Polarization formula. Let $V$ be a $\mathbb{R}$-vector space.
Let $\Phi:V^n\to\mathbb{R}$ a multilinear symmetric operator.
Is it true and how do we show that for any $v_1,\ldots,v_n\in V$, we have:
$$\Phi[v_1,\ldots,v_n]=\frac{1}{n!} \sum_{k=1}^n \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-k}\phi (v_{j_1}+\cdots+v_{j_k}),$$
where $\phi(v)=\Phi(v,\ldots,v)$.
My question come from that, I have seen this formula when I was reading about mixed volume, and also when I was reading about mixed Monge-Ampère measure. The setting was not exactly the one of a vector space $V$ but I think the formula is true here and I am interested by having this property shown out of the specific context of Monge-Ampère measures or volumes. I have done some work in the other direction, i.e. starting from an operator $\phi:V\to\mathbb{R}$ satisfying some condition and obtaining a multilinear operator $\Phi$ ; bellow are the results I have seen in this direction.
I already know that if $\phi':V\to\mathbb{R}$ is such that for any $v_1,\ldots,v_n\in V$, $\phi'(\lambda_1 v_1+\ldots+\lambda_n v_n)$ is a homogeneous polynomial of degree $n$ in the variables $\lambda_i$, then there exists a unique multilinear symmetric operator $\Phi':V^n\to\mathbb{R}$ such that $\Phi'(v,\ldots,v)=\phi'(v)$ for any $v\in V$. Moreover $\Phi'(v_1,\ldots,v_n)$ is the coefficient of the symmetric monomial $\lambda_1\cdots\lambda_n$ in $\phi'(\lambda_1 v_1+\ldots+\lambda_n v_n)$ (see Symmetric multilinear form from an homogenous form.).
I also know that if $\phi'(\lambda v)=\lambda^n \phi'(v)$ and we define
$$\Phi''(v_1,\ldots,v_n)=\frac{1}{n!} \sum_{k=1}^n \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-k}\phi' (v_{j_1}+\cdots+v_{j_k}),$$
then $\Phi''(v,\ldots,v)=\frac{1}{n!} \sum_{k=1}^n (-1)^{n-k} \binom{n}{k} k^n \phi'(v)=\phi'(v)$ (see Show this equality (The factorial as an alternate sum with binomial coefficients).). It is clear that $\Phi''$ is symmetric, but I don't know if $\Phi''$ is multilinear.
Formula for $n=2$:
$$\Phi[v_1,v_2]=\frac12 [\phi(v_1+v_2)-\phi(v_1)-\phi(v_2)].$$
Formula for $n=3$:
$$\Phi[v_1,v_2,v_3]=\frac16 [\phi(v_1+v_2+v_3)-\phi(v_1+v_2)-\phi(v_1+v_3)-\phi(v_2+v_3)+\phi(v_1)+\phi(v_2)+\phi(v_3)].$$
 A: This is true, here's a proof.
I'm going to use the polynomial notation $\Phi\left(v_{1},\ldots,v_{n}\right)=v_{1}\cdots v_{n}$
- note that the multilinearity and symmetry of $\Phi$ means that
manipulating these like polynomials (i.e. commuting elements, distributing
``multiplication'') is completely legitimate. Let the RHS of your
proposed equation be $\frac{1}{n!}F\left(n\right)$.
Using the multinomial expansion, we have
$$
F\left(n\right)=\sum_{k=1}^{n}\left(-1\right)^{n-k}f\left(n,k\right)
$$
where
$$
f\left(n,k\right)=\sum_{1\le j_{1}<\cdots<j_{k}\le n}\ \sum_{l_{1}+\cdots+l_{k}=n}{n \choose l_{1},\ldots,l_{k}}v_{j_{1}}^{l_{1}}\cdots v_{j_{k}}^{l_{k}}.
$$
Let's try to compute the coefficient of $v_{j_{1}}^{l_{1}}\cdots v_{j_{k}}^{l_{k}}$
in $F\left(n\right)$. The most obvious contribution is from $f\left(n,k\right)$,
which gives $\left(-1\right)^{n-k}{n \choose l_{1},\ldots,l_{k}}.$
But there are more contributions: for every $K>k$ we have terms where
$K-k$ of the $l$s are zero. The contribution from $f\left(n,K\right)$
is
$$
\left(-1\right)^{n-K}\sum\left\{ {n \choose l_{1},\ldots,l_{k},0,0,\ldots}:j_{k+1},\ldots,j_{K}\textrm{ distinct from }j_{1},\ldots,j_{k}\right\} .
$$
All we need to do to compute this is count the number of choices of
$K-k$ of the $n-k$ remaining indices, so we get
$$
\left(-1\right)^{n-K}{n \choose l_{1},\ldots,l_{k}}{n-k \choose K-k}.
$$
The coefficient of $v_{j_{1}}^{l_{1}}\cdots v_{j_{k}}^{l_{k}}$ in
$F\left(n\right)$ is thus
$$
\sum_{K=k}^{n}\left(-1\right)^{n-K}{n \choose l_{1},\ldots,l_{k}}{n-k \choose K-k}=\sum_{K=k}^{n}\left(-1\right)^{n-K}\frac{n!}{l_{1}!\cdots l_{k}!}{n-k \choose K-k}.
$$
We want to show that this is $n!$ when $k=n,l_{j}=1$ and zero otherwise.
