I am trying to show
Consider $F\colon C[a,b] \rightarrow R$ ; $F(f)=f(t_0)$ where $a<t_0< b$ . Show that this linear functional is not continuous under $\|\cdot\|_1$ on C[a,b]
I have tried a proof that is apparently wrong and I would like help finding out where the flaw is. I try to show the functional is discontinuous by constructing a sequence of functions such that $f_n \rightarrow f$ under $ \|.\|_1$ but $F(f_n)\not\to F(f)$ where $f_n$ and $f$ are continuous on [a,b].
$f_n(x)= n(x-t_o)+1, \, x\in [-1/n +t_0,t_0]$
$f_n(x)= n(-x+t_o)+1, \, x\in [t_0,t_0 +1/n]$
$f_n(x)= 0$, otherwise
Then, $\|f_n-0\|_1=\|f_n\|_1=(2/n)\times 1/2=1/n \to 0$ as $n\to \infty$ So $f_n \to f$ under $\|.\|_1$, where $f=0$ on $[a,b]$.
However $F(f)=f(t_0)=0$ and $F(f_n)=f_n(t_0)=1 $ for all $n$. Thus $|F(f_n)-F(f)|=1$. Therefore $f_n \rightarrow f$ under $ \|.\|_1$ but $F(f_n)\not\to F(f)$.