# Example of a discontinuous multilinear map whose symmetrization is continuous

Let $X,Y$ be normed spaces over $\mathbb{K}$ and $T:X^n\rightarrow Y$ be an $n$-multilinear map.

Let the symmetrization of $T$ be defined as $ST(x):=\frac{1}{n!}\sum_{\sigma\in S_n} T(x_{\sigma(1)},...,x_{\sigma(n)})$ for all $x\in X^n$.

Is there a case $ST$ is continuous but $T$ is discontinuous?

Define $\phi(x):=T(x,...,x)$ for each $x\in X$. Invoking multilinear algebra fact, one can show that if $\phi$ is continous, then $ST$ is continuous. I am curious if the conclusion can be generalized to $T$.

(Since $T$ is always continuous if $X$ is finite-dimensional, if there is a counterexample for this, it must be the case $X$ is infinite-dimensional.)

Let $X=C^1(\mathbb R^n;\mathbb R)$, $Y=C^0(\mathbb R^n;\mathbb R)$, both with the uniform norm, and $\partial_i=:\frac{\partial}{\partial x^i}:X\to Y$ the usual partial derivative. Define $T=\partial_1\otimes\cdots\otimes\partial_n:X^n\to Y$ by $T(f^1,\dots,f^n)=\partial_1f^1\cdots\partial_nf^n$. Let $A$ be the alternation of $T$. Then $A$ is discontinuous, but $SA=0$. ($A$ is a kind of Poisson bracket.)