Anoter question on vector spaces an rationnal numbers

[matt grime]

a linear functional is not the same as a linear function from H to itself.

 

[seratend]

a linear functional is not the same as a linear function from H to itself.

Ok, I am sorry for my bad mathematical terms (sometimes they are direct translations).
I mean a linear operator (call it f), if you prefer, "acting" on the hilbert space H (an endomorphism if I am right):
f: H --> H
x$\in H$ --> f(x) $\in H$

such that f(x1+k.x2)=f(x1)+k.f(x2)

we choose f as a bounded compact operator on H. My question stays: can we say that there isn't any injective bounded compact linear operator on a non separable hilbert space?

 

[seratend]

We can either restrict this question to the normal subset of these linear operators (i.e. an operator such that ff*=f*f).

Seratend.

 

[seratend]

I doubt that that is true. Can you save me the effort and give me some examples of non-separable hilbert spaces?

having said that, you mean functional by application don't you?

In order to help the members of this forum to answer to this question (and also to believe that the answer is probably yes), we can already say that there isn't any bijective compact bounded operator on infinite dimensional Hilbert spaces (separable or not).
Here is the demonstration:
We know that the closed unit ball cannot be a compact set in an infinite dimensional Hilbert space. However, if we have a compact isomorphism between Hilbert spaces that are continuous, the image of closed set is a closed set and thus because we have a compact operator, the image of the closed unit ball is a compact set (the closure of the set is equal to the set, because the set is closed). Thus, the image of this compact set under the inverse continuous operator is a also compact set. Thus, the unit ball is a compact set for a compact continuous isomorphism. QED. So it is easy to demonstrate the impossibility of compact isomorphisms on infinite dimensional Hilbert spaces (separable or not).

However, if we reduce the case to the injective compact continuous operators case, we just have the property that the closure of the image of a closed unit ball is a compact set. Therefore, the unit ball is not a compact set if the image is not compact and the question is still valid.

Seratend.

 

[seratend]

The answer:

There is no compact (continuous) injective operator on a non separable hilbert space.

Quick demo: H the hilbert space. T the compact injective operator => Closure[T(H)] is a separable space (compacity property) => there is no injection between a non separable hilbert space and the separable one.

Moreover, we have a weaker formulation of this impossibility: there is no injective compact operator between a non separable Hilbert space and any banach space. (i.e. we can have an injective compact operator between 2 banach spaces, even if the banach spaces are non separable).

Seratend.