Vtyjkyuk

I am reading the book: Fully nonlinear elliptic equations of Caffarelli and Cabre. In page 8 (Prop 1.2) they prove that if function $u$ in a convex domain locally has at least one paraboloid touching from above and one touching from below then $u$ is $C^1,1$.

The proof basically is by proving $u$ is differentiable, and then it has weak second derivative which is bounded (I can understand this part). Now, let assume that $u$ is differentiable and has weak second derivative which is bounded, the remaining part is the following:

Since $u_i:=partial_i u in W^1,infty(B)$ and $B$ is convex, we have that $u_i$ is continuous and:

$$u_i(x)-u_i(y)=int_0^1 fracddtu_i(tx+(1-t)y),dt=sumlimits_jint_0^1 partial_iju(tx+(1-t)y),dt(x_j-y_j)$$

for any $x,y in barB$.

Then using the boundedness of weak second derivative to have the Lipschitz continuity of first derivative.

My question is just they start from the assumption that $u_i in W^1,infty(B)$ and I do not know why we have this? My problem is I do not know why $u_i in L^infty(B)$. Could anyone help me please. Thank you very much!

With no loss of generality, we may directly assume that $B_varepsilon(x)subsetOmega$ for any $xinbar B.$ Otherwise, we can choose an $varepsilon'in(0,varepsilon)$ which is sufficient small, such that $B_varepsilon'(x)subsetOmega$ for any $xinbar B.$ So we have $Theta(x,varepsilon')leqTheta(x,varepsilon)leq K$ for any $xinbar B,$ and we replace $varepsilon'$ by $varepsilon.$

We note that $Theta(x,varepsilon)leq K$ implies that there is a convex paraboloid $P$ with opening $K$ such that $P$ touches $u$ by above at $xinbar B$ in $B_varepsilon(x)$, and also $nabla P(x)=nabla u(x)$. A straightforward calculation yields that $$P(y)=fracK2left|y-x+fracnabla u(x)Kright|^2+u(x)-frac^22K, yin B_varepsilon(x).$$ By using that $uleq P$ in $B_varepsilon(x),$ we obtain that $$|nabla u(x)|leqfrac_L^infty(bar B),mathrmdiam(bar B))varepsilon.$$ Thus $nabla uin L^infty(bar B).$