Let $G$ be a compact Lie group and $u\in C^{0}\left(G\right) $. If $\int_{G} u\left( g \right)v \left(g \right)dg= 0$ for every $v\in V $, a subset which is dense in $C^{0}\left(G\right)$, then $u=0$.
How to proof the above statement? It seems intuitive true but I can't convince myself in mathematical rigor.
This is true for any compact Hausdorff topological space $X$ equipped with a Borel measure $\mu$. First, note $\left|\int_X f g\ \text{d}\mu\right|\leq \|f\|_\infty \|g\|_\infty\ \mu(X)$ for all $f,g\in C^0(X)$, so $(f,g)\mapsto \int_X f g\ \text{d}\mu$ is continuous in both variables with respect to the $\|-\|_\infty$-norm. Hence, if $f\in C^0(X)$ satisfies $\int_X f g\ \text{d}\mu=0$ for all $g$ within a dense subset of $(C^0(X),\|-\|_\infty)$, then $\int_X f g\ \text{d}\mu=0$ for all $g\in C^0(X)$. Now if $f$ was non-zero at some $x\in X$, you could pick for $g$ a non-negative function with support inside a sufficiently small neighbourhood of $x$, forcing $\int_X f g\ \text{d}\mu\neq 0$.