Jtdylktuy

So, I am trying to prove that if $X$ is a normed and reflexive space, its unity ball, that is $B$=$lbrace xin X : parallel xparallel =1rbrace$, is weakly compact.

For that matter, I have already proven that:

1. $J:X longrightarrow X''$ the natural embedding is continuous provided that $X$ has the $tau_{d}$ topology (that is, the initial topology of $X'$) and $X''$ with $tau_{d}^{''}*$




2. If $X$ is reflexive, $J$ is an homeomorphism (with respect the previous topologies).

My idea is to use Alaoglu-Bourbaki's theorem,which stablishes that the unity ball in $X'$ is weakly* compact, and use that the image of a compact is a compact through a continuous function.

Therefore, my problem is: if $B$=$lbrace xin X : parallel xparallel =1rbrace$ is a subset of $X'$, how do I relate it? I mean, I just do not know how to finish the argument, which I feel to be very close to me.

Also, I have checked Using the Banach-Alaoglu Theorem to show that if $X$ is reflexive then $B_X$ is weakly compact. but I think it uses the algebraic dual, or I just don't fully understand his proof.

Thank you in advance.

Here is a rather detailed proof.

If $X$ is reflexive, the natural embedding $J:Xto X''$
is an isometric isomorphism. In particular, the weak topology on $X$, that is, the topology $sigma(X,X')$, is identical to the weak topology on $X''$, that is, the topology $sigma(X'',X''')$. Let $B_X$ denote the unit ball of $X$. Then $J(B_X)$ is the unit ball of $X''$. By Alaoglu's theorem (Bourbaki had nothing to do with it), the unit ball of $X''$ is compact in the $sigma(X'',I(X'))$ topology, where $I$ is the natural embedding of $X'$ into $X'''$. But $I:X'to X'''$ is an injective isometry, so if a set is compact in the $sigma(X'',I(X'))$ topology then clearly it is also compact in the larger topology $sigma(X'',X''')$. Therefore, the unit ball of $X''$ is compact in the $sigma(X'',X''')$ topology, and therefore its inverse image under the homomorphism $J$, namely, $B_X$, is compact in the weak topology of $X$.