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Let $u, v \in L^2 (\mathbb R)$ such that $$ \int_{\mathbb R} u \varphi'' = \int_{\mathbb R} v \varphi \quad \forall \varphi \in C^\infty_c (\mathbb R). $$

I think that it is not necessarily true that $u \in H^2 (\mathbb R)$.

Could you provide an example of such $u$ with $u \notin H^2 (\mathbb R)$?

Thank you so much for your elaboration!

Akira
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1 Answers1

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If $u$ and $v$ are both in $L^{2}$ which means that $u''$(weak derivative) belongs to $L^{2}$. If we know $u'$ is also in $L^{2}$ thus we can get $u\in H^{2}$.

If ALSO $u\in C_{0}^{\infty}(\mathbb{R})$, $\int_{\mathbb{R}}u'u'dx=\int_{\mathbb{R}}u'du=-\int_{\mathbb{R}}uu''dx$. Thus by Cauchy inequality, we have $\|u'\|_{L^{2}}\lesssim \|u''\|_{L^{2}}+\|u\|_{L^{2}}$.

monotone operator
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  • Your post does not answer my question. I don't assume the existence of $u'$. Btw, is your last inequality trivially true? – Akira Nov 04 '23 at 13:11
  • In the theory of distribution(generalized function theory), if you have $u\in L^{p}$, then as a distribution, $u$ has any order distribution derivative(which does not same as weak derivative explicitly speaking). What we just do not know is whether the first order distribution derivative $u'$ is a $L^{2}$ function. We know that every distribution can be approximated by smooth function with compact support. – monotone operator Nov 04 '23 at 13:16
  • I got it. So my question is whether there is such $u$ with $u' \notin L^2$. – Akira Nov 04 '23 at 13:24
  • There is always an inclusion $H^2 \hookrightarrow H^1$ whatever the domain – see here: https://math.stackexchange.com/questions/625263/sobolev-spaces-inclusion – giobrach Nov 04 '23 at 13:38
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    @giobrach his question is whether $u'\in L^{2}$, in his discription of this problem, we don't know whether $u\in H^{2}$. – monotone operator Nov 04 '23 at 13:45