Let $E$, $F$ complex Banach spaces and $p,q\in \mathbb{N}$ with $p+q\geq 1$. I will denote by $\mathcal{L}_a(^{p,q}E;F)$ the subspace of all $(p+q)$-linear mappings $A\in \mathcal{L}_a(^{p+q}E_\mathbb{R};F_\mathbb{R})$ such that:
$$ A(\lambda x_1, \ldots,\lambda x_{p+q}) = \lambda^p \overline{\lambda}^qA(x_1,\ldots,x_{p+q})$$
for all $\lambda\in\mathbb{C}$ and $x_1,\ldots,x_{p+q}\in E$
I am having problems to prove the following assertion:
If $j\neq k$, then $\mathcal{L}_a(^{m-j,j}E;F) \bigcap \mathcal{L}_a(^{m-k,k}E;F)=\{0\}$
I tried the following. Given $(x_1,\ldots,x_m)$ and $\lambda \in \mathbb{K}$, we have: $$A(\lambda x_1,\ldots,\lambda x_m) = \lambda^{m-j}\overline{\lambda}^{j}A(x_1,\ldots,x_m)= \lambda^{m-k}\overline{\lambda}^{k}A(x_1,\ldots,x_m) $$
without loss of generality, $j<k$. Hence:
$$\lambda^{m-k}\overline{\lambda}^j (\lambda^{k-j} - \overline{\lambda}^{k-j})A(x_1,\ldots,x_m)=0$$
I couldn't find an scalar that makes me conclude $A(x_1,\ldots,x_m)=0$.
$A(\lambda x_1,\ldots,\lambda x_m) = \lambda^{m-j}\overline{\lambda}^jA(x_1,\ldots,x_m)= \lambda^{m-k}\overline{\lambda}^k A(x_1,\ldots,x_m)$
From the above, since $\lambda \neq 0$, we have:
$$\left[ \left(\frac{\lambda}{\overline{\lambda}}\right)^{k-j} - 1\right] A(x_1,\ldots,x_m) = 0$$
Thus: $\left[ \left(\frac{\lambda}{|\lambda|}\right)^{2(k-j)} - 1\right] A(x_1,\ldots,x_m) = 0$
Then, writing $\lambda = |\lambda|e^{i\phi}$ and choosing $|\lambda|=1$ and $\phi = \frac{\pi}{4(k-j)}$, we have:
$$(i-1)A(x_1,\ldots,x_m) = 0$$
Hence, $A(x_1,\ldots,x_m)=0$ for all $(x_1,\ldots,x_m)$ and $A=0$.