A nice conjecture about convexity

Question

Let $L : C([0,1]) \rightarrow \mathbb R$ linear continuous, with $L(id)=0$, $L(1)=0$ and $L(c)>0$

($c:x \rightarrow x^2$)

Is it true that $\forall f\in C([0,1])$ convex, $L(f) \geq 0$ ?

$C([0,1])$ with the uniform norm.

Answer 1

Take $L(f)=27f(0)-72f(1/3)+63f(2/3)-18f(1)$. You get $$L(1) = L(\mathrm{Id}) = 0, \quad L(x\mapsto x^2) = 2, \quad L(x\mapsto x^3) = -2.$$

Answer 2

By defining $L(1)=0$, $L(x)=0$, $L(x^2)=1$, and $L(x^3)=-1$; this gives a linear functional $L$ on the span of $\{1,x,x^2,x^3\}$ that is continuous on the span. The Hahn-Banach theorem provides a continuous extension. The function $x\mapsto x^3$ is convex, but its image is negative.