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As above, I'm trying to find all linear operators $L: C([0,1]) \to C([0,1])$ which satisfy the following 2 conditions:

  • I) $Lf \, \geq \, 0$ for all non-negative $f\in C([0,1])$.
  • II) $Lf = f$ for $f(x)= 1$, $f(x)=x$, and $f(x)=x^2$.

I'm honestly not sure where to start here - I'm struggling to use these conditions to pare down the class of linear operators which could satisfy the conditions significantly. Could anyone help me get a result out of this? Thank you!

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Do you require your operators to be continuous? If so, with respect to what topology? – Chris Eagle Jan 8 '11 at 2:27
@Chris and @Arturo: Positivity and $L1 = 1$ imply $\|L\| = 1$ in the $\sup$-norm because $L(\|f\|\cdot 1 \pm f) \geq 0$, hence $\|Lf\| \leq \|f\|$. – t.b. Jan 8 '11 at 2:34
Why do you want to do this? – Mariano Suárez-Alvarez Jan 8 '11 at 2:56
So you should use the tag (homework) and ask for hints. – AD. Jan 8 '11 at 5:54
@Stephen: Have fun! Just post a comment to my answer in case you're stuck (then I will be notified). In case of desperation I'll give you full details, but not just now, ok? – t.b. Jan 8 '11 at 15:10

2 Answers 2

The answer is: $L$ must be the identity of $C([0,1])$.

Let $f: [0,1] \to \mathbb{R}$ be continuous. We want to prove that $Lf = f$.

Since $[0,1]$ is compact, $f$ is uniformly continuous, so for fixed $\varepsilon \gt 0$ we can choose $\delta$ such that $|x - y| \leq \sqrt{\delta}$ implies $|f(x) - f(y)| \lt \varepsilon$. Observe that \[ |f(x) - f(y)| \leq \varepsilon + \frac{2\|f\|}{\delta}(x - y)^{2} \] for all $x,y \in [0,1]$.

If $|x - y| \leq \sqrt{\delta}$ this is clear and otherwise we have $(x - y)^{2} \gt \delta$, so $\varepsilon + \frac{2\|f\|}{\delta}(x - y)^{2} > \varepsilon + 2\|f\| \gt |f(x) - f(y)|$.

In other words, for $C = \frac{2\|f\|}{\delta}$ we have \[ -\varepsilon - C(x - y)^{2} \leq f(x) - f(y) \leq \varepsilon + C(x-y)^{2}. \] Keep $y$ fixed (so we regard $y$ and $f(y)$ as constants), consider the three parts of these inequalities as functions of $x$ and apply $L$. Using that $L$ is monotone and that $L(ax^{2} + bx + c) = ax^{2} + bx + c$ by hypothesis, we get \[ -\varepsilon - C(x - y)^{2} \leq (Lf)(x) - f(y) \leq \varepsilon + C(x-y)^{2} \qquad \text{for all $x,y \in [0,1]$}. \] In particular, setting $x = y$ yields $|Lf(y) - f(y)| < \varepsilon$. Since $\varepsilon$ and $y$ were arbitrary, we conclude $Lf(y) = f(y)$ for all $y \in [0,1]$.

The argument given here may be strengthened with only little effort:

Korovkin's Theorem. Let $L_{n}: C([0,1]) \to C([0,1])$ be positive operators such that $\|L_{n}g_{i} - g_{i}\|_{\infty} \xrightarrow{n \to \infty} 0$ for $g_{i}(x) = x^{i}$, $i = 0,1,2$. Then $\|L_{n}f - f\|_{\infty} \to 0$ for all $f \in C([0,1])$.

Its proof (as well as the argument above) is a variant of the usual proof of the Weierstrass approximation theorem using Bernstein polynomials. One may take \[ L_{n}f(x) = \sum_{k = 0}^{n} \begin{pmatrix} n \\ k \end{pmatrix}x^{k}(1-x)^{n-k} f(k/n), \] in Korovkin's theorem and verify directly that $L_n g_{i} \to g_{i}$ for $i = 0,1,2$, so Korovkin's theorem yields the Weierstrass approximation theorem.

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Here are some hints:

  • You don't have much choice for $L$.
  • If $f: [0,1] \to \mathbb{R}$ is continuous then it is uniformly continuous, so for all $\varepsilon > 0$ you can choose $\delta > 0$ such that $|x - y| < \sqrt{\delta}$ implies $|f(x) - f(y)| < \varepsilon$. This gives \[ |f(x) - f(y)| < \varepsilon + C (x - y)^{2} \quad \text{for ALL $x,y \in [0,1]$} \] with $C = \frac{2\|f\|_{\infty}}{\delta}$.
  • Use this to estimate $f(x) - f(y)$ from above and below by a quadratic polynomial in $x$ (view $y$ and $f(y)$ as constants).
  • If $p(x) = ax^{2} + bx + c$ then $Lp = p$ and if $g \leq h$ then $Lg \leq Lh$.
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I'm sorry, I don't think i'm quite there yet - I can see how we can use your second point to get $|Lf(x)-Lf(y)|<\epsilon+C(x-y)^2$, but what can we make use of this bound for? If we treat y as constant we can say $|Lf(x)-f(y)|<\epsilon+C(x-y)^2$ too but I'm not sure how that helps. As we take a smaller epsilon, our C will become arbitrarily large, so presumably we can't really guarantee much accuracy here? Sorry about this... – Stephen Gregg Jan 8 '11 at 15:39
You're almost there. Once you've fixed $\varepsilon$ and $\delta$, the inequality in the second point holds for all $x,y \in [0,1]$! To prove this you need to to consider two cases: $|x - y| < \sqrt{\delta}$ and $|x - y| \geq \sqrt{\delta}$. – t.b. Jan 8 '11 at 15:48
Oh, yes, I just got there and then saw your post! In fact, can't we just evaluate my second inequality in the previous post at x=y? At which point we'd get $|Lf(y)-f(y)|< \epsilon$? – Stephen Gregg Jan 8 '11 at 15:53
Yes, of course, but you need to know that the inequality holds for all $x \in [0,1]$ (so that it is an inequality of continuous functions on $[0,1]$ and you can use the positivity of $L$). – t.b. Jan 8 '11 at 15:55
@Stephen: You're welcome, that was fun! Could you accept one of my posts as answer, please? Then the question will stop showing up as unanswered. – t.b. Jan 8 '11 at 23:55

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