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Previous answer was wrong as it required scalar $W_i$ (i.e. $n=1$) but more than on term in product.
F_G
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Function $W \mapsto f(W)$ is not convex when $n \ge 2$.

Take for example when $n=2$ and fix $x=1$.

If function $f(W_1,W_2)$ is convex then it must also be convex when restricted to a linear subspace.

Restricting to $W_1 = \begin{pmatrix} x & 0 \\ 0 & 0 \end{pmatrix}$ and $W_2 = \begin{pmatrix} y & 0 \\ 0 & 0 \end{pmatrix}$ then gives $f(W_1,W_2) = x^2 y^2$, which is not convex in (x,y) (this can be checked by computing its Hessian matrix).

F_G
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