Let $1<p<\infty.$
Let $U$ be a bounded open subset in $\mathbb R^n$.
Let ${u_k}$ and $u$ be functions in $L^p(U).$
Suppose $u_k\to u$ weakly in $L^p(U).$
There is a statement saying that the weak convergence does not imply almost everywhere pointwise convergence, i.e. $u_k\to u$ a.e. doesn't hold.
The author gives a short explanation: it is quite possible that for instance that $u_k$ although bounded in $L^p(U)$, are oscillating more and more wildly as $k\to \infty$.
Intuitively I feel like that weak convergence only controls the behavior of integral (by Riesz Representation Theorem), which cannot affect pointwise structure. But I don't know how to construct such example as mentioned by the author in the explanation.
Can someone give me such an example or explain the technic of construction of such an example? Thanks.
Edit: The inequality in the following second question doesn't always hold and so please ignore my second question. For keeping the consistency of previous discussion, I don't delete the second question.
And if possible, I want to ask another question about weak convergence too.
We know by definition, if $L$ is a bounded linear functional on $L^p(U)$, then $$\lim_{k\to \infty} L(u_k)=L(u).$$
Statement: If $L$ is merely a functional on $L^p(U),$ neither linear nor bounded, then we still have that $$\liminf_{k\to \infty} L(u_k)\ge L(u).$$
My second question is how to prove the statement.
There is a condition called coercivity condition on $L$, I don't know if which is needed for the proof of the statement.
Suppose $u\in W^{1,p}(U).$ Then we say $L$ satisfies a coercivity condition if $$\exists \gamma \ge 0 \text{ and }\delta >0\text{ s.t. } L(u)\ge\delta \| Du\|_{L^p}^p - \gamma$$
Even norm convergence in $L^{p}$ does not imply a.e. convergence. There is standard example of a sequence of measurable in $(0,1)$ (with Lebesgue measure) converging in measure , uniformly bounded, but not a.e. convergent. [Arrange characteristic functions of the intervals $(\frac {i-1} {2^{n}}, \frac i {2^{n}})$ in a sequence with increasing denominators].
If the inequality in your second question is true then, by changing the signs of $u$ and $u_k$ we can conclude that $L(U_k) \to L(U)$. In particular this would be true whenever $u_k \to u$ in the norm. So thie make $L$ necessarily continuous.