Show that if $f,g\in\mathcal{S}_r (E;F)$ and $f(v,v,...,v)=g(v,v,...,v), \forall v \in E $ then $f=g$. Give $E, F$ vectorial spaces
, where  $\mathcal{S}_r (E;F)$ is the vectorial space the all applications r-linear symmetrics$f$, this is, the all  applications $f:E \times E \times....\times E \rightarrow F$ such that $ f (v_ { \omega(1)}, v_{\omega(2)},...,v_{\omega(r)})= f(v_1,v_2,...,v_r)$ for all permutations $\omega \in S_r$ and $v_1,v_2,....,v_m \in E$.
Show that if $f,g\in\mathcal{S}_r (E;F)$ and $f(v,v,...,v)=g(v,v,...,v), \forall v \in E $ then $f=g$.
OBS: Sorry for some mistakes in English that may exist.
 A: In all that follows, I'll be using $h=f-g$. $h$ is also $r$-linear and symmetric, and for any $x\in E$, we have $h(x,x,...,x)=0$. The goal is to prove that $h=0$. I'll also assume that the field over which $E$ is defined has characteristic 0. The first two cases ($r=2,3$) are not necessary for the proof, but might give better insight.

$r=2$
Let $x,y\in E$. $0=h(x+y,x+y)=h(x,x)+2h(x,y)+h(y,y)=2h(x,y)$, thus $h=0$.

$r=3$
In the same fashion, let $x,y,z\in E$, and $S=x+y+z$.
\begin{align*}
 0&=h(x+y,x+y,x+y)\\
  &=h(x,x,x)+3h(x,x,y)+3h(x,y,y)+h(y,y,y)\\
  &=3\left(h(x,x,y)+h(x,y,y)\right)\\[10pt]
 0&=h(S,S,S)\\
  &=3h(x+z,x+z,y)+3h(x+z,y,y)\\
  &=6h(x,y,z)+3h(x,x,y)+3h(x,y,y)+3h(z,z,y)+3h(z,y,y)\\
  &=6h(x,y,z)
\end{align*}
Notice how $h(S-x,S-x,S-x)$, etc., is used when developing $h(S,S,S)$.  

$r\geq1$
Let $x_1,....,x_r\in E$, $S=\sum_1^r x_k$. Like before, we want to evaluate $h(S,S,...,S)$. Let $F=\{\sigma :[\![1,r]\!]\mapsto [\![1,r]\!]\}$, then we can write:
$$
 h(S,S,...,S)=\sum_{\sigma\in F} h(x_{\sigma(1)},...,x_{\sigma(r)}).
$$
Now fix $k\in [\![1,r]\!]$, and let $F_k=\{\sigma :[\![1,r]\!]\mapsto ([\![1,r]\!]\backslash\{k\})\}$.
$$
 h(S-x_k,S-x_k,...,S-x_k)=\sum_{\sigma\in F_k}h(x_{\sigma(1)},...,x_{\sigma(r)})
$$
The key is to notice that $F=\left( \bigcup_{k=1}^r F_k \right) \bigcup \mathfrak{S}_r$, where $\mathfrak{S}_r$ is the set of permutations of $[\![1,r]\!]$, so that we can write
\begin{align*}
 h(S,S,...,S)&=\sum_{\sigma\in F} h(x_{\sigma(1)},...,x_{\sigma(r)})\\
  &=\sum_{k=1}^r \sum_{\sigma\in F_k} h(x_{\sigma(1)},...,x_{\sigma(r)}) + \sum_{\sigma\in\mathfrak{S}_r} h(x_{\sigma(1)},...,x_{\sigma(r)})\\
  &= r!\, h(x_1,x_2,...,x_r) + \sum_{k=1}^r h(S-x_k,...,S-x_k)\\
  &= r!\, h(x_1,x_2,...,x_r),
\end{align*}
since $h$ is symmetric. Since the characteristic is 0, this ends the proof.
