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Denote By $V$ the real vector spaces of all real polynomials in one variable, and let $P : V \rightarrow \mathbb{R}$ be a linear map. Suppose that $\forall$ $f,g \in V$ with $P(fg) = 0$ we have $P(f) = 0$ or $P(g) = 0$. Prove that there exists real numbers $x_0, c$ such that $P(f) = cf(x_0)$ for all $f$.

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up vote 8 down vote accepted

It's enough to show is true for $f(x)=x^k , \ \ k\in \mathbb{N}\cup\{0\}.$

Define $c=P(1)$ and assume $c\neq 0$. Then define $P\left(\dfrac{x}{c}\right)=x_0$ and assume that $\ x_0 \neq 0$. Now show, using induction, that $P(x^k)=cx_0^k, \ \forall k \in \mathbb{N}$.

For example for $k=2$:
$P(x^2)=r \in \mathbb{R} \Rightarrow P(r\frac{x}{cx_0}-x^2)=0 \Rightarrow P(x)=0 \text{ or } P(\frac{r}{cx_0}-x)=0$. Since $P(x) \neq 0$ we get $P(\frac{r}{cx_0})=P(x) \Rightarrow \frac{r}{cx_0}P(1)=c x_0 \Rightarrow r=c x_0^2$. Thus $P(x^2)=r=cx_0^2$.

Next we consider the cases $c=0$ or $x_0=0.$

  • $c\neq0$ and $x_0=0$:
    We have $P(x+1)=c$. For $k\geq 2, \ \text{if} \ \ P\left((x+1)^k\right)=r \Rightarrow P\left((x+1)^k-\frac{r}{c}(x+1)\right)=0.$
    Since $P\left((x+1)\right)\neq0 \Rightarrow P\left((x+1)^{k-1}-\frac{r}{c}\right)=0 \Rightarrow P\left((x+1)^{k-1}\right)=\frac{r}{c}P(1)=r$.
    Now use induction to show that $P\left((x+1)^k\right)=c \ \forall k \in \mathbb{N}$.
    Then use induction to show that $P\left(x^k\right)=0 \ \forall k \in \mathbb{N}$ (use that $x^k=(x+1)^k-kx^{k-1}-\ldots-kx-1$).
  • $c=0$:
    If $P(x^k)=r\neq 0$ for some $k \in \mathbb{N}$ then $P(x^{2k})=s\neq0$. Since $P\left(x^{2k}-\frac{s}{r}x^k\right)=0 \Rightarrow P\left(x^k\right)=0 \ \text{or} \ P\left(x^k-\frac{s}{r}\right)=0$. In either case we get $P\left(x^k\right)=0$↯.
    Therefore $P\left(x^k\right)=0 \ \forall k \in \mathbb{N}$.
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How can you solve the cases c = 0 or x_0 = 0? – ILoveMath Nov 24 '12 at 7:14
@LJym89 I have updated my answer. – P.. Nov 24 '12 at 11:26

I read this question just today. After I wrote down a proof, I read Pambos' and found that the two proofs are very similar. So I think I'd better not repeat Pambos' argument here, but only write down the part of my proof that differs the most from Pambos'.

Specifically, consider the second case in Pambos' proof, where $P(1)\not=0$ and $P(x)=0$. By linearity of $P$, we may assume that $P(1)=1$. Let $a_k=P(x^k)$ (hence $a_1=0$). Then for $k>1$, $$P\left((x-a_k)(x^{k-1}+1)\right)=P(x^k)-a_kP(x^{k-1})+P(x)-a_k=-a_ka_{k-1}.$$ Hence by mathematical induction, we can show that $a_k=0$ for all $k>1$. Thus $P(f)=f(0)$.

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