For, $1\le p < \infty$, let $X=L^p[-\pi,\pi]$ and define
$$Y := H^p = \left\{x \in L^p : \widehat{x}(n) = \frac{1}{2\pi}\int_{-\pi}^{\pi} x(s)e^{ins}\ \mathrm ds = 0, n=-1,-2,\ldots\right\}$$ $$H_{0}^p = \left\{ x \in L^p : \widehat{x}(n) =0, n=0,-1,-2,\ldots \right\} \subseteq H^p$$ $$(H^p)^{\perp} = \{ f \in (L^p)^{*} \mid f(x)=0 \,\,\forall \, x \in H^p\}$$
I want to show $H_0^q$ is isometrically isomorphic with $(H^p)^{\perp}$, where $1/p+1/q=1$.
My attempt: Consider the map $T : H_0^q \to (H^p)^{\perp}$ defined by $T(x) = T_x$ where $\displaystyle T_x(y) = \int_{-\pi}^{\pi} x(s)y(s)\ \mathrm ds$ for $y \in L^p$. The map $T$ is an onto linear isometry (or isometrically isomorphic map) between $L^q$ and $(L^p)^{*}$. I am having trouble showing that this map is onto and well defined. First, we need to show that $T_x$ is indeed an element of $(H^p)^{\perp}$, i.e $T_x(y) = 0$ for every $y \in H^p$ or, given $\displaystyle\int_{-\pi}^{\pi} x(s)e^{ins}\ \mathrm ds = 0 $ for $n \in \mathbb{Z}_{\le 0}$ and $\displaystyle\int_{-\pi}^{\pi} y(s)e^{-ins}\ \mathrm ds = 0$ for $n \in \mathbb{Z}_{> 0}$, we should have $\displaystyle \int_{-\pi}^{\pi} x(s)y(s)\ \mathrm ds = 0$. For proving that the map $T$ is onto, we need to show that there exists an $x \in H_0^q$ for every $f \in (H^p)^{\perp}$ such that $f = f_x$. Any hints?