$ induced by some permutation of $B$. We have $||Tb||=||b||$ for every $b\in B$, but there's no reason that should imply $||Tx||=||x||$ for all $x$, or even that they should be comparable. (Because if $a,b\in B$ we haven't given any connection between $||T(a+b)||$ and $||a+b||$, for example.) That's not a proof that the answer is no, of course. But my psychic powers tell me you don't have a proof that it's yes - if you think it's yes be more specific about "the same logic shows"... $\endgroup$ – David C. Ullrich Dec 29 '15 at 1:14
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    $\begingroup$ The existence of inequivalent complete norms does indeed require AC (and dependent choice, DC, is not enough). If every set of reals has the Baire property (which is consistent with ZF+DC, by a result of Shelah) then every linear map between two Banach spaces is continuous. In particular the identity map from $(X, \|\cdot\|_1)$ to $(X, \|\cdot\|_2)$ is continuous. By the open mapping theorem, its inverse is also continuous, which means the two norms are equivalent. $\endgroup$ – Nate Eldredge Dec 29 '15 at 2:01

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