For a given space $\mathcal{X}$, consider the Banach space $L_p(\mathcal{X},\mu)$ for Lebesgue distribution $\mu$, where $p \neq 2$. It is straightforward that the dual space to this is $L_q(\mathcal{X},\mu)$ where $1/p + 1/q = 1$.
I'm trying to find kernels $\mathcal{K}: \mathcal{X} \times \mathcal{X} \to C$ and $G: \mathcal{X} \times \mathcal{X} \to C$ such that
$G(x,\cdot) \in L_p(\mathcal{X},\mu)$ and $\mathcal{K}(\cdot, x) = (G(x,\cdot))^*$ for all $x \in \mathcal{X}$
$f(x) = (f, G(x, \cdot))_{L_p}$, $f^∗(x) = (K(x, ·), f )_{L_p}$, $f \in L_p, f^∗ \in L_p, x \in \mathcal{X}$.
One can consider a semi-inner product in this case where $(f,g)_{L_p} = \frac{\int_{\mathcal{X}} f\bar{g}|g|^{p-2}}{||g||^{p-2}_{L_p}}$ for $f,g \in L_p$. It is clear that the dual map in this case has the following form: $f^* = \frac{\bar{f}|f|^{p-2}}{||f||^{p-2}_{L_p}}$. But it is not clear to me what could be valid reproducing kernels. Can anyone help understand this or provide some idea as to how to construct one?
PRELIMINARY REMARK. In the following, $p\in [1, \infty)$. The extreme case $p=\infty$ is not covered by the question, as the duality pairing $(\cdot, \cdot)_p$ defined there makes no sense in that case.
Let us consider the case $\mathcal{X}=\mathbb R$. Aiming for a contradiction, suppose that a function $G$ satisfying point 1 existed. Note that, by Hölder's inequality, for arbitrary $f, g\in L^p$ we have that $$ \lvert (f, g)_{L^p}\rvert\le \lVert f\rVert_p \lVert g \rVert_p, $$ so applying this with $g=G(x, \cdot)$ we obtain that, for each $x\in\mathbb R$, there is $C(x)>0$ such that $$\tag{1} \lvert f(x)\rvert\le C(x) \lVert f\rVert_{L^p(\mathbb R)}.$$ Indeed, $C(x)$ can be taken to be $\lVert G(x, \cdot)\rVert_p$.
This inequality (1) is clearly absurd, since $f$ is only defined up to almost everywhere equivalence, hence $f(x)$ can take any value without altering $\lVert f\rVert_{L^p(\mathbb R)}$.
Even if we insisted on trying to prove (1) for continuous $f$ only, we would find it cannot hold unless $p=\infty$. The scaling is off: indeed, considering $$ f_\lambda(y):=\lambda^\frac{1}{p}f(\lambda(y-x)+x), \quad \text{for }\lambda>0, $$ we have that $\lVert f_\lambda\rVert_{L^p(\mathbb R)}=\lVert f\rVert_{L^p(\mathbb R)}$, while $\lvert f_\lambda(x)\rvert =\lambda^\frac1p\lvert f(x)\rvert$. So the left-hand side of (1) can be scaled to be arbitrarily big, while the right-hand side remains constant. This proves that (1) is absurd, even for continuous $f$ only.
NOTE TO THE READER: This scaling argument is easier to understand in the case $x=0$.