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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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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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    @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.. Bye2011-09-05