Let $G$ be finite group and $H$ its subgroup. We assume that everywhere the base field is $\mathbb{C}$. For any characters $\phi$ of $G$-representation and $\psi$ of $H$-representation Frobenius reciprocity states that: $$ \langle Ind^G_H(\psi), \phi \rangle_G = \langle \psi, Res^G_H(\phi) \rangle_H \qquad (1) $$ In particular, this means that multiplicity of $G$-representation in induction of $H$ is equal to the multiplicity of $H$-representation in the restriction of $G$-representation (correct me please if I am wrong somewhere).
Now, let $G$ be reductive Lie group and $H$ its Lie subgroup. Since any $G$-representation is also an $H$-representation, we may define restriction of $G$-representation to subgroup $H$, which will decompose into direct sum of $H$-irreducibles with some multiplicities.
Question: Is there an induction process that builds $G$-representation by given $H$-representation (1)? Does this works?
Yes, there is. Given a representation $\rho: H\to GL(V)$ you can define $Ind_H^G(\rho)$ to be the space of continuous functions $f:G\to V$ such that $f(gh) = \rho(h^{-1})f(g)$ for $h\in H$ and $g\in G$.
$G$ acts on this space via $(g\cdot f(g'))=f(g^{-1}g').$
Even in this setting, the statement of Frobenius reciprocity holds.