Let $G$ be a semisimple simply connected group over an algebraically closed field $k$ of characteristic zero, $B$ a Borel and $T$ a maximal torus.
Let $\lambda,\mu,\nu$ be dominant characters of $T$.
Let $V(\lambda)$ be the irreducible representation of highest weight $\lambda$.


If $\lambda=\mu+\nu$, then we know that $V(\lambda)$ is the direct factor of multiplicity one inside $V(\mu)\otimes V(\nu)$.


Is it possible to characterize $V(\lambda)$ inside this tensor product ? 



highest weight representations inside tensor product
To expand my comment (and Robert Bryant's) a little further, the main point here is that tensor products of *finite dimensional* irreducible representations behave nicely. In the equivalent Lie algebra setting, the finite dimensional irreducibles are those whose highest weights are *dominant integral* in the dual of a Cartan subalgebra $\mathfrak{t}:= \mathrm{Lie} \;T$ of $\mathfrak{g} := \mathrm{Lie} \;G$, relative to some fixed choice of positive (or simple) roots. 


In particular, $V(\mu) \otimes V(\nu)$ has highest weight $\lambda:= \mu + \nu$ with multiplicity 1. The complete reducibility of this tensor product depends mainly on the fact that the center $Z(\mathfrak{g})$ of the universal enveloping algebra of $\mathfrak{g}$ (generated by "Casimir operators") acts by distinct central characters on the distinct finite dimensional representations. In other words, the sum of all copies of a typical $V(\pi)$ in the tensor product decomposition is a single "eigenspace" for the action of the center, where the action is given by a "central character" $\chi\_\pi$. 


Since the summand $V(\lambda)$ occurs only once in the tensor product (the weight space for $\lambda$ being one dimensional), $V(\lambda)$ may be characterized as the set of all vectors on which $Z(\mathfrak{g})$ acts by the central character $\chi\_\lambda$. (There may of course be many isomorphic summands of smaller highest weights $\pi$, where the center acts by the single character $\chi\_\pi$.) In different language this is essentially Robert's comment. 


