Prove $$\|Du\|_{L^{2p}(U)} \le C\|u\|_{{L^\infty}(U)}^{1/2} \|D^2 u\|_{L^p(U)}^{1/2}$$ for $1 \le p < \infty$ and all $u \in C_c^\infty(U)$.
This is PDE Evans, 2nd edition: Chapter 5, Exercise 10(b).
Here is what I did so far: \begin{align} \int_U |Du|^{2p} \, dx &= \sum_{n=1}^\infty \int_U u_{x_i} |Du|^{2p-1} \, dx \\ &= -\sum_{i=1}^n \int_U u(x_i)(2p-1)|Du|^{2p-2} |D^2 u| \, dx \\ &\le C\int_U u|Du|^{2p-2} |D^2 u| \, dx \\ &\le C\left(\int_U u|D^2 u|^{p} \right)^{\frac 1{p}} \left(\int_U \left||Du|^{2p-2}\right|^{\frac{2p}{2p-2}} \, dx\right)^{\frac{2p-2}{2p}} \\ \end{align} where the last step is trying to use Hölder's inequality. The $2p-1$'s (or $2p-2$'s) in the exponent of the second integrand are supposed to cancel, but they aren't. Where can I go from here?
Assuming my previous work is correct so far (probably not), we divide both sides of our inequality by $\left(\int_U u|D^2 u|^{2p} \right)^{\frac 1{2p}}$ to obtain \begin{align} \left(\int_U |Du|^{2p} \, dx \right)^{\frac{2p-1}{2p}} &\le C \left(\int_U \left||Du|^{2p-2}\right|^{\frac{2p}{2p-2}} \, dx\right)^{\frac{2p-2}{2p}} \\ &= C \left(\int_U |Du|^{2p} \, dx\right)^{\frac{2p-2}{2p}} \end{align}
By the way, since this is part (b) of Exercise 10 in the 2nd edition book, I suspected this was similar to part (a) of the exercise. So I suspected this method I utilized here should be similar to the method presented in part (a).
Anyways, how can I show that $$\left(\int_U |Du|^{2p} \, dx \right)^{\frac 2{2p}} \le C\left(\text{ess sup}_U |u| \right) \left(\int_U |D^2 u| \, dx \right)^{\frac 1p},$$ from which the result follows by taking the square root of both sides?
Notice that rewriting what you have on the second to last line of teh first chain of inequalities (you applied Holder wrong in the last line): $$ \| Du\|_{2p}^{2p} \leq C\int |u||D^2 u||Du|^{2p-2} \leq C\| u\|_{\infty} \| D^2 u\|_{p}\| Du\|_{2p}^{2p-2}, $$ since $2p/(2p-2)=p/(p-1)$ is the conjugate exponent of $p$. In other words, $$ \| Du\|_{2p}^2 \leq C\| u\|_{\infty} \| D^2 u\|_p. $$