Consider the Dirichlet problem with boundary data $f(x)\in L^{\infty}(\mathbb{R}^{d-1})$ on the halfspace $\mathbb{R}^{d}_{+}$, where $y>0$ and $x\in\mathbb{R}^{d-1}.$
One can prove that $$u(x,y)=\int_{\mathbb{R}^{d-1}}P_{y}(t)f(x-t)\;dt=\int_{\mathbb{R}^{d-1}}\frac{c_{n}y}{(|t|^{2}+y^{2})^{d/2}}f(x-t)\;dt$$ converges to $f(x_{0})$ non-tangentially for a.e. $x_{0}\in\mathbb{R}^{d-1}$ (the perpendicular approach is already covered by the well-known approximation to the identity convergence theorems using $y\to0$).
I have two questions, the first I think is straight-forward, but I would like my proof to be reviewed. The second one is something I can only prove under the assumption $f$ is continuous, so would like help generalizing the reuslt for general $L^{\infty}$ functions.
Q1. Prove that $f$ matches $u$ on the boundary $\mathbb{R}^{d-1}$ in the sense of distributions.
Q2. Suppose $v$ is weakly harmonic (harmonic in the sense of distributions) in $\mathbb{R}^{d}_{+}$, $L^{\infty}$ on $\mathbb{R}^{d-1}$, and equal to $f$ in a distributional sense on $\mathbb{R}^{d-1}$. Show that $u=v$ in the sense of distributions, where $u$ is as above.
Q1 Proof. Since $f\in L^{\infty}$, $Mf\in L^{\infty}$ since $M(\cdot)$ is bounded on $L^{p}$ for $p>1$. Thus, using a well-known estimate for approximations to the identity (i.e. that the "non-tangential" maximal function is bounded by the Hardy-Littlewood maximal function), we get $$\sup_{y>0}|u(x,y)|=\sup_{y>0}|P_{y}*f|(x)\leq A(Mf)(x)\leq A||Mf||_{\infty}<\infty,$$ and this implies that $$\sup_{y>0}||u(x,\cdot)||_{\infty}<\infty.$$ Fixing a test function $\phi\in C_{c}^{\infty}(\mathbb{R}^{d-1})$, the dominated convergence theorem now implies (since $|u(x,y)\phi(x)|\leq A||Mf||_{\infty}|\phi(x)|,$ which is integrable) $$\lim_{y\to0}\int_{\mathbb{R}^{d-1}}u(x,y)\phi(x)\;dx=\int_{\mathbb{R}^{d-1}}\lim_{y\to0}u(x,y)\phi(x)\;dx=\int_{\mathbb{R}^{d-1}}f(x)\phi(x)\;dx.$$ It follows that $u(x,0)=f(x)$ in the sense of distributions (the $0$ is just to be symbolic; $u$ is not defined by convolution with $P_{y}$ when $y=0$ and must be obtained by a limiting process).
NOTE. In retrospect, the distributional equivalence is almost obvious since basically $u$ and $f$ are equal a.e. on $\mathbb{R}^{d-1}$. The only difficulty is setting up the use of the dominated convergence theorem, since there is no tangible definition for $u$ on the boundary without first passing to a limit $y\to0$.
Q2 Proof.
For test functions $\phi\in C_{c}^{\infty}(\mathbb{R}^{d-1})$ and $\psi\in C_{c}^{\infty}(\mathbb{R}^{d}_{+})$, we have $$(1) \int_{\mathbb{R}^{d-1}}v(x,0)\phi(x)\;dxdy=\int_{\mathbb{R}^{d-1}}f(x)\phi(x)\;dxdy$$ and $$(2) \int_{\mathbb{R}^{d}_{+}}v(x,y)\Delta\psi(x,y)\;dxdy=0.$$
(Again, the setting $y=0$ is just symbolic and means we are integrating $u$ and $v$ on $\mathbb{R}^{d-1}.$)
By Q1, $(1)=\int_{\mathbb{R}^{d-1}}u(x,0)\phi(x)\;dx$ and so we see already that $u=v=f$ in the sense of distributions on $\mathbb{R}^{d-1}.$
Now, I do not know how to show that $u$ and $v$ are equal in the sense of distributions in $\mathbb{R}^{d}_{+}$. As a first observation, we clearly have $(2)=\int_{\mathbb{R}^{d}_{+}}u(x,y)\Delta\psi(x,y)\;dxdy$ since $u$ is obviously weakly harmonic.
As a second observation, since weakly harmonic implies classically harmonic when the weakly harmonic function is assumed continuