Is the symmetrization of $f\in L^2(\mathbb R)$ equal to the orthogonal projection of $f$ on $\hat L^2(\mathbb R)$? Let $f:\mathbb R^n\to\mathbb R$ be a square integrable function, and consider it's symmetrization
$$\tilde{f}(x_1,...,x_n)=\frac 1 {n!}\sum_{\sigma\in\mathcal P_n} f\left(x_{\sigma(1)},...,x_{\sigma(n)}\right).$$
Then let $L^2(\mathbb R)^{\widehat{\otimes} n}\cong \widehat{L}^2(\mathbb R^n)$ be the space of square inegrable symmetric functions, it is true that $\tilde{f}$ equals the orthogonal projection of $f$ on this subspace?
I am aware that this could be a simple question, but I am not sure how to proceed.
 A: Define the operator $S{f}(x_1,...,x_n):=\frac 1 {n!}\sum_{\sigma\in\mathcal P_n} f\left(x_{\sigma(1)},...,x_{\sigma(n)}\right)$. First of all, note that for $\pi \in \mathcal{P}_n$  we have $Sf(x_{\pi(1)},\dots,x_{\pi(n)}) = Sf(x_1,\dots,x_n)$, since $\mathcal{P}_n \circ \pi = \mathcal{P}_n$ as $\mathcal{P}_n$ is a group and
$$Sf(x_{\pi(1)},\dots,x_{\pi(n)})=\frac 1 {n!}\sum_{\sigma\in\mathcal P_n} f\left(x_{\sigma\circ \pi(1)},...,x_{\sigma\circ \pi(n)}\right) = \frac 1 {n!}\sum_{\sigma'\in\mathcal P_n \circ \pi} f\left(x_{\sigma'(1)},...,x_{\sigma'(n)}\right) = Sf(x_1,\dots,x_n). $$
In particular, it follows that
$$ S^2f(x_1,\dots,x_n) = \frac 1 {n!}\sum_{\sigma\in\mathcal P_n} Sf\left(x_{\sigma(1)},...,x_{\sigma(n)}\right) = \frac 1 {n!}\sum_{\sigma\in\mathcal P_n} Sf\left(x_{1},...,x_{n}\right) = Sf(x_1,\dots,x_n).$$
Which shows that $S$ is a projection. To show that $S$ is an orthogonal projection, take $f,g \in L^2$ and calculate
$$\begin{align}\langle Sf, g\rangle &= \frac{1}{n!}\sum_{\sigma \in \mathcal{P}_n}\int f\left(x_{\sigma(1)},...,x_{\sigma(n)}\right)g(x_1,\dots,x_n)~dx_1,\dots,dx_n \\
&= \frac{1}{n!}\sum_{\sigma \in \mathcal{P}_n}\int f\left(x_{1},...,x_{n}\right)g(x_{{\sigma^{-1}}(1)},\dots,x_{{\sigma^{-1}}(1)})~dx_1,\dots,dx_n= \langle f, Sg \rangle,\end{align}
$$
where we used the multivariate substitution rule and the fact that summing over $\sigma^{-1}$ for all $\sigma$ in $\mathcal{P}_n$ is the same as summing over all $\sigma$ in $\mathcal{P}_n$ since $\mathcal{P}_n$ is a group. We have shown that $S$ is an orthogonal projection, and since $Sf = f$ for symmetric $f$ and $Sf$ is symmetric for any $f \in L^2$ we have shown that its range is the space of symmetric functions, which must therefore be closed.
