M^{\gamma}}$$ for $\gamma$ small enough.

Now suppose the directional derivative is replaced by the full gradient. By (\[ibp2\]), we have that$$h(\eta)=h(\eta)\cdot\bigg(1-\prod_{j=1}^{3}\varphi_{0}(2\epsilon_{1}^{-1}\partial_{j}\Phi(\eta))\bigg),$$ so we only need to bound the integral with $h$ replaced by $h_{1}=h\cdot\varphi_{0}(2\epsilon_{1}^{-1}\partial_{1}\Phi)\varphi_{0}(2\epsilon_{1}^{-1}\partial_{2}\Phi)$ (the other similar terms are bounded in the same way).

Since (\[ibp2\]) now holds with $\partial_{\eta}$ replaced by $\partial_{1}$ at each point where $h_{1}$ or one of its derivative is nonzero (possibly with different constants), we only need to show that $h_{1}$ also verifies the bound (\[ibp1\]), but with $\lambda$ replaced by the maximum of $\lambda$ and each $\epsilon_{j+1}/\epsilon_{j}$ (again, let $\epsilon_{n+1}=1$). Using Leibniz rule, we can further reduce to proving the same result for $\varphi_{0}(2\epsilon_{1}^{-1}\partial_{2}\Phi)$ but with $\lambda$ replaced by the maximum of $\epsilon_{j+1}/\epsilon_{j}$, the $L^{1}$ norm replaced by the $L^{\infty}$ norm, and restrict to the subset where $h$ or one of its derivatives is nonzero (note that $\varphi_{0}(2\epsilon_{1}^{-1}\partial_{1}\Phi)$ is estimated in the same way).

Now, using (\[induct\]) we have $$\partial_{\eta}^{\mu}(\varphi_{0}(2\epsilon_{1}^{-1}\partial_{2}\Phi))=\sum_{r=0}^{N}(2\epsilon_{1}^{-1})^{r}\sum_{\mathcal{T}}\Omega(\mathcal{T})\cdot(\partial_{j_{1}}\cdots\partial_{j_{r}}\varphi_{0})(2\epsilon_{1}^{-1}\partial_{2}\Phi)\cdot\prod_{q=1}^{r}\partial_{x}^{\mu_{q}}\partial_{j_{q}}\partial_{2}\Phi,$$ here the summation is taken over all tuples$$\mathcal{T}=(\mu_{1},\cdots,\mu_{d},j_{1},\cdots,j_{d}),$$ and we have $$\sum_{i=1}^{r}|\mu_{i}|=N-r,\qquad|\Omega(\mathcal{T})|\leq (N+1)!.$$Let, for each $0\leq j\leq n-2$, the number of $q$’s such that $|\mu_{q}|=j$ be $r_{j}$, then we will have $$\rho=r-\sum_{j=0}^{n-2}r_{j}\geq0,\qquad\sum_{j=0}^{n-2}jr_{j}\leq N-r-(n-1)\rho.$$ Therefore, since we are in the set where the second part 