Let $\{\phi_n\}$ be a sequence in $L^p(X)$. Assume that $\phi_n\to \phi$ weak* under the natural identification $L^p\cong (L^q)^*$. Of course, it is not true in general that $\phi_n\to \phi$ in $L^p$ and we can't even hope for $\|\phi_n\|_p \to \|\phi\|_p$.
My question is whether something can be said about the relation between $\|\phi\|_p$ and $\lim_n \|\phi_n\|_p$.
I guess that the correct relation should be something like
$$ \|\phi\|_p\le \liminf \|\phi_n\|_p $$
at least in some cases. (If it helps, feel free to consider special cases for $1\le p\le\infty$ and/or $X$).
Thanks!
Note first that we generally have the identification $L^p \cong (L^q)^\ast$ only for $q < \infty$. For $q = \infty$, $L^1$ is usually a proper subspace of $(L^\infty)^\ast$. And I'm not 100% sure that $L^\infty \cong (L^1)^\ast$ if the measure is not $\sigma$-finite, so let us assume a $\sigma$-finite measure. Also, let's exclude the trivial measure. We then have an isometric embedding $L^p \hookrightarrow (L^q)^\ast$ whenever $\frac{1}{p} + \frac{1}{q} = 1$.
Indeed, the inequality is
$$\lVert\phi\rVert_p \leqslant \liminf_{n\to\infty} \lVert \phi_n\rVert_p.\tag{1}$$
For any $\varepsilon > 0$, there is a $\psi \in L^q$ with $\lVert\psi\rVert_q = 1$ and
$$\int \psi\cdot \phi \,d\mu \geqslant \lVert\phi\rVert_p - \varepsilon.$$
But then we have
$$\lVert\phi_n\rVert_p = \lVert\phi_n\rVert_p\lVert\psi\rVert_q \geqslant \int \phi_n\cdot\psi\,d\mu \to \int \phi\cdot\psi\,d\mu \geqslant \lVert\phi\rVert_p-\varepsilon,$$
which implies
$$\liminf_{n\to\infty} \lVert\phi_n\rVert_p \geqslant \lVert\phi\rVert_p - \varepsilon.$$
Since that holds for all $\varepsilon > 0$, $(1)$ follows.
That is not particular to $L^p$ spaces, whenever we have a weakly (or weak$^\ast$) convergent sequence, the norm of the weak (weak$^\ast$) limit is less than or equal to the $\liminf$ of the norms. The argument is mutatis mutandis the same.
It may be worth mentioning that for $1 < p < \infty$, the $L^p$ spaces are uniformly convex, and in uniformly convex spaces, if $x_n$ converges weakly to $x$ and also $\lVert x_n\rVert \to \lVert x\rVert$, then in fact $x_n$ converges to $x$ in the norm topology.