# Inequivalent Hilbert norms on given vector space

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 ?

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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.