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Let $Q:C^\infty(\mathbb R)\to\mathbb R$ be a linear function. Let us suppose that $Q(f)\geq 0$ for any $f\in C^\infty(\mathbb R)$ such that $f(0)=0$ and the set $\{x\in\mathbb R\; \colon\; f(x)\geq 0\}$ is a neighborhood of $0$. Prove that there exist constants $a,b,c\in\mathbb R$ such that $$Q(f)=af''(0)+bf'(0)+cf(0),\qquad \forall f\in C^\infty(\mathbb R).$$

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up vote 3 down vote accepted

The assertion is equivalent to saying that $Q$ is zero on all functions $f$ with $f(0)=f'(0)=f''(0)=0$. Assume there is such a function $f$ with $Q(f)\neq0$, and consider $g(x)=s(x)+\lambda f(x)$ with $s(x)=x^2$. We have $Q(g)=Q(s)+\lambda Q(f)$, and we can choose $\lambda$ such that $Q(g)<0$. But since $f''(0)=0$, there is a neighbourhood of $0$ where $g$ is non-negative, so $Q(g)$ should be non-negative. The contradiction shows that there is no such function, and the assertion follows.

[Edit in response to the comments:]

Let $Q(f)=0$ for all $f\in C^\infty(\mathbb R)$ with $f(0)=f'(0)=f''(0)=0$, and let $a=Q(f_2)$, $b=Q(f_1)$ and $c=Q(f_0)$ with $f_2(x)=\frac12x^2$, $f_1(x)=x$ and $f_0(x)=1$. Let $g$ be any function in $C^\infty(\mathbb R)$, and consider $h=g-g''(0)f_2-g'(0)f_1-g(0)f_0$. Then $h(0)=h'(0)=h''(0)=0$, so $Q(h)=0$, so $Q(h)=Q(g)-ag''(0)-bg'(0)-cg(0)=0$, so $Q(g)=ag''(0)+bg'(0)+cg(0)$.

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I'm afraid I can't see how your proof kills the problem Joriki. I mean.. I came to the same point as you did however I can't convince myself the problem is killed.. could someone explain me in details how to finish following this argument Joriki explained? Thanks.. – uforoboa Sep 5 '11 at 8:25
@user15453: Could you point out more specifically what part you want filled in? – joriki Sep 5 '11 at 8:31
i mean... how can you conclude from the fact that for any function with $f(0)=f'(0)=f''(0)=0$, $Q(f)=0$ then $Q$ has exactly the form mentioned in the problem? Thanks – uforoboa Sep 5 '11 at 8:58
@user15453: Done. Somewhat less formally, if $f(0)=f'(0)=f''(0)=0$ implies $Q(f)=0$, then $Q$ is the same for all functions with the same values of $f(0)$, $f'(0)$, $f''(0)$, so $Q$ can't depend on anything but those values, and thus, since it's linear, it must be linear combination of those values. – joriki Sep 5 '11 at 9:16
@Joriki... Thanks... So i was blinfolded in not seeing the conclusion... I should work more :) However really thanks for your kindness and patience. Really grateful.. Bye – uforoboa Sep 5 '11 at 9:22

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