$n$-linear mappings and homogenous polynomials

Question

Given $E$ and $F$ real vector spaces, I would like to prove that if $\varphi:E^n\to F$ is a $n$-linear symmetric mapping such that $$\varphi\cdot h^{(n)}=0_F, \text{ for every } h\in E,$$ then $$\varphi(x_1,\dots, x_n)=0_F, \text{ for every } x_1,\dots, x_n\in E,$$ where $h^{(n)}:=(\underbrace{h,\dots, h}_{n \text{ times}})$ and $\varphi\cdot x:=\varphi(x)$, for each $x\in E^n$.

I've tried to prove that by fixing $v_1,\dots, v_n\in E$ and considering the mapping $P:\mathbb{R}^n\to F$ given by $$P(x_1,\dots, x_n):=\varphi\cdot (x_1v_1+\dots+x_nv_n)^{(n)}, \text{ for all } x_1,\dots, x_n\in \mathbb{R},$$ and noticing that $P(x)=0_F$, for all $x\in \mathbb{R}^n$. I've tried to calculate the coefficient of $x_1\dots x_n$, but all I got was $x_1\cdot \dots \cdot x_n \cdot n! \varphi(v_1,\dots, v_n)$. How may I proceed to prove that it is, in fact, zero?

Answer 1

Here an idea.

Following your notation $x^p = \underbrace{(x,\cdots, x)}_{p \ times}$, we may write $$(x^i,y^j) = (\underbrace{x,...,x}_{i \ times}, \underbrace{y,...,y}_{j \ times}).$$

Thus, for very $t\in\mathbb{R}$, we get

$$0=\varphi\cdot (x+ty)^n = \sum_{p=0}^n {n \choose p}t^p \varphi\cdot (x^{n-p}, y^p) $$ (use: induction + symmetry)

The above expression is a vanishing polynomial with coeficients on the vector space $F$. It is not hard to prove that such a polinomial can only vanish when all the (vectorial) coefficients also vanish (apply all funtionals $\xi\in F^*$ to make things real). Therefore

$$\varphi\cdot (x^p, \ y^{n-p})=0_F, \ \ \forall x,y\in E, \ \forall p=0,1,...,n$$

Now, one can easily prove the result by induction on $n$.

Fixed $x\in E$, we define

$$\varphi_x (y_1,...,y_{n-1}) = \varphi \ (x,y_1,...,y_{n-1})$$

Then $$\varphi_x \cdot y^{n-1} = 0_F, \ \ \forall y\in E,$$ and we can use induction.