TL;DR:  

Given representations $D,\Lambda$ of subgroups $K,Q$ of a Lie group $G$, is it true that every intertwining operator $T$ between the resulting induced representations of $G$ can be written
$$
(T\varphi)(g) = \int\_G t(g^{-1} g')\varphi(g')\,dg'
\tag1
$$
for some function $t$ on $G$?




---


LONG VERSION:


I placed this question at the [Mathematics](https://math.stackexchange.com/q/3415897) StackExchange. Although I presumed that the question is simple if not trivial, no one was able to help me there. So I am now elevating it to the MathOverflow. 


Consider the space $\,{\cal{L}}^G\,$ of all continuous functions $\,G\longrightarrow{\cal{L}}\,$ mapping a Lie group $\,G\,$ into a vector space $\,{\cal{L}}\,$.


Assume that $\,G\,$ has two proper subgroups:
 $$
 K\,,~Q~<~G~~,
 $$
 whose representations, $\,D(K)\,$ and $\,\Lambda(Q)\,$, are acting in $\,{\cal{L}}^G\,$. 


Consider two subspaces of $\,{\cal{L}}^G\,$. One subspace,
 $$
 {\mbox{Map}}\_K(G,\,{\cal{L}})\,=\,\left\{\,\varphi\,\right\}~~,
 $$
 comprises the vector functions $\,\varphi\,$ obeying the equivariance condition
 $$
 \varphi(g\, k)~=~D^{-1}(k)~\varphi(g)~,~~~k\,\in\, K~~.
 $$  

 In this subspace, $\,D(K)\,$ is induced to a representation of $\,G\,$, denoted by
 $$
 U^{(D)}\,\equiv\,D(K)\,\uparrow\, G
 $$
 and implemented with 
 $$
 U^{(D)}\_g\,\varphi(g^{\,\prime})~=~\varphi(g^{-1}\, g^{\,\prime})~~.
 $$


Another subspace,
 $$
 {\mbox{Map}}\_Q(G,\,{\cal{L}})\,=\,\left\{\,\psi\,\right\}
 $$
 will comprise the functions $\,\psi\,$ satisfying 
 $$
 \psi(g\, q)~=~\Lambda^{-1}(q)~\psi(g)~,~~~q\,\in\, Q~~.
 $$
 In this subspace, $\,\Lambda(Q)\,$ is induced to a representation of $\,G\,$, denoted by
 $$
 U^{(\Lambda)}\,\equiv \,\Lambda(Q)\,\uparrow\, G
 $$
 and implemented with 
 $$
 U^{(\Lambda)}\_g\,\psi(g^{\,\prime})~=~\psi(g^{- 1}\, g^{\,\prime})~~.
 $$


While both $\,U^{(D)}\,$ and $\,U^{(\Lambda)}\,$ are realised via left translations, they are different representations, as they are
 acting in subspaces defined by different subsidiary conditions.


For convenience, we summarise this in the table:
 $$
 \varphi(g\, k)=D^{-1}(k)\,\varphi(g)\,,~k\in K \quad \quad \psi(g\, q)=\Lambda^{-1}(q)\,\psi(g)\,,~q\in Q
 $$
$$
 U^{(D)}\,\equiv\,D(K)\,\uparrow\, G \qquad \qquad \qquad U^{(\Lambda)}\,\equiv \,\Lambda(Q)\,\uparrow\, G
 $$
$$ 
 U^{(D)}\_g\,\varphi(g^{\,\prime})~=~\varphi(g^{-1}\, g^{\,\prime}) \quad \quad \qquad U^{(\Lambda)}\_g\,\psi(g^{\,\prime})~=~\psi(g^{- 1}\, g^{\,\prime})~
 $$


Our goal is to describe the space $\,\left[\, D(K)\,\uparrow\, G\,,~\Lambda(Q)\,\uparrow\, G \,\right]\,$ of the morphisms $\,\psi\,=\,\hat{T}\,\varphi\,$. 


QUESTION:


How to prove that the most general form of a morphism is
 $$
 \psi(g)~=~(\hat{T}\,\phi)(g)~=~\int\_G t(g^{-1}\, g^{\,\prime})\,\varphi(g^{\,\prime})\,dg^{\,\prime}~~,\qquad\qquad\qquad(1)
 $$
 where $\,dg\,$ is an invariant measure on $\,G\,$.


PS.


As an aside, I would mention that for the equivariance conditions to be satisfied the kernel must obey one more condition:
$$
t(qgk) = \Lambda(q) t(g) D(k)~~,~~~q\in Q\,,~~g\in G\,,~~k\in K~~.
$$
This, however, is the next theorem; and I don't want to go there until the basic property (1) is proven.



A morphism intertwining two induced representations
I have now found a remarkably simple proof of the fact that the variables $\,g\,$ and $\,g^{\,\prime}\,$ enter $\,t(g\,,\, g^{\,\prime})\,$ through convolution:
 $$
 t(g\,,\, g^{\,\prime})\;=\;t(g^{-1}\, g^{\,\prime})~~.\qquad\qquad\qquad (\*)
 $$ 
To prove this, we should use the definition of an intertwiner,
$$
U\_{g\_1}^{(A)}\,T\,\varphi~=~T\,U^{(D)}\_{g\_2}\,\varphi~~.\qquad\qquad\qquad (\*\*)
$$ 
Its left-hand side can be written down as 
$$
 (\, U\_{g\_1}^{(A)}\, T\,\varphi)\,(g\_2)\,=\,(T\,\varphi)\,(g\_1^{-1}g\_2)
 \,=\,\int t(g\_1^{-1}g\_2\,,\,g^{\,\prime})\,\varphi(g^{\,\prime})\,dg^{\,\prime}\,~.
 $$
Its right-hand side can be processed as
 $$ 
(T\,U\_{g\_1}^{(D)}\,\varphi)\,(g\_2)$$
$$=\,\int t(g\_2\,,\,g^{\,\prime\prime})\,(U\_{g\_1}^{(D)}\varphi)\,(g^{\,\prime\prime})\,dg^{\,\prime\prime}
 \,=\,\int t(g\_2\,,\,g^{\,\prime\prime})\,\varphi(g^{-1}\, g^{\,\prime\prime})\,dg^{\,\prime\prime}
$$
$$ =\int t(g\_2,\,g\_1(g\_1^{-1}\, g^{\prime\prime}))\,\varphi(g^{-1}g^{\prime\prime})\,dg^{\prime\prime}
 =\int t(g\_2,\,g\_1 g^{\prime})\,\varphi(g^{\prime})\, dg^{\prime}~~,
 $$ 
where we defined $\,g^{\,\prime}=\, g\_1^{-1}\, g^{\,\prime\prime}\,$ and assumed the measure left-invariant, $\,dg^{\,\prime}=\, d(g\_1^{-1} g^{\,\prime\prime})
 \,=\, dg^{\,\prime\prime}\,$.


If we plug the right-hand sides of the former and latter formulae in equation (\*\*), we shall arrive at 
 $$
 t(g\_1^{-1}g\_2\,,\,g^{\,\prime})~=~t(g\_2\,,\,g\_1 g^{\,\prime})~~.
 $$ 
If we now set $\,g\_2=1\,$ and $\,g=g\_1^{-1}\,$, the above formula will shape into
 $$
 t(g\,,\,g^{\,\prime})~=~t(1\,,\,g^{-1} g^{\,\prime})~~.
 $$ 
We now can rename $\,t(1\,,\,g^{-1} g^{\,\prime})\,$ as $\,t(g^{-1} g^{\,\prime})\,$ and regard (\*) proven: $\,t(g\,,\, g^{\,\prime})\,=\, t(g^{-1}\, g^{\,\prime})\,$. Accordingly, 
 $$
 \psi(g)\,=\,(\hat{T}\,\phi)(g)\,=\,\int t(g^{-1}\,g^{\,\prime})\,\varphi(g^{\,\prime})\,dg^{\,\prime}\;\;.
 $$ 
My understanding is that this integral may, in principle, diverge *even when the representations are defined on nuclear functions*. (Please correct me if I am wrong.) In such situations, one has to establish analytic continuation, i.e. to define this integral in the complex domain, where is does converge to an analytic function.


