I am confused with the following argument concerning dual spaces of Lebesgue spaces. Let $\Omega \subset \mathbb R^n$ be a bounded measurable set and consider $L^p = L^p(\Omega)$ for $p \geq 1$. We know the dual $(L^p)^*$ of $L^p$ space is identified with $L^{p'}$ when $1 \leq p < \infty$.
For simplicity, consider $L^3$ so that $L^3 \subset L^2$ and $L^2 = (L^2)^* \subset (L^3)^*$.
Let $T \in (L^3)^*$. By Hahn-Banach extension theorem, there exists $\tilde T \in (L^2)^*$ such that $\tilde T \restriction_{L^3} = T$, and by Riesz' representation theorem, there exists a unique $f \in L^2$ such that $\tilde T(g) = \int_{\Omega} f(x) g(x) \,dx$ for all $g \in L^2$.
Hence, for all $g \in L^3$, we have
\begin{equation*} T(g) = \tilde T(g) = \int_{\Omega} f(x) g(x) \,dx. \end{equation*}
This means $(L^3)^* \subset L^2$ and hence $(L^3)^* = L^2$. But in this is argument, there must be a mistake since $(L^3)^* = L^{3/2}$, though, I do not know what is wrong.
Please any help.
Finally, I understand which step is wrong, it is the Hahn-Banach extension theorem.
For simplicity, let us consider $n=1$ and $\Omega = [0,1]$. Then, the linear functional $T$ on $L^3$ defined by
$$T(g) = \int_{\Omega} x^{-1/2} g(x) \,dx, \quad g \in L^3$$
is continuous on $(L^3, \| \cdot \|_{L^3})$, but not continuous on $(L^3, \| \cdot \|_{L^2})$. Hence, the Hahn-Banach extension theorem is not applicable.