@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!
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.
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!
its theorem 24.5 by BV Limaye
This appears a bit abstract
what is it trying to say?
Lets see why $j_{y}$ is linear?
$j_{y} : H' \rightarrow \Bbb{K}$ is linear as say $j_{y}(f+g) = (f+g)(y) = f(y)+g(y)=j_{y}(f) + j_{y}(g)$ and also $j_{y}(\alpha f) = (\alpha f)(y) = \alpha f(y) = \alpha j_{y}$ so $j_{y}$ is alinear functional! i think this proof is correct?
The theorem says that $H$ is reflexive. So that's exactly what the theorem is trying to say.
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.
@BAYMAX Yes, it's a different result, but somewhat similar.
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 you look at WP article Dual space, they call it algebraic dual and continuous dual.
Briefly:
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 $X\to\mathbb K$.
So they are different things, but as you can see, they are somewhat analogous constructions.
And if you have been taught dual spaces in linear algebra, you have certainly seen also the canonical map $V\to V^{**}$. This is an isomorphism for finite dimensional spaces.
So the situation here is somewhat similar: We have a similar construction (also called dual) for normed spaces. And for a large class of well-behaved spaces you have that there is a canonical isomorphism between $X$ and $X''$.
But I'll repeat again, that I mentioned this mainly because if you have been taught dual vector spaces in linear algebra, then it's worth looking at them and comparing with this notion - to see that there is some kind of analogy.
If you never heard of duals of vector spaces you can simply ignore the above remarks.
