In a previous post (see continuity of $L^p$ norms with respect to $p$) it is shown that in a measure space $(\Omega,\Sigma,\mu)$, if $1\leq p_0\leq p\leq p_1\leq+\infty$, then the function $\Phi\colon L^{p_0}(\Omega)\cap L^{p_1}(\Omega)\times [p_0,p_1]\rightarrow \mathbb{R}$ defined as
$$ (f,p)\mapsto \|f\|_{p}, $$
where $\|f\|_p$ denotes the norm of $f$ in $L^p(\Omega)$, is continuous with the respect to the metric $d((f_1,q_1),(f_2,q_2)):=\|f_1-f_2\|_{p_0}+\|f_1-f_2\|_{p_1}+|q_1-q_2|$ on the product space $L^{p_0}(\Omega)\cap L^{p_1}(\Omega)\times [p_0,p_1]$.
Is it true that it is sequentially weakly lower semicontinuous as well, i.e., is it true that if $ f_j\rightharpoonup f$ weakly in $L^{q}(\Omega)$ for any $q\in[p_0,p_1]$ and $p_j\rightarrow p$ then
$$\|f\|_{p}\leq \liminf_{j\to\infty} \|f_j\|_{p_j}?$$
It would be enough to prove that the function defined above is convex, but I did not manage doing so. I also tried using Mazur's Lemma to exploit the convexity of the function in the first variable and the fact that it is continuous with respect to the strong convergence in $L^{p_0}(\Omega)\cap L^{p_1}(\Omega)\times [p_0,p_1]$, but it did not help much.
I am not able to produce simple counterexamples with periodic functions either (nor more complex), so I am tempted to say that it is true. Do you have any references, hints or an actual proof (or counterexample) for this?
I managed to positively answer to the question in the case where $(\Omega,\Sigma,\mu)$ is $\sigma$-finite (which was the case I was truly interested in).
Consider $1\leq p_0<p_1\leq\infty$ and assume first that $\mu(\Omega)=1$, where $\mu$ and $\Omega$ are the (non-negative) reference measure and the reference set, respectively. Suppose that $u_j\rightharpoonup u$ weakly in $L^{p_1}(\Omega)$ and $p_j\to p$. Note that, since $\mu(\Omega)<+\infty$, this implies that $u_j\rightharpoonup u$ weakly in $L^{q}(\Omega)$ for any $q\in[p_0,p_1]$.
If $p_j \searrow p$, then by Holder's inequality and the usual lower semicontinuity at fixed exponent we have
$$\liminf_j \|u_j\|_{p_{j}} \geq \liminf_j \|u_j\|_{p} \geq \|u\|_p.$$
If $p_{j}\nearrow p$, upon passing to a proper subsequence (not relabelled), we may assume that
$$\exists\, L:= \lim_j \|u_j\|_{p_{j}} = \liminf_j \|u_j\|_{p_j} <\infty.$$
By Mazur's lemma, for every $j\in\mathbb{N}$ we find natural numbers $C(j)$ and non negative terms $c_{j,n}$ with naturals $n\in\{j,...,C(j)\}$ such that
$$\sum^{C(j)}_{n={j}}c_{j,n}=1$$
and the functions
$$v_j:=\sum^{C(j)}_{n={j}} c_{j,n} u_{n}, \quad j\in\mathbb{N}$$ strongly converge to $u$ in in $L^q(\Omega)$ for every $q\in[p_0,p_1]$.
Fix $\eta>0$, by the triangular inequality and Holder's inequality we have \begin{eqnarray*} \nonumber \|v_j\|_{p_{j}} \leq && \sum^{C(j)}_{n={j}}c_{j,n} \|u_{n}\|_{p_{j}} \\ \nonumber \leq && \sum^{C(j)}_{n={j}}c_{j,n} \|u_{n}\|_{p_{n}} \\ \nonumber \leq && L+\eta \end{eqnarray*} for every $j$ eventually large. Passing to the limit in $j$, we get \begin{equation*} L+\eta \geq \lim_j \|v_j\|_{p_{j}}\\ = \|u\|_{p}, \end{equation*} where the last equality follows by the strong convergence (the answer in the question posted above). We conclude by the arbitrariness of $\eta$.
If $1\neq\mu(\Omega)<\infty$, we repeat the above argument changing the reference measure into $\mu':=\mu/\mu(\Omega)$ and the claim follows.
If $\mu(\Omega)=\infty$, assume that $\mu$ is $\sigma$-finite, so that there exists a countable family of measurable sets $\{B_R, R\in \mathbb{N}\}$ such that $\mu(B_R)<\infty$ for every $R$ and $\chi_{B_R} \nearrow 1$ pointwise $\mu-$a.e. in $\Omega$; then define the finite measures $\mu_R$ as the restriction of $\mu$ to $B_R$ and apply the above result to get
$$\liminf_j \|u_j\|_{L^{p_{j}}_{\mu}} \geq \liminf_j \|u_j\|_{L^{p_{j}}_{\mu_R}} \geq \|u\|_{L^p_{\mu_R}}. $$ Finally take the supremum among $R\in\mathbb{N}$ to conclude.