I need help on the problem 8.9 at page 238 of the book "Functional Analysis, Sobolev Spaces and Partial Differential Equations" by Haim Brezis.

Set $I=(0,1)$.
Let $u \in W^{2,p}(I)$ with $1<p<\infty$.
Assume that $u(0)=u'(0)=0$.
Show that $$\frac{u(x)}{x^2}\in L^p(I)\quad\text{and}\quad \frac{u'(x)}{x}\in L^p(I),$$ with $$\left|\left|\frac{u(x)}{x^2}\right|\right|_{L^p(I)}+ \left|\left|\frac{u'(x)}{x}\right|\right|_{L^p(I)} \leq C_p\left|\left|u''\right|\right|_{L^p(I)}.$$

Thank you in advance for any help.


Not exactly sure I am correct. At first, take a look at 8.8.1 which may be helpful.

First substitute u(x)/x as u(x), you will get: ||(u(x)/x)/x||<=p/(p-1)||(u(x)/x)'|| =p/(p-1)||u'/x-u(x)/x^2||<=p/(p-1)(||u'(x)/x||+||u(x)/x^2||). It gives: ||u(x)/x^2|| <= Cp1 ||u'(x)/x||.

It means we only need to show that: ||u'(x)/x||<=Cp2||u''||.

Now apply 8.8.1 once again, with: u' as u, now: ||u'(x)/x||<=p/(p-1)||u''||. Done.

I will appreciate it if somebody could formalize this proof/my typing (seems the command is a little bit different with Tex).

  • $\begingroup$ Ghosh, I made a mistake in the first step. Let me think of it... $\endgroup$ – hsc Jul 1 '14 at 1:31

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