Is the space of E-spectral functions of Lp(G) Banach? By the space of $E$-spectral functions of $L^p(G)$ where $G$ is an infinite compact abelian group, $E \subseteq \Gamma$ where $\Gamma$ is the dual group of $G$, and $1<p<\infty$, I mean the set
$L_E^p(G) = \{f \in L^p(G) : \widehat{f}(\gamma) = 0, \forall \gamma \notin E\}$
Is this set a Banach subspace of $L^p(G)$?
Since the Fourier transform is linear on $L^1(G)$ and $L^p(G) \subseteq L^1(G)$ since $G$ is compact, therefore $L_E^p(G)$ is a subspace of $L^p(G)$. I have read what seem to be implicit references to it also being a complete space, but I can't find an actual reference for it.
 A: I think I found a solution.
I claim that indeed $L_E^p(G)$ is a Banach subspace of $L^p(G)$.
First, note that trivially $0 \in L_E^p(G)$, and that since the Fourier transform is linear, therefore $L_E^p(G)$ is clearly closed under addition and scalar multiplication and hence is a linear subspace of $L^p(G)$.
To show that $L_E^p(G)$ is moreover a complete subspace of $L^p(G)$, since $L^p(G)$ is complete itself it is therefore sufficient to show that $L_{E}^{p}\left(G\right)$ is a closed subset of $L^{p}\left(G\right)$. Let $f_{n}\in L_{E}^{p}\left(G\right)$ such that $f_{n}\rightarrow f\in L^{p}\left(G\right)$. It is sufficient to show that $\widehat{f}\equiv0$ on $\Gamma-E$. Since $G$ is compact, therefore $L^{p}\left(G\right)\subseteq L^{1}\left(G\right)$. Also since $1<p$ and $f_{n}-f\in L^{p}\left(G\right)$ and that the Haar measure $\lambda$ on $G$ is normalized, therefore
$\left\Vert f_{n}-f\right\Vert _{1}\leq\left\Vert f_{n}-f\right\Vert _{p}$
Hence $f_{n}\rightarrow f$ in $L^{1}\left(G\right)$ too. Since the Fourier transform as a mapping $L^{1}\left(G\right)\rightarrow L^{\infty}\left(\Gamma\right)$ has operator norm less than or equal to $1$, therefore it is bounded and hence
$\left\Vert \widehat{f_{n}}-\widehat{f}\right\Vert _{\infty}=\left\Vert \widehat{f_{n}-f}\right\Vert _{\infty}\leq1\cdot\left\Vert f_{n}-f\right\Vert _{1}$
Consequently $\widehat{f_{n}}\rightarrow\widehat{f}$ in $L^{\infty}\left(\Gamma\right)$. Since for any $\gamma\in\Gamma$
${\displaystyle \left\Vert \widehat{f_{n}}-\widehat{f}\right\Vert _{\infty}=\sup_{\tau\in\Gamma}\left|\widehat{f_{n}}\left(\tau\right)-\widehat{f}\left(\tau\right)\right|\geq\left|\widehat{f_{n}}\left(\gamma\right)-\widehat{f}\left(\gamma\right)\right|}$
therefore $\widehat{f_{n}}\rightarrow\widehat{f}$ pointwise. Let $\gamma\in\Gamma-E$. Now $\forall\varepsilon>0$, $\exists N$ such that $\forall n\geq N$,
$\left|\widehat{f}\left(\gamma\right)\right|=\left|\widehat{f_{n}}\left(\gamma\right)-\widehat{f}\left(\gamma\right)\right|<\varepsilon$
since $f_{n}\in L_{E}^{p}\left(G\right)$. This implies $\widehat{f}\left(\gamma\right)=0$.
