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In the $1$st edition, this was question $5.9$. The question is:

Integrate by parts to prove: $$\int_{U} |Du|^p \ dx \leq C \left(\int_{U} |u|^p \ dx\right)^{1/2} \left(\int_{U} |D^2 u|^p \ dx\right)^{1/2}$$ for $ 2 \leq p < \infty$ and all $u \in W^{2,p}(U) \cap W^{1,p}_{0}(U)$.

Hint: $$\int_{U} |Du|^p dx = \sum_{i=1}^{n}\int_{U} u_{x_i}u_{x_i}|Du|^{p -2}\ dx$$

I am trying to prove this holds for $u \in C^{\infty}_{c} (U)$, and then by density conclude the theorem. However, this integration does not appear to be working.

The question says to use the integration by parts formula. Is it possible to use integration by parts in the following integral? $$\int_{U} u_{x_i}u_{x_i}|Du|^{p -2}\ dx$$

My friend used it here, but he doesn't know if it is valid...

Can someone please give a hint for how I could start?

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  • $\begingroup$ in the case p=2 , I proved first for smooth u with compact suport , and by density I proved the question $\endgroup$ Apr 23, 2013 at 3:34
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    $\begingroup$ In case $p=2$, the solution can be found in page 6 of columbia.edu/~la2462/Problems%20from%20Evans.pdf But I have some confusion when comparing with the general case (I mean $p\ne 2$). In $p=2$ case, one needs to approximate by smooth function...(see bottom half of page 6), but general $p$ we do not need this step, could you please explain? @math student or anyone can help... $\endgroup$
    – math101
    Jul 10, 2015 at 13:16

1 Answer 1

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As you guessed, the first step is an integration by parts: $$\int_U |Du|^p dx = \int_U \nabla u \cdot \nabla u |Du|^{p-2} dx = - \int_U u \nabla \cdot ( \nabla u |Du|^{p-2}) dx = - \int_U u \left( \Delta u |Du|^{p-2} + (p-2) (\nabla u^T D^2 u \nabla u) |Du|^{p-4})\right) \leq C \int_U u |Du|^{p-2} |D^2 u| dx $$ Now the next step is an invocation of the Holder inequality, with conjugate exponents $\frac{p}{2}$ and $\frac{p}{p-2}$ (notice that this step requires $p \gt 2$, if $p=2$, it is unnecessary): $$\int_U u |Du|^{p-2} |D^2 u| dx \leq \left(\int_U |u|^\frac{p}{2} |D^2u|^\frac{p}{2}\right)^\frac{2}{p} \left(\int_U |Du|^p \right)^\frac{p-2}{p} $$ So, dividing the original left hand side by the gradient term and invoking Holder on the remaining term with exponent 2, we get the desired result. $$ \int_U |Du|^p dx \leq C \left(\int_U |u|^p dx \right)^\frac{1}{2} \left( \int_U |D^2u|^p dx \right)^\frac{1}{2} $$

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  • $\begingroup$ why i can use integration by parts in the integral ( the second integral in the first line of your solution) ? thank you $\endgroup$ Apr 23, 2013 at 13:17
  • $\begingroup$ oops, that was a typo, since corrected. Is it clearer now? $\endgroup$
    – Ray Yang
    Apr 23, 2013 at 13:22
  • $\begingroup$ @RayYang: May I ask why the differential of $|Du|^{p-2}|$ is so complicated..? $\endgroup$ Apr 24, 2013 at 6:46
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    $\begingroup$ It's not that complicated. $\frac{\partial}{\partial x_i} |Du|^{p-2} = (p-2) |Du|^{p-4} \sum_j \frac{\partial^2 u}{\partial x_i \partial x_j} \frac{\partial u}{\partial x_j}$. Which turns into the form I've written above after sorting terms. $\endgroup$
    – Ray Yang
    Apr 24, 2013 at 10:28
  • $\begingroup$ @RayYang I have difficulties in this calculation, I would appreciate if you could check my question at math.stackexchange.com/questions/1355970/… $\endgroup$
    – math101
    Jul 10, 2015 at 3:41

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