Comparing different kinds of convergence

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I want to gain a better understanding of the convergence of a sequence of functions $(u_n:I\to \mathbb R)_{n\in \mathbb N}$ in different norms, I know some results, for example that if $u_n(x)\to u(x)$ pointwise a.e. and $u_n\to v$ in $L^1(I)$, then $v(x)=u(x)$ a.e.

However, I don't know if there is a rule of thumb to compare the limits in different conditions, specially when dealing with weak topologies. In particular, I am interest the most in these two scenarios:

  1. Suppose $(u_n)_{n\in \mathbb N}$ is a sequence of continuous functions in $L^\infty(I)$ such that $u_n(x)\to u(x)$ a.e. and $u_n\rightharpoonup v$ weakly in $L^1(I)$, is it true that $v(x)=u(x)$ a.e. ?

  2. If $(u_n)_{n\in \mathbb N}\subset L^p(I)\cap L^q(I)$, where $\frac 1 p+\frac 1 q = 1$ and $(u_n)$ converges strongly in $L^p(I)$ and weakly in $L^q(I)$, does the limits coincide?

In general, if some of these kind of results are false, what kind of 'illness' need the sequence have to have different limits in different norms.

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  1. It suffices to treat the case where $I$ is a finite interval; if we treat the case of $I\cap [-N,N]$ for any $N$ we will get $u=v$ almost everywhere on this interval. We can also assume that $u=0$ since the assumption of weak convergence implies that $\lVert u_n\rVert_1$ is bounded which implies, by Fatou's lemma, that $u$ belongs to $L^1$. using Egoroff's theorem, we split for each integer $k$ the set $I$ into $U_k$ and $R_k$, where $U_k$ is such that $\sup_{x\in U_k}\lvert u_n(x)\rvert\to 0$ as $n$ goes to infinity and $\lambda(R_k)<1/k$. Then for each measurable subset of $I$, $$\int_{A}u_n=\int_{U_k\cap A}u_n+\int_{R_k\cap A}u_n.$$ Taking the limit in $n$, we get $\int_A v=\int_{R_k\cap A}v$ and since $v$ is integrable, we get $\int_A v=0$ hence $v=0$ almost everywhere.

  2. The limits coincide, as you can test weak convergence on indicator functions of sets of finite measure.