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Let $H$ be a compact subgroup of a locally compact topological group $G$ and $$ L^1(G:H)=\{f\in L^1(G): R_h f=f\;\text{ a.e. }\; \forall h \in H\}$$ and $\widehat{(G:H)}=\{\xi\in \hat{G}:\xi|_H=1\}$($\hat{G}$ is the character group of $G$).

We know that corresponding to each $\xi \in \widehat{(G:H)}$ there is a $\varphi_\xi \in \text{Spectrum} (L^1(G:H)).$ Does every member of $\text{Spectrum} (L^1(G:H))$ arise as $\varphi_\xi$ where $\xi \in \widehat{(G:H)}$?

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    $\begingroup$ I think I'm confused about the multiplication in $L^1(G:H)$. If you want to use the standard convolution from $L^1(G)$, don't you need $R_h f = f$ and $L_h f= f$? $\endgroup$
    – Ben Willson
    Commented Feb 24, 2015 at 13:32
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    $\begingroup$ Nevermind... I was thinking of something else. $\endgroup$
    – Ben Willson
    Commented Feb 24, 2015 at 13:40
  • $\begingroup$ When we are going to prove the multiplcativity of $\varphi$ on $L^1(G:H)$, we need that if φ(f)≠0 for some f∈L^1(G:H), then we can conclude that φ(L_xf)≠0 for every x∈G, which this true in $L^1(G)$. Because $g\ast L_xf=\Delta(x^{-1})R_xg\ast f$ and we know that $R_xg \in L^1(G)$ where $f,g\in L^1(G)$. But, since $R_xg$ dosen't lie in $L^1(G:H)$ when $f,g \in L^1(G:H)$, hence this is don't hold about of $L^1(G:H)$. I can refer to Hewwit and ross's book about of L^1(G). $\endgroup$
    – B.Gillan
    Commented Feb 25, 2015 at 18:29

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