 Solution

Let $M$ bound the operator norm of $D^{2}F$; the bound on individual second derivatives supplies such an $M$ depending also on dimension. The Taylor theorem with integral remainder givesThus quadratic growth from a bounded Hessian givesSince $\Omega$ is bounded it has finite measure, and a function in the Sobolev space $H^1(\Omega )$ has $Du\in L^2(\Omega )$. The given lower bound also implies that the integrand is nonnegative. ConsequentlyFor later use the same Hessian bound yields $|DF(p)|\le |DF(0)|+M|p|$, hence $DF(Du)\in L^2(\Omega )$.