weak convergence⇒weak-∗ convergence

 
$f_n\overset{w*}⇀f$ if $f_n(x)→f(x)∀x∈X$ i.e. if $ι_x(f_n)→ι_x(f)$ i.e. $T(f_n)→T(f)∀T∈ι(X)⊆X^{**}$ $f_n\overset{w}⇀f$ if $T(f_n)→T(f)∀T∈X^{**}$ Example that weak convergence-∗ doesn't imply weak convergence Try to find a sequence $l_n∈X^*,l_n\overset{w*}⇀l,l_n\overset{w}↛l$. as $∃T∈X^{**}:T(l_n)↛T(l)$ while $∀T∈ι(X),T(l_n)→T(l)$. Try $X=L^1([0,1]),X^*=L^∞([0,1])$ so want sequence $f_n⊆L^∞$ s.t. $∫gf_n→∫gf ∀g∈L^1$ but $T∈(L^∞)^*,Tf_n↛0$ since $Tf_n=1∀n$. Take $f_n=1_{A_n}$ e.g. $A_n=[0,\frac{1}{n}]$ or anything $f(A_n)→0$ and $A_n⊇A_{n+1}…$ Then $|∫f_ng|≤∫|g|1_{A_n}→0$ since $1_{A_n}→0$ a.e. Try to find $T:Tf_n=1$. would like something like $f↦f(0)$ only ok on $C^0([0,1])$ $Y=C^0([0,1])⊆X$ $T_0:Y→ℝ$ $T_0(f)=f(0)$ Since $C^0⊆L^∞$ can extend $T_0$ by Hahn-Banach gives $T∈(L^∞)^*$. so $T(f)=\lim_{n→0}n\int_0^{\frac1n}f$