A question about weak convergence
Let $X$ be a Banach space and $X'$ its dual. Let ${x_n}\subset X$ weak convergent to $x$ and ${f_m}\subset X'$ weax* convergent to $f$. Is it true that $f_n(x_n)\rightarrow f(x)$?
Risposte
ok ...
You arrived second 
Take a Hilbert space $H$ and an orthonormal base $(e_n)$. It is well known that $e_n\to o$ weakly.
Moreover you can consider $e_n$ as an element of the dual by setting $f_n( v)= \forall v$.
Again $f_n\to0$ weakly (or weakly star, which is the same in this case).
But $f_n(e_n)= = 1$, which doesn't tend to zero..
That's it
Moreover you can consider $e_n$ as an element of the dual by setting $f_n( v)=
Again $f_n\to0$ weakly (or weakly star, which is the same in this case).
But $f_n(e_n)=
That's it
The book of elgiovo seems to be correct about this 
Take $l^2$.
And the ususal orthonormal basis $e_n$. It is weakly convergent to zero.
But $e_n * e_n = 1$, which seems not to be willing to converge to zero
I leave to the interested reader(s) the translation into the slang of Banach spaces, duals, crochet, etc.
Take $l^2$.
And the ususal orthonormal basis $e_n$. It is weakly convergent to zero.
But $e_n * e_n = 1$, which seems not to be willing to converge to zero
I leave to the interested reader(s) the translation into the slang of Banach spaces, duals, crochet, etc.
"elgiovo":
Anyway, my book suggests to look for a counterexample on a Hilbert space.
What book?
If we have the following further hypothesi: $f_n$ bounded (or also $||f_n||=1$); and if we search only a convergent subsequence?
"Camillo":
el giovo : I cann't read yor third and fifth line
Maybe you have problems with the angle brackets indicating the scalar product in duality (langle - rangle)? Here is my former post, written down verbatim:
No. Anyway, the following are true:
1) \$x_n\$ weak convergent to \$x\$ and \$f_n\$ convergent to \$f\$ (i.e. \$||f_n-f||_(X^**) to 0\$) implies \$langle f_n , x_n rangle to langle f, x rangle\$;
2) \$f_n\$ weak* convergent to \$f\$ and \$x_n\$ convergent to \$x\$ (i.e. \$||x_n-x||_(X) to 0\$) implies \$langle f_n , x_n rangle to langle f, x rangle\$.
"ubermensch":
Those are trivial! Do you have a counter-example (si dice così?!) to my question?
I should think a little bit about that. Anyway, my book suggests to look for a counterexample on a Hilbert space.
el giovo : I cann't read yor third and fifth line 
Those are trivial! Do you have a counter-example (si dice così?!) to my question?
No. Anyway, the following are true:
1) $x_n$ weak convergent to $x$ and $f_n$ convergent to $f$ (i.e. $||f_n-f||_(X^**) to 0$) implies $langle f_n , x_n rangle to langle f, x rangle$;
2) $f_n$ weak* convergent to $f$ and $x_n$ convergent to $x$ (i.e. $||x_n-x||_(X) to 0$) implies $langle f_n , x_n rangle to langle f, x rangle$.
1) $x_n$ weak convergent to $x$ and $f_n$ convergent to $f$ (i.e. $||f_n-f||_(X^**) to 0$) implies $langle f_n , x_n rangle to langle f, x rangle$;
2) $f_n$ weak* convergent to $f$ and $x_n$ convergent to $x$ (i.e. $||x_n-x||_(X) to 0$) implies $langle f_n , x_n rangle to langle f, x rangle$.
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