Can we determine symmetric multilinear map from this information? Let $X,Y$ be normed spaces and $T:X^n\rightarrow Y$ be a continuous symmetric $n$-multilinear map.
Then, is $T$ determined by $T(x,...,x)$'s where $x\in X$?
That is, can $T(x_1,...,x_n)$ be expressed by $T(x,...,x)$'s?
 A: Going off of Bazin's answer from the link: we have
$$
T(x_1,\dots,x_k) = 
\frac{1}{2^kk!}\sum_{\epsilon_j = \pm 1}
\epsilon_1 \cdots \epsilon _k T\left( 
\sum_{j=1}^k \epsilon_j x_j, \sum_{j=1}^k \epsilon_j x_j, \dots, \sum_{j=1}^k \epsilon_j x_j
\right)
$$

Proof that this works: note that it is equivalent to prove the polynomial identity
$$
x_1\cdots x_k = \frac{1}{2^kk!}\sum_{\epsilon_j = \pm 1}
\epsilon_1 \cdots \epsilon _k \left( \sum_{j=1}^k \epsilon_j x_j\right)^k.
$$
With that in mind: for any exponents $p_1,\dots,p_k$, each non-negative with $p_1 + \cdots + p_k = k$, we find that the coefficient of the term $x_1^{p_1} \cdots x_k^{p_k}$ in this product is given by
$$
\sum_{\epsilon_j = \pm 1} \epsilon_1 \cdots \epsilon_k \binom{k}{p_{1},\dots,p_k}
\epsilon_1^{p_1} \cdots \epsilon_k^{p_k}, \tag{*}
$$
where $\binom{k}{p_{1},\dots,p_k}$ is a multinomial coefficient. I claim that if any of the exponents $p_k$ is zero, then the coefficient above is zero. To see that this is the case, suppose without loss of generality that $p_1 = 0$. It follows that
$$
\begin{align}
\sum_{\epsilon_j = \pm 1} \epsilon_1\epsilon_2 \cdots \epsilon_k \binom{k}{p_{1},\dots,p_k}
\epsilon_1^{p_1}\epsilon_2^{p_2} \cdots \epsilon_k^{p_k} &= 
\sum_{\epsilon_j = \pm 1} \epsilon_1\epsilon_2 \cdots \epsilon_k \binom{k}{0,p_2,\dots,p_k}
\epsilon_2^{p_2} \cdots \epsilon_k^{p_k}
\\ & = 
\sum_{\epsilon_j = \pm 1} (1)\cdot\epsilon_2 \cdots \epsilon_k \binom{k}{0,p_{2},\dots,p_k}
\epsilon_2^{p_2} \cdots \epsilon_k^{p_k}
\\ & \quad + \sum_{\epsilon_j = \pm 1} (-1)\cdot\epsilon_2 \cdots \epsilon_k \binom{k}{0,p_{2},\dots,p_k}
\epsilon_2^{p_2} \cdots \epsilon_k^{p_k}
\\ & = 0.
\end{align}
$$
Thus, the expansion of $\sum_{\epsilon_j = \pm 1}\epsilon_1 \cdots \epsilon _k \left( \sum_{j=1}^k \epsilon_j x_j\right)^k$ is simply of the form $Cx_{1}\dots x_k$. From there, it suffices to find $C$. To that effect, we can apply (*) to find
$$
\sum_{\epsilon_j = \pm 1} \epsilon_1 \cdots \epsilon_k \binom{k}{1,\dots,1}
\epsilon_1^{1} \cdots \epsilon_k^{1} = 
\sum_{\epsilon_j = \pm 1} k! = 2^k k!,
$$
which leads to the desired result.