The first case is easy - there is only a single term in the sum and
all of $n,k,K$ are just $n$, so it falls out immediately. Let's
try the zero case. Factoring out the $K$-independent terms gives
$$
\left(-1\right)^{n}\frac{n!}{l_{1}!\cdots l_{k}!}\sum_{K=k}^{n}\left(-1\right)^{K}{n-k \choose K-k}.
$$
Making a change of variables $j=K-k$ turns the sum to
$$
\left(-1\right)^{k}\sum_{j=0}^{n-k}\left(-1\right)^{j}{n-k \choose j}.
$$
This is the alternating sum of the binomial coefficients, which vanishes
as required.
A: SKETCH OF THE PROOF : Your big sums are always sums of (sums of sums of) terms of the form $\Phi(v_{k_1},v_{k_2},\ldots ,v_{k_n})$ for some tuples of indices $(k_1,k_2, \ldots ,k_n)$. Thanks to the symmetry of $\Phi$, we can always rearrange and put the tuple in increasing order.
You are then left with a simpler sum with less terms, where exactly one term is multilinear (the term $\Phi(v_1,v_2, \ldots ,v_n)$) and all the others are not. So this
is a Rambo-like situation of one against a hundred. But fortunately for us, the combinatorial property (let us call it $P$) that for any finite set $X$, the sum
$\sum_{B \subseteq X}(-1)^{|B|}$ is zero except when $X$ is empty, allows us to show that all the non-multilinear terms have zero coefficient in the sum.
THE DETAILS : Given a tuple $(k_1,\ldots ,k_n)$, denote by
$\rho(k_1,k_2,\ldots,k_n)$ the rearranged tuple according to increasing order.
(Thus, $\rho(1,3,2)=(1,2,3)$).
It is more convenient here to view tuples as functions, so we shall speak of $u$ and
$\rho u$ where $u$ and $\rho u$ are maps $\lbrace 1,2, \ldots, n\rbrace \to \lbrace 1,2, \ldots, n\rbrace$ and $\rho u$ is increasing. Also, we put
$\psi(f)=\Phi(v_{f(1)},\ldots,v_{f(n)})$.
For an arbitrary increasing tuple $i$, denote by $w(i)$ the number of tuples $j$
satisfying $\rho(j)=i$. For any $A \subseteq \lbrace 1,2, \ldots ,n \rbrace$, denote by $I(A)$ the set of all increasing maps $\lbrace 1,2, \ldots ,n \rbrace \to A$. Also, let
$I=I(\lbrace 1,2, \ldots, n \rbrace)$ and
$V(f)=\lbrace A \subseteq \lbrace 1,2, \ldots ,n \rbrace | f\in I(A) \rbrace$ .
Note that if
$K(f)=\lbrace 1,2, \ldots ,n \rbrace \setminus Im(f)$,
then there is a natural bijection between ${\cal P}(K(f))$ and $V(f)$, given
by $B \mapsto Im(f) \cup B$.
Let
$$
\lambda (A)=\phi\bigg(\sum_{a\in A}v_a\bigg) \tag{2}
$$
Then, expanding $\lambda(A)$ completely shows that
$$
\lambda (A)=\sum_{f\in I(A)} w(f) \psi(f) \tag{3}
$$
Then, the RHS (call it $\Phi''$) of the desired equality can be rewritten as
$$
\begin{eqnarray}
\Phi'' &=& \sum_{A\subseteq \lbrace 1,2, \ldots ,n \rbrace}(-1)^{n-|A|}\lambda(A)\\
&=& \sum_{A\subseteq \lbrace 1,2, \ldots ,n \rbrace}(-1)^{n-|A|}\sum_{f\in I(A)} w(f) \psi(f) \\
&=& \sum_{f\in I}w(f)\psi(f)\sum_{A\in V(f)}(-1)^{n-|A|} \\
&=& \sum_{f\in I}w(f)\psi(f)\sum_{B\subseteq K(f)}(-1)^{n-|Im(f)|+|B|} \\
&=& w({\mathsf{id}})\psi(\mathsf{id}) \ \text{by property } P. \\
&=& n! \Phi(v_1,v_2, \ldots ,v_n)
\end{eqnarray}
$$
which concludes the proof.
A: This in not an answer, but an incomplete attempt of induction proof.
First, we will consider the following notation:
$$\Phi_v[v_1,\ldots,v_{n-1}]=\Phi[v_1,\ldots,v_{n-1},v],$$ so $\Phi_v:V^{n-1}\to\mathbb{R}$ is the multinear symmetric operator we obtain when we fix a variable in $\Phi$.
We then of course note $\phi_w(v)=\Phi_w[v,\ldots,v]=\Phi[v,\ldots,v,w]$.
In this 'answer', I show that the formula is proved if 
$$\phi(v_1+\cdots+v_n)
=\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-1-k}\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k}),$$
which is probably not more easy then the polarization formula itself, but that where lead me my attempt of induction.
We also write
$$\Phi_v[v_1,\ldots,\hat{v_i},\ldots,v_n]=\Phi_v[v_1,\ldots,v_{i-1},v_{i+1},\ldots,v_n].$$
Let assume the formula is true for multilinear symmetric operator $V^{n-1}\to\mathbb{R}$.
Since $\Phi[v_1,\ldots,v_n]=\Phi_{v_i}[v_1,\ldots,\hat{v_i},\ldots,v_n]$ by symmetry, we have:
$$\Phi[v_1,\ldots,v_n]=\frac1n \sum_{i=1}^n \Phi_{v_i}[v_1,\ldots,\hat{v_i},\ldots,v_n].$$
By the induction we have 
$$\Phi_{v_i}[v_1,\ldots,\hat{v_i},\ldots,v_n]
=\frac{1}{(n-1)!} \sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n\ ;\ j_l\neq i} (-1)^{n-1-k}\phi_{v_i} (v_{j_1}+\cdots+v_{j_k}).$$
So
$$\Phi[v_1,\ldots,v_n]
=\frac1{n!}\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-1-k} 
\sum_{\{i\mid i\neq j_l \forall l\leq k\}}\phi_{v_i} (v_{j_1}+\cdots+v_{j_k}).$$
But
\begin{align}
\sum_{\{i\mid i\neq j_l \forall l\leq k\}}\phi_{v_i} (v_{j_1}+\cdots+v_{j_k})
&={\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k})-\phi_{v_{j_1}+\cdots+v_{j_k}}(v_{j_1}+\cdots+v_{j_k})}\\
&={\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k})-\phi(v_{j_1}+\cdots+v_{j_k})}.
\end{align}
So
\begin{align}
&\Phi[v_1,\ldots,v_n]\\
&=\frac1{n!}\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-1-k} 
(\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k})-\phi(v_{j_1}+\cdots+v_{j_k}))\\
&=\frac1{n!}\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-k} 
\phi(v_{j_1}+\cdots+v_{j_k})\\
&\qquad +\frac1{n!}\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-1-k}\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k}).\\
\end{align}
In this last expression the first part is almost the R.H.S. of the polarisation formula ; the sum goes until $n-1$ instead of $n$. But when $k=n$,
$\sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-k} \phi(v_{j_1}+\cdots+v_{j_k})
=\phi(v_1+\cdots+v_n)$.
Hence we will have prove the polarization formula if 
$$\phi(v_1+\cdots+v_n)
=\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-1-k}\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k}).$$
Unfortunately I don't see how to show that, it is maybe as difficult as the polarization formula. 
A: (This is just the summary of the answers by Ewan and Anthony, but a lot simpler.)
Let $S_n$ denote the symmetric group. Also, let us write $X=\{1,\dots,n\}$. We compute $$\begin{eqnarray}\sum_{k=1}^{n}\sum_{1\leq j_{1}<\cdots<j_{k}\leq n}(-1)^{k}\phi(v_{j_{1}}+\cdots+v_{j_{k}})&=&\sum_{A\subset X}(-1)^{|A|}\phi(\sum_{a\in A}v_{a})\\&=&\sum_{A\subset X}(-1)^{|A|}\sum_{f:X\to A}\Phi(v_{f(1)},\dots,v_{f(n)})\\&=&\sum_{f:X\to X}\Phi(v_{f(1)},\dots,v_{f(n)})\sum_{f(X)\subset A\subset X}(-1)^{|A|}\\&=&\sum_{f\in S_{n}}\Phi(v_{f(1)},\dots,v_{f(n)})(-1)^{n}\\&=&(-1)^{n}n!\Phi(v_{1},\dots,v_{n}),\end{eqnarray}$$
where the fourth equality follows from the binomial formula.
