Just a simple curious remark on how “rare” are Complete normed structures Topologies on Vector Spaces :Every Banach Space is isometrically isomorphic to a closed subspace of a space of continuous function defined on a Compact Hausdorff Space .If one introduces a continuous multiplicative structure , making it a Banach Algebra and if it is self adjoint , one has the Gelfand Theorem that the whole Banach Space is isometrically isomorphic to a space of continuous function on a Hausdorff Compact Space (defined as the Maximal ideal space of the Algebraic structure endowed with the Gelfand Topology ).This line may be useful to pursue !!. ]]>

The linear bijection used in my construction shows that the two Banach spaces constructed there are indeed isomorphic. On the other hand, one can take two nonisomorphic Banach spaces of the same linear dimension (e.g. and for some and use Hamel bases and the axiom of choice to construct a linear bijection between them to create two Banach space norms on the same space that are not isomorphic.

]]>(without knowing about this post at that time).

Question: Let V be a vector space provided with two inequivalent Banach norms. Terence explained how to do this using Hamel bases.

Is it possible that (V, first norm) and (V, 2nd norm) are isomorphic as

Banach spaces? Never? Always? Sometimes? We know of course that the identity map does not provide for us an isomorphism.

If E is an infinite dimensional Banach space, then there exists a discontinuous linear functional φ: E → R. Take u ∈ E such that φ(u) = 1. Then Sx := x − 2φ(x)u defines a linear, discontinuous map S : E → E, and

S o S = Id. Thus |||x||| := ||Sx|| defines a complete norm on E which is not equivalent to the given norm || · ||

This is taken from first lines of the introduction to the article W.Arendt, R. Nittka: Equivalent complete norms and positivity. Arch. Math. 92 (2009), 414–427,

where it is also shown that on each Banach space there is always in infinite number of mutually non-equivalent complete norms. They might all be useless, of course.

]]>For instance, my answer for the non-existence of an F-space topology on c_c would be that infinite dimensional F-spaces cannot have countable dimension. (This is a standard exercise in basic courses on FA, for Banach spaces, as an application of Baire’s catagory theorem. And the proof works in the same way for F-spaces.) The same proof works for the proof that C_c(\R) (or C_c^\infty(\R)) cannot carry an F-space topology stronger than the topology of pointwise convergence. These examples are cases where the space is ‘too small’.

For the case that the space is too large the method has been indicated – to my taste – by Will Sawin in his comment given above. To expand this, let us look at the question why \R^\N, with product topology, cannot carry a stronger Banach space topology: in this case, the product topology is a Fr\’echet space topology which is easily seen not to be defined by a norm. So, the closed graph theorem gives the negative answer. (However, I do not see an immediate answer to the question whether \R^\R, with product topology, carries a stronger F-space topology, although I have the feeling that the answer is negative.)

After this discussion, I think that I have given some answer to the question. First decide whether the space is too small or too large. In the first case, use Baire catagory. In the second case find a topology on the space which allows the application of a closed graph theorem and which is not Banach, Fr\’echet or F-space, respectively.

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