Finding norm of a bounded linear functional

Question

Let $X=\{x\in C[0,1]: x(0)=0\}$ with sup norm and let $f:X\to \mathbb{K}$ be defined by $$f(x)=\int\limits_0^1 x(t)dt \text{ for all }x\in X.$$ I want to show that $\|f\|=1$. It is easy to show that $\|f\|\leq 1$ also I have shown that there does not exist ant $x\in B_X$ such that $\|f\|=|f(x)|$. Thus I have to find a sequence $(x_n)\subset B_X$ such that $|f(x_n)|\to \|f\|$. I am stuck here. Any help to find such a sequence is appreciated.

Answer 1

Consider $f_n(x)=1,$ if $x>{1\over n}$, $f_n(0)=0$ and $f_n(x)=nx$ on $[0,{1\over n}]$, $\|f\|=1$, and $\int_0^1f_n(x)\geq 1-{1\over n}\geq\|f\|$. This implies that $\|f\|\geq 1$. Since you have remarked that $\|f\|\leq 1$, $\|f\|=1$.