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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?

JustAnAmateur
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  • What do you mean by subsequence principle? – Kandinskij Sep 13 '23 at 14:50
  • @Kandinskij https://math.stackexchange.com/q/397978/589373 it is this Lemma. Isn't my argument precisely an application of this or am I too tired and I messed it up somehow? – JustAnAmateur Sep 13 '23 at 14:54
  • @MW why do you need the topology to be metrizable? See the link above, this works in any topological space if I understand it correctly. – JustAnAmateur Sep 13 '23 at 14:57
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    @JustAnAmateur You are correct subsequence principle is fine for general topology, unless we are both missing something, apologies for jumping the gun on that explanation. The issue is that pointwise ae convergence doesn’t come from a topology at all. – M W Sep 13 '23 at 15:18
  • @MW thank you very much! I wasn't aware of this, but I looked it up now and you are correct. Now I see why my argument doesn't work. – JustAnAmateur Sep 13 '23 at 15:22
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    @JustAnAmateur I would add that if there were such a topology, then your argument would already show that $L^2$ convergence implies ae convergence, which is of course false since you can take $f_n =\chi_{I_n}$ where $I_n$ enumerates the diadic intervals in $[0,1]$. – M W Sep 13 '23 at 15:34

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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?

Kandinskij
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  • I tend to believe that your counterexample works, but could you please fill in some details so that it is more clear why this is a counterexample? – JustAnAmateur Sep 13 '23 at 16:36
  • Is it clear to you that this sequence converges to $0$ in $H^1$ ? – Kandinskij Sep 14 '23 at 12:19
  • I don't know how to easily see that rigorously, but I feel like it does converge to $0$ in $H^1$, yes. – JustAnAmateur Sep 14 '23 at 13:55
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    Clearly the maximum of $F_{[c,d]}$ is reached in the middle point of $[c,d]$ and it's $(d-c)/2=\mathcal{L}[c,d]/2$ so clearly $|F_{J_n}|\leq \frac{\mathcal{L}(J_n)}{2}\chi_{J_n}$. This implies that $||F_{J_n}||{L^2}\leq ||\frac{\mathcal{L}(J_n)}{2}\chi{J_n}||{L^2}=\frac{\mathcal{L}(J_n)^2}{2}$. Clearly as $n\to \infty$ the RHS converges to 0. So $F{J_n}$ converges to $0$ in $L^2$. A similar reasoning applies to the weak derivative of $F_{J_n}$ (which is equal to the classical derivative almost everywhere since $F_{J_n}$ is a continous piecewise $C^1$ function). Is it now more clear? – Kandinskij Sep 15 '23 at 08:09
  • yes, it is, I can see it now, thanks. – JustAnAmateur Sep 15 '23 at 14:06
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    And can you see that derivatives $F_{J_n}'$ don't converge almost everywhere to $0$? – Kandinskij Sep 15 '23 at 14:28
  • yes, I can, thank you for your help! – JustAnAmateur Sep 15 '23 at 21:17