I have an exercise which seems to be missing some information. Or it could be that I really don't need that information at all. Please let me know what you think and give a solution if possible. Thank you in advance.
"A linear functional $f$ on $X = C[0,1]$ is called positive if $f(x) \geq 0$ for all nonnegative functions $x(t)$. Prove that $f \in X'$."
So here, $X'$ is the dual space where all linear bounded functionals $f:X \to \mathbb{R}$ since $X$ is just the continuous real functions. I'm thinking that the norm on $X$ is the usual maximum norm $||x(t)||_{\textrm{max}} := \textrm{max}_{t \in [0,1]}{x(t)}$ and that since $f(x(t)) \in \mathbb{R}$, the norm on it is just the absolute value norm.
So far, since we know that $f$ is a linear functional, I've been trying to show it is bounded but haven't gotten anywhere substantial.
Best Answer
Actually, you can even give an explicit bound.
Let $C := \langle f,1 \rangle$, where I abused notation and wrote $1$ for the constant function equal to $1$.
Then, for any $g$ with $\|g\| \leq 1$, you must have $|\langle f,g\rangle| \leq C$. Indeed, as both $1+g$ and $1-g$ are nonnegative, you must have $C \pm \langle f,g\rangle \geq 0 $.
Hence $\|f\| = C$.
Edit : Here I used implicitly the uniform norm on $\mathcal{C}^0[0,1]$ and the functional norm on its dual, so you have to read $\|g\|$ as $\sup_{x\in [0,1]} |g(x)|$, and $\|f\|$ as $\sup_{\|g\| \leq 1} |f(g)|$.
I also noted $\langle f,g\rangle$ for the evaluation $f(g)$ as I thought it made the linearity of $f$ clearer.