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Let $P$ denote the subspace of $C^0([0,1])$ defined by polynomials restricted to [0,1]. Suppose that $l:P\rightarrow \mathbb{R}$ is a linear function with the property that

$p(x)\geq 0$ in $x\in [0,1]$ implies $l(p)\geq 0$.

Then how can we show that $l$ can be extend to define a linear function $\hat{l}$ on $C^0([0,1])$ satisfying an estimate of the form $|\hat{l}(f)|\leq C||f||_{\infty}$?

I may need the Hahn-Banach theorem, but I think the set {p: $p(x)\geq 0$ }is not a linear subspace then I don't know how to do it. Thanks for any hint!

violin
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2 Answers2

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First you should extend $l$ from $P$, to $\bar l:U\to \mathbb R$ where $U$ is the space of all polynomials (hint: note that every monomial is a non-negative polynomial in $[0,1]$, and extend linearly).

$U$ is clearly a subspace of $C^0([0,1])$, and the extended $\bar l:U\to \mathbb R$ can be checked to be sub-linear (actually it is linear). Apply Hahn-Banach to $\bar l$, to get the desired $\hat l$.

Note: I am not sure, but it may be possible to use Weierstrass theorem to get an explicit extension.

Shay Ben Moshe
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  • It makes sense. I think I understand what you say. But how we find the dominated function so we can apply Hahn Banach theorem? Thanks! – violin Dec 20 '14 at 15:30
  • I guess I know, combining your answer with the answer below, I think the dominating function is the norm of P. – violin Dec 20 '14 at 15:42
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The polynomials are dense in $C^{0}[0,1]$. You need to show that $l$ is bounded; then you can extend by continuity. The norm of $l$ is $M=l(1)$, which is non-negative by assumption. Indeed, if $p$ is a real polynomial, then $\|p\|\pm p \ge 0$ which gives you $$ 0 \le l(\|p\|1)\pm l(p),\\ \mp l(p) \le \|p\|l(1),\\ |l(p)| \le M\|p\|. $$

Disintegrating By Parts
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  • So we don't need to use Hahn Banach theorem? – violin Dec 20 '14 at 15:31
  • I think we can also use the Hahn Banach theorem so we don't need to extend by continuity. Is that right? Thanks! – violin Dec 20 '14 at 15:43
  • ||p|| is a number, then what how you say ||p||+p is non negative, since p is a polynomial? – violin Dec 20 '14 at 16:35
  • @piano : The norm $|p|=\sup_{x\in[0,1]}|p(x)|$ which means $|p(x)|\le |p|$ for all $x\in[0,1]$. So $-p(x) \le |p|$ and $p(x) \le |p|$ which means $|p|\pm p(x) \ge 0$ for all $x$. And, Hahn Banach is the wrong thing here because of continuity. – Disintegrating By Parts Dec 20 '14 at 18:21
  • Yeah, that makes sense. I still don't understand why Hahn Banach is wrong here. The norm can serve as a dominating function on the subspace of polynomial. So apply Hahn Banach thm, we can extend this property to the whole space, which is the space of all continuous function. Why is it wrong? – violin Dec 20 '14 at 18:49
  • @piano : I should have been more specific. It's the wrong technique because extension by continuity is unique from the dense subspace of polynomials, and extension by continuity is a much simpler thing. No use in sending a man to do a boy's job. :) – Disintegrating By Parts Dec 20 '14 at 18:57
  • Oh, okay. I see, so you mean Hahn Banach thm is still right but no need to do so since we can extend it by continuity, right? – violin Dec 20 '14 at 19:03
  • @piano : That's it. – Disintegrating By Parts Dec 20 '14 at 19:04
  • I think I totally understand. Thank you very much for the explanation! – violin Dec 20 '14 at 19:10