Norm of the linear functional $f(u(x)) = \int_{0}^{1} (2x-1) u(x) dx$ Let  $ f(u(x)) = \int_{0}^{1} (2x-1) u(x) dx $ be a linear functional defined for all $ u \in (C[0,1], ||.||_\infty) $. Show that $ ||f|| = \frac{1}{2} $.
I have been able to show that $ || f ||  \le \frac{1}{2} $ without much trouble.
Usually I'm also able to prove by setting $ u(x) = 1 $ and using $ |f(u)| \le ||f|| \space ||u|| $ that the above inequality is actually an equality.
In this case, however (abusing notation a bit):
$ | f (1)  |  = | \int_{0}^{1} (2x-1) dx| = 0 $
Therefore I'm only able to prove in the end that $ 0 \le ||f|| \le \frac{1}{2} $ which is not quite a solution.
I tried playing around with different functions with sup-norm of 1 but I could not quite get the desired result. Hopefully I'm not missing something too obvious.

For $\varepsilon > 0$, sufficiently small, consider the function $u_\varepsilon(x)$ defined as follows:

*

*$u_\varepsilon(x) = -1$ for $0 \leq x \leq \frac{1}{2}-\varepsilon$

*$u_\varepsilon(x)=1$ for $\frac{1}{2}+\varepsilon \leq x \leq 1$

*$u_\varepsilon(x)$ is the equation of the line connecting the points $u_\varepsilon(1/2-\varepsilon)$ and $u_\varepsilon(1/2+\varepsilon)$ for $\frac{1}{2}-\varepsilon < x <  \frac{1}{2}+\varepsilon$
It's easy to see that $u\in C[0,1]$ and $\lVert u \rVert_\infty = 1$, and with a few calculations we can verify that
$$|f(u(x))| = \frac{1}{2} - \frac{2\varepsilon^2}{3}$$
which tends to $1/2$ as $\varepsilon \to 0^+$. This proves that $\lVert f \rVert \geq \frac{1}{2}$.
