Let $\mathbb{R}_{+}^n = \{(x',x_n): x' \in \mathbb{R}^{n-1}, x_n > 0 \}$ be the half space.
It is known that if $u \in H_0^1(\mathbb{R}_{+}^n)$ is a weak solution to $-\Delta u = f$ with $f \in L^2(\mathbb{R}_{+}^n)$ then in fact $u \in H^2(\mathbb{R}_{+}^n)$.
Furthermore if $f \in H^1$ then $u \in H^3$, etc.
This second part can be shown as follows: first show that for $u\in H_0^1\cap H^2$, the tangential derivatives $\partial_j u \in H_0^1(\mathbb{R}_+^n)$ (ie for $j\neq n$) is a weak solution to poisson equation, thus by first part $\partial_j u \in H^2$. Secondly, we need to just show $\partial_n^2 u\in H^1$, which comes from $-\partial_n^2 u = \partial_1^2 u + \cdot \cdot \cdot + \partial_{n-1}^2u + f \in H^1$.
Now suppose we only have that $f \in L^2$ with $\partial_j f \in L^2$ for some $j$. I would like $\partial_j u \in H^2$. In the case when the derivative is tangential to boundary, ie $j\neq n$ then this extra smoothness is carried over to $u$ as explained above. But what happens if $j=n$, ie the extra smoothness is normal to the boundary? The argument above requires tangent derivatives to control the normal derivative so it doesn't work.
Is there some other way to show that that the additional normal smoothness on $f$ carries over to $u$, or is this in fact false?