Function defined by diagonalized L2-function of two variables is in L2? Lets take a complex-valued function $f\in L^{2}(G^{2},\mathbb{C})$, where $G$ is some compact Lie group and where the $L^{2}$-space has to be understood with respect to the normalized Haar measure on $2$-copies of the group. By definition, $f$ fulfills
$$\Vert f\Vert_{L^{2}}^{2}=\int_{G^{2}}\,\mathrm{d}g_{1}\mathrm{d}g_{2}\,\vert f(g_{1},g_{2})\vert^{2}<\infty$$
where $\mathrm{d}g_{i}$ is just a short-hand notation for the (normalized) Haar measure with respect to the ith variable. Normalized here means that $\int\mathrm{d}g_{i}=1$ for all $i$. If I define a function $k$ via
$$k(g):=f(g,g)$$
is it then true that $g\in L^{2}(G,\mathbb{C})$?
By definition, it has to satisfy
$$\Vert k\Vert_{L^{2}}^{2}=\int_{G}\mathrm{d}g\,\vert k(g)\vert^{2}=\int_{G}\mathrm{d}g\,\vert f(g,g)\vert^{2}<\infty$$
In the end, we can rewrite
$$\Vert k\Vert_{L^{2}}^{2}=\int_{G}\mathrm{d}g\,\vert k(g)\vert^{2}=\int_{G}\mathrm{d}g\,\vert f(g,g)\vert^{2}=\int_{G^{2}}\mathrm{d}g_{1}\mathrm{d}g_{2}\,\delta(g_{1}g_{2}^{-1})\vert f(g_{1},g_{2})\vert^{2}$$
but I do not know how to continue...
 A: As said in the comments, this does not really make sense since the diagonal is a measure 0 set and elements of $L^p$ spaces are equivalence classes of functions that are equal a.e., so given any $F\in L^2$ we can define $F'$ being equal to $F$ everywhere except the diagonal and $0$ on the diagonal and then we trivially get the result. However, if we are seeing elements as equal only when they are equal in the usual sense and not a.e., things change:
I am not sure about this since I am not at all familiar with Lie groups and such, but here is an argument I've come across:
The circle $\mathbb{S}^1$ is a Lie group and the Haar measure is the Lebesgue measure on $[0,1]$ after identifying $0$ with $1$. Now on $\mathbb{S}^1\times\mathbb{S}^1$ consider the function $F$ defined by
$$F(\zeta_1,\zeta_2)=\begin{cases}0,\text{ if }\zeta_1\ne\zeta_2\\ \infty,\text{ if }\zeta_1=\zeta_2\end{cases}$$
Then since the diagonal $\{(\zeta_1,\zeta_2)\in\mathbb{S}^1\times\mathbb{S}^1:\zeta_1=\zeta_2\}$ has measure $0$, we have that $F=0$ a.e. and thus $F\in L^2(\mathbb{S}^1\times\mathbb{S}^1)$. However, the map $f(\zeta):=F(\zeta,\zeta)$ is constantly $\infty$, so $f\not\in L^2(\mathbb{S}^1)$.
If there is something wrong with this potential counter-example, please let me know so I delete this potential answer.
