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Can one show that the composition of two bounded linear operators $T_1, T_2: L_2(\Omega) \to H^1(\Omega)$ maps $L_2$ functions to $H^2$, i.e., $T_3 = T_1 T_2$ and $T_3: L_2(\Omega) \to H^2(\Omega)$?

What I tried to do is to bound the $H^2$ norm of a function $\varphi = T_1 T_2 u$. For that I tried to bound $\| \partial_{x_i} \partial_{x_j} \varphi\|_{L_2}$. I tried to do it for $u \in L_2$ and $u \in C^\infty$ (possibly coupled with an argument that $C^\infty$ is dense in $H^2$), but nothing seemed to lead to anywhere.

A second thing I tried to do, is estimating the operator norm of $\| D_{x_i} D_{x_j} T_1 T_2 \|_{L_2}$, where $D_{x_l}$ represents the differentiation operator. However, what I would need for that is that $D_{x_j}$ and $T_1$ commute within the operator norm, which I couldn't see why.

bheinzek
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  • What did you try? – LL 3.14 Jun 12 '23 at 08:37
  • I tried to bound the $H^2$ norm of a function $\varphi = T_1 T_2 u$. For that I tried to bound $| \partial_{x_i} \partial_{x_j} \varphi|_2$. I tried to do it for $u \in L_2$ and $u \in C^\infty$, but nothing seemed to lead anywhere. – bheinzek Jun 12 '23 at 08:41
  • @bheinzek Could you please add that to your post? That would be greatly appreciated. Thank you. – Cheese Cake Jun 12 '23 at 08:42
  • That's a good point. I have done it. Thanks – bheinzek Jun 12 '23 at 08:46
  • Added it to the question – bheinzek Jun 13 '23 at 07:55
  • @stange: The problem with that is that the inclusion map $L_2 \to H^1$ is not well-defined as far as I know. – bheinzek Jun 13 '23 at 08:38
  • I don't think this is true in general: Take $f\in H^1\setminus H^2$, and define $P$ to be the $L^2$-orthogonal projection onto the span of $f$. There exists $C=C(f)>0$ such that $| f|{H^1}\leq C| f|{L^2}$, and therefore $| P u|{H^1} \leq C| P u|{L^2}\leq C| u|_{L^2}$, i.e. $P:L^2\to H^1$. Now take $T_1=T_2=P$. – Jose27 Jun 13 '23 at 09:08
  • I see, that makes sense. Thanks so much! – bheinzek Jun 13 '23 at 15:24

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