Does $H^1(a, b)$ convergence imply pointwise convergence a.e. for the weak derivative?

Question

Let $(q_n)_n$ be a sequence in $H^1(a, b)$ that converges to some $q\in H^1(a, b)$. I would like to know if it is true that $q_n'(t)\to q'(t)$ for a.e. $t\in (a, b)$.

I believe it is and here is what I tried to do. Consider a subsequence $(q_{n_k})_k$ of $(q_n)_n$. Then, we also have $q_{n_k}\to q$ in $H^1(a, b)$, so, in particular, $q_{n_k}'\to q'$ in $L^2(a, b)$, which says that there is some subsequence $(q_{n_{k_l}})_l$ of $(q_{n_k})_k$ that converges pointwise for a.e. $t\in (a, b)$ to $q'(t)$. Therefore, by the subsequence principle I believe that we may conclude that $q_n'(t)\to q'(t)$ for a.e. $t\in (a, b)$. Is this correct?

Answer 1

This is not true. Let $[c,d]\subseteq [a,b]$ and let $F_{[c,d]}:[a,b]\to \mathbb{R}$ be the function defined as follows:

1. Constantly zero for $x\leq c$ and for $x\geq d$
2. It grows linearly with slope $1$ in the first half $[c,d]$. It decreases linearly with slope $-1$ in the second half of $[c,d]$

Now I define the family of subintervals $(I_{ij})_{i\in \mathbb{N},j\leq i}$ to be the subinterval of $[a,b]$ obtained as follows

• Partition $[a,b]$ into $i$ subintervals of measure $(b-a)/i$
• $I_{ij}$ is the $j$-th interval in this partition.

Now I define the following sequence of intervals $$I_{11},I_{21},I_{22},I_{31},I_{32},I_{33},I_{41},...$$ and I indicate with $J_n$ the $n$-th interval of this sequence. Can you see that $q_n:=F_{J_n}$ is a counterexample?