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Let $1\le p \le \infty$. Prove that for all $\epsilon >0$ there exists a constant $C>0$ such that $$\|u'\|_{L^p(\mathbb{R})}\le \epsilon \|u''\|_{L^p(\mathbb{R})}+C\|u\|_{L^p(\mathbb{R})} \;\; \; \forall u\in W^{2,p}(\mathbb{R}).$$ This is a special case of an interpolation inequality, see here http://conteudo.icmc.usp.br/pessoas/andcarva/sobolew.pdf , theorem 2.

I need help to prove this. In any case the first step is to choose a continuous representative function for u. Then, I have the following 2 ideas:

1) To do it similarly as to prove theorem 2. Apply the mean value theorem to u to receive an estimation for $|u'|$. But the first problem I have here is, that the mean value theorem is local. But here the domain is $\mathbb{R}$. Therefore, i don't know how to proceed.

2) To use the fundamental theorem of anlysis applied to $u'$, to receive something like $$u'(\delta+x)=u'(x)+\int_x^{x+\delta} u''(z)dz.$$ Furthermore, convexity of $|\cdot |^p$ and Jensens inequality https://de.wikipedia.org/wiki/Jensensche_Ungleichung might be important, but I am not completely sure how to math that.

The second idea may be more promising. I appreciant any help.

Edit: There are other variants of this inequality available and meanwhile I have discovered Interpolation inequality and Proving an interpolation inequality for $C^2_b$ functions which seem to be a variant of the inequality above (however, the norms are different).

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  • $\begingroup$ Should it not rather be $W^{2,p}$ otherwise the summands don’t Need to exist? $\endgroup$ – Idun E. Jan 16 at 2:59
  • $\begingroup$ thank you. I fixed it $\endgroup$ – TheAppliedTheoreticalOne Jan 16 at 10:32
  • $\begingroup$ The first pdf you link gives a proof (they even prove it first for the one-dimensional case). Is there anything you don't understand? $\endgroup$ – MaoWao Jan 16 at 15:42
  • $\begingroup$ @MaoWao in the meantime I have been succesful proving it with idea 2! I am going to post my solution on MSE as soon as possible $\endgroup$ – TheAppliedTheoreticalOne Jan 16 at 16:15
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In fact, I think you only need to prove for $u\in C_c^{\infty}(\mathbb{R})$ which is a dense subset of $W^{1,2}(\mathbb{R})$ (But NOT dense in $W^{1,2}[0,100]$). Then the desired result holds by pushing limit. I hope I remembered it right. :)

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  • $\begingroup$ Could you explain how you would push the Limits? Since $C^\infty_c(\mathbb{R})$ is dense, we finde a converving subsequence with regards to $W^{2,k}$. But Since $\mathbb{R}$ is not bounded, this does not imply convergence in $L^p$, if I remeber it correctly. So how does ist work? $\endgroup$ – Idun E. Jan 16 at 11:51
  • $\begingroup$ @IdunE. $C_c^\infty(\mathbb{R})$ is dense in $W^{2,p}(\mathbb{R})$. Since both sides of the inequality are continuous w.r.t. to the $W^{2,p}$-norm, it suffices to prove it on a dense subset. The case $p=\infty$ has to be treated separately, I think (if it's even true at all). $\endgroup$ – MaoWao Jan 16 at 15:44

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