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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.

inkievoyd
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  • you might want to check this page http://en.wikipedia.org/wiki/Riesz%E2%80%93Markov%E2%80%93Kakutani_representation_theorem – daw Oct 13 '14 at 15:07
  • Thanks but I'll like to try a more elementary method since I haven't learned about any of that. – inkievoyd Oct 13 '14 at 15:12

3 Answers3

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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.

Chocosup
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  • What is the definition of the inner product that you're using here? – inkievoyd Oct 13 '14 at 15:26
  • This is not an inner product, it's the evaluation of the functional $f$ on the function $g$. If you prefer I can note it $f(g)$, but as it's linear I prefer a bracket notation. Also, I used implicitly the uniform norm on the function space $\mathcal{C}^0[0,1]$ as it is the natural norm of this space as a Banach. – Chocosup Oct 13 '14 at 15:27
  • So usually we say $||f||= \textrm{sup}_{||g||=1}{||f(g)||}$. You are saying that your $||f||$ is equal to this?

    Also, why must $\langle f,g \rangle \leq C$ if $||g|| \leq 1$? We are only told that $\langle f,g \rangle \geq 0$ for nonnegative $g$.

     – inkievoyd Oct 13 '14 at 15:42
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    Well, I think that the third line of my answer is pretty clear as to why we must have $\langle f,g\rangle \leq C$. As $1-g$ and $1+g$ are nonnegative you have $0 \leq \langle f,1-g\rangle = C - \langle f,g\rangle$ and $0 \leq \langle f,1+g\rangle = C + \langle f,g\rangle$, so $-C \leq \langle f,g\rangle \leq C$. – Chocosup Oct 13 '14 at 15:57
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    I see. Thank you very much. – inkievoyd Oct 13 '14 at 16:03
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Let $L$ be the linear functional on $C(X)$. If $f$ is real, then $\|f\|\pm f \ge 0$ gives $$ L(1)\|f\|\pm L(f) \ge 0 \implies |L(f)| \le L(1)\|f\|. $$ If $f = h+ik$ where $h$, $k$ are real, then $|L(f)|\le L(1)(\|h\|+\|k\|) \le 2L(1)\|f\|$ (a better estimate is possible, but all you need is continuity.)

Disintegrating By Parts
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From the given, X is a Banach space with Sup norm (from ||x|| defined as x(t) is a continuous function on compact metric space [0,1], so it is bounded). Since, f is already linear functional into another normed space R, with Uniform Boundedness Principle, f is bounded. (I believe this works...)

MA94554
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