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?
Best Answer
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. $$ \int_U |Du|^p dx \leq C \left(\int_U |u|^p dx \right)^\frac{1}{2} \left( \int_U |D^2u|^p dx \right)^\frac{1}{2} $$