Flaw in proof that a functional is not continuous

Question

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 $F(f_n)\not\to F(f)$.

Answer 1

Your proof is correct, and it is the natural idea to attack this problm. One could criticize that you don't explicitly show that the $f_n$ are continuous, but that's a minor point that is saved with a sentence noting that the only points where there could be any doubt are $a,t_0, b$ and that in the three cases the side limits agree.