Let $C([-1,1])$ be the set of continuous functions $f:[-1,1]\to \mathbb{R}$. Consider the functional, $F:C([-1,1])\to \mathbb{R}$ given by, $$F(x)=\int_{-1}^1 x(t)dt-x(0).$$ Show that $F$ defines a bounded linear functional and find $\lVert F\rVert$.
My attempt:
I've already shown that $F$ is a bounded linear functional such that $\lVert F\rVert\leq 3$. My idea for proving that $\lVert F\rVert\geq 3$ is to define a sequence of functions $x_n:[-1,1]\to \mathbb{R}$ satisfying $\lVert x_n\rVert=1$ and $|f(x_n)|>3-\frac{1}{n}$, for all $n\in \mathbb{N}$, but so far I couldn't figure out how to define such sequence. Any suggestion?
Best Answer
Take $x_n(t)=1$ for $\frac 1 n \leq |t| \leq 1$ and $x_n(t)=1+2n(|t|-\frac 1 n)$ for $|t| \leq \frac 1 n$. Then $\|x_n\| = 1$ and $F(x_n) \to 3$.