4:39 AM
@MartinSleziak you may see this when you are around! - I am a bit understanding the proofs of the theorems in Functional analysis but when asked the proof after some days or recalling,i dont remember how they approached the steps ,like how they constructed sucha function for giving counterexample? any advice you want to give,for which i shall be grateful!
2 hours later…
7:01 AM
This is definitely not something specific to functional analysis, and others can you probably give a better answer than I can. Have you tried searching for posts about this on the main site.
Google returns quite a few posts, for example, Are there any tricks to remembering proofs of mathematical theorems?, Recalling Proofs, How to remember all the proofs in mathematics, How to stop forgetting proofs - for a first course in Real Analysis?
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4 hours later…
1:45 PM
For $y \in H$ define $j_{y}:H' \rightarrow K$ by $j_{y}(f) = f(y),f \in H'$ ,then $j_{y}$ is acontinuous linear functional on $H'$ and the map $J$ from $H $ to $H'$ defined by $J(y) = j_{y},y \in H$ is a surjective linear isometry that is $H$ is reflexive!
Re: What is it trying to say? Exactly what is says on the tin: Every Hilbert space is a reflexive space.
It probably helps if you have some experience with reflexive space and duals. But if not and if this is your first encounter with reflexive space, it's perfectly ok. You learn some stuff about Hilbert spaces, duals (and double duals) and maybe later you'll appreciate that some Banach spaces have similar properties.
1:59 PM
3:03 PM
say $j_{x+y}(f) = ?$ so it is $j_{x+y} (f) = f(x+y) = f(x) + f(y) = j_{x}(f) + j_{y}(f) \forall f $ so $j_{x+y} = j_{x} + j_{y}$ so it is linear
4:28 PM
I am not sure whether this way of looking at reflexive spaces is useful or not, but maybe is worth mentioning.
If you learned about dual spaces in linear algebra, then you have certainly learned about canonical embedding of a space into the double dual.
This is in fact isomorphism for finite dimensional spaces, but not for infinite dimensional vector spaces.
This canonical embedding is the same map as in your theorem, but now you are looking at different type of dual. (You require also continuity.)
2 hours later…
6:08 PM
I mentioned it because I thought that if you learned about duals of vector spaces before, you could see how this result for duals of normed spaces is analogous.
If I have a vector space $V$, I can define (algebraic) dual $V^*$ as the vector space of all linear maps $V\to\mathbb K$.
If I have a normed space $X$, I can define (continuous) dual $X'$ as the normed space consisting of all linear continuous maps $X\to\mathbb K$.
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