Finding all linear operators $L: C([0,1]) \to C([0,1])$ which satisfy 2 conditions 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! 
 A: 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.
A: 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 $Lg \leq Lh$.

