Lets take a complex-valued function $f\in L^{2}(G^{d},\mathbb{C})$, 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 $d$-copies of the group. By definition, $f$ fulfills
$$\Vert f\Vert_{L^{2}}^{2}=\int_{G^{d}}\,\bigg (\prod_{i=1}^{d}\mathrm{d}g_{i}\bigg )\,\vert f(g_{1},\dots,g_{d})\vert^{2}<\infty$$
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 $g$ via
$$g(g_{1},g_{2},g_{3}):=f(g_{1},\dots,g_{d})$$
for some fixed values $g_{4},\dots,g_{d}$, is it then true that $g\in L^{2}(G^{3},\mathbb{C})$? I tried several times to find some estimates, but I was not able to prove it.
By definition, we have that
$$\Vert g\Vert_{L^{2}}^{2}=\int_{G^{3}}\,\bigg (\prod_{i=1}^{3}\mathrm{d}g_{i}\bigg )\,\vert f(g_{1},\dots,g_{d})\vert^{2}$$
but I am not sure how I can "insert" the integration for the other variables..
This is not true. Consider
$$ f(x,y) = \frac{e^{-y^2e^{2x^2}}}{x^2 + 1}.$$
This is less than $ e^{-y^2e^{2x^2}} $, which is an $L^2$ function since its square integral with respect to y with x fixed is at most some constant proportion of $ e^{-x^2} $ (via the Gaussian integral), whose integral converges.
But if you fix $y = 0$, the function with respect to $x$ is proportional to a Cauchy distribution.
This counterexample is over $\mathbb{R}^2$, but you should be able to construct an analogous counterexample for a compact Lie group: just have the function’s square integral diverge in the first 3 variables for a specific choice of the other variables, but converge for all other choices and decay sufficiently fast in those other variables so that the whole thing is still $L^2$.