Let $B[0,1]$ be the space of bounded functions with supremum norm. Let $\{f_n\}\subset B[0,1]$, $f_n(x)=nx$ for $x\in[0,1/n]$ and $f_n(x)=0$ otherwise. Is this sequence weakly convergent to zero function in $B[0,1]$ ?
My try is following: We want to prove that $f_n\rightharpoonup 0$ in $B[0,1]$. Define $\{g_n\}\subset B[0,1]$ for $n\geq 2$, $g_n(x)=n(1-n)(x-\frac{1}{n-1})$ for $x\in(\frac{1}{n},\frac{1}{n-1}]$ and $g_n(x)=0$ otherwise. Then $f_n+g_n\in C[0,1]$ and hence $f_n+g_n$ converges weakly to zero in $C[0,1]$ and so in $B[0,1]$. It remains to prove that $g_n\rightharpoonup 0$ in $B[0,1]$. Choose one $F\in B^*[0,1]$ and denote $h_n=\sum_{i=1}^{n}sgn(F(g_i))g_i$. Then clearly $|h_n|\leq 1$ and $ \sum_{i=1}^{n}F(g_i)=|F(h_n)|\leq\|F\| $. The constant $\|F\| $ is finite and does not depend on $n$. Hence $ \sum_{i=1}^{\infty}F(g_i)<\infty $ and so $F(g_n)\to 0$ as $n\to\infty$.
I am not sure about this proof.
