# Finding operator norm of functional

I am given a functional $$F:C([0,1],\|x\|_\infty) \to \mathbb C$$ by formula

$$F(f)=2\int_0^{1/2}f(t)dt \text{ - } \int_{1/2}^1f(t)dt$$

I should find its operator norm. I've found that $$\|F\| \le 3/2$$ using some basic inequalities, but am unable to prove the equality here, not to mention that I could be wrong and $$3/2$$ is not the optimal bound.

Any help?

Approximate the function $$f(x)=\begin{cases}1,&x\leq 1/2\\ -1,&x>1/2 \end{cases}$$ with continuous functions.
For this, we take the sequence of norm $$1$$ continuous functions: $$f_n(x)=\begin{cases} 1,&x\leq 1/2-1/n\\ -n(x-1/2),&1/2-1/n I leave to you to verify $$F(f_n)\uparrow 3/2$$