# strong topology

• Mar 3rd 2011, 01:17 AM
ebi120
strong topology
Hi all,
i have in my lecture notes this statement :
" the strong topology is equivalent to the topology given by the norm ||w||=sup |(w|v)|
which makes the E' (dual of E ) a Banach space , and the space E is the subspace of E''
"

why the strong topology is equal to this norm topology ?
and why the space E is the subspace of E'' ?
• Mar 3rd 2011, 06:53 AM
girdav
$E$ is not "really" a subspace of $E''$ . It is the case after an identification. We define $J : E\rightarrow E''$ by the following way. Let $\varphi_x$ defined by $\varphi_x(f):= f(x)$. $\varphi_x\in (E')'$ and $J(x) =\varphi_x$ makes $J$ linear.
• Mar 3rd 2011, 06:57 AM
ebi120
thanks , i dont get u mean , is it possible to clrify more ?
• Mar 3rd 2011, 11:18 AM
girdav
$J(E)$ is a subspace of $E''$. We identify $E$ with a subspace of $E''$.
• Mar 7th 2011, 08:30 AM
ebi120
thanks, but when i see my lecture notes again i saw same experssion that :
" because A is a subspace of A'' ,the elements a belongs to the A can be viewed as functions on X(A) ( characters of A ) by setting :
a(x):=X(a)
"
and still i am confusing about the double dual , how space is the subspace og that , and also i can't understand the meaninig of this phrase that i wrote .