Let $G$ be a discrete group and $LG$ the corresponding group von Neumann algebra. For a set $F\subset G$, let $P_{F}:\ell^{2}(G)\rightarrow\ell^{2}(F)$ denote the orthogonal projection onto $\ell^{2}(F)$. Let $x\in LG$.
My question is: why does $\|P_{F}(x)\|_{2}$ makes sense? That is, how do you interpret $P_{F}(x)$ and why does it belong to $L^{2}(LG)$ $\;$($=LG$)? It doesn't appear to me that $P_{F}$ belongs to $LG$ in general, so I don't think it can simply be $P_{F}x$.
I prefer different symbols for the objects presented in OP.
Let $VN(G)$ denote the von Neumann algebra of $G,$ i.e. the space of all operators $A:\ell^2(G)\to \ell^2(G)$ which commute with right translations $\varrho(g)$ given by $\varrho(g)f(x)=f(xg),$ for $x\in G.$ Let $A\in VN(G).$ Then $a=A\delta_e$ is a function in $\ell^2(G),$ i.e. the quantity $\|A\delta_e\|_{\ell^2}$ is well defined.
Concerning the question whether a projections onto a subset of $F\subset G,$ of the operators in $VN(G)$ belong to $VN(G),$ the answer is negative even for $G=\mathbb{Z}.$ The convolution operator on $\ell^2(\mathbb{Z})$ is bounded if and only if the Fourier transform $\widehat{a}$ is a bounded function. Let $F=\mathbb{N}_0.$ It is well known that the mappings $$\sum_{-N}^N a_ne^{int}\longmapsto \sum_{n=0}^Na_ne^{int}$$ are not uniformly bounded on $L^1(\mathbb{T})$ and therefore on $L^\infty(\mathbb{T}).$ By the Banach-Steinhaus uniform bounded principle there exists a $f\in L^\infty(\mathbb{T})$ such that $\sum_{n=0}^\infty f_ne^{int}$ is not bounded.
Additional explanations