By multilinearity of $f:\ell^\infty(\mathbb{R})\to\mathbb{R}$ here I mean that, for every $(x_1,\cdots,x_n,\cdots)\in\ell^\infty$ and $i\in\mathbb{N}^*$, we have \begin{align*} f(x_1,\cdots,x_{i-1},x_i+x,x_{i+1},\cdots)=f(x_1,\cdots,x_{i-1},x_i,x_{i+1},\cdots)+f(x_1,\cdots,x_{i-1},x,x_{i+1},\cdots),\quad\forall x\in\mathbb{R};\\ f(x_1,\cdots,x_{i-1},\lambda x_i,x_{i+1},\cdots)=\lambda f(x_1,\cdots,x_{i-1},x_i,x_{i+1},\cdots),\quad\forall \lambda\in\mathbb{R}. \end{align*} If we suppose that $f$ is multilinear and (pointwise) continuous, must we have $f\equiv 0$?
Of course, if $x=(x_1,\cdots,x_n,\cdots)\in\ell^\infty$ has $x_i=0$ for some $i\in\mathbb{N}^*$, then $f(x)=0$. By continuity we obtain that $f(x)=0$ if $\displaystyle\inf_{i\in\mathbb{N}^*}|x_i|=0$. What else can be said about $f$? Any help appreciated.
No, even with just the second requirement. Assume $f(x) \neq 0$ for some $x$.
Let $y^n = \left((1 + \frac 11)\cdot x_1, (1 + \frac 12)\cdot x_2, (1 + \frac 13)\cdot x_3, \ldots, (1 + \frac 1n)\cdot x_n, x_{n+1}, x_{n+2}, \ldots\right)$. Then $y^n$ converges in $l^\infty$, but $f(y^n) = f(x) \cdot \prod\limits_{i=1}^n \left(1 + \frac 1n\right)$ diverges.