Let $(X,\Sigma,\mu)$ be a (finite) measure space $\kappa \in L^2(\mu)$ and $p \in (1,2)$. Furthermore, suppose that there is a linear subspace (not necessarily dense) $V \subseteq L^p(\mu)$ such that the mapping \begin{equation} T: \omega \mapsto \int_X \kappa \, \omega \, \mathrm{d} \mu \end{equation} is $L^p(\mu)$-continuous. Then, it is clear that $T$ extends uniquely and continuously to the closure of $V$. However, it is not clear (to me) if this extension is also given by an integral with kernel $\kappa$.
Is there some general statement about when this holds? e.g. it should hold whenever $V$ contains all simple functions. But what if $V$ is much smaller? Canonically, we only get a unique representation in $L^q(\mu) / \mathrm{Ann}\, V$ but since we do not know whether $\kappa \in L^q(\mu)$, this appears non-trivial to me.
Let us consider $X = (-1,1)$ equipped with the Lebesgue measure. Let $\kappa \in L^2$ be an odd function with a singularity at $0$, e.g., $$\kappa(x) = x^{-\alpha}$$ for some $\alpha \le 1/2$. Next, $$ V := \{ v \in L^p \mid \text{$v$ is even and there exists $\varepsilon > 0$ with $v = 0$ on $(-\varepsilon,\varepsilon)$}\}.$$ Then, $T$ is the zero operator on $V$, thus it is clearly bounded by the $L^p$ norm. The closure of $V$ is the set of all even functions from $L^p$ and the extension of $T$ is still zero. However, $v \kappa$ will not be integrable for all even $v \in L^p$ as long as $\alpha > 1/q$.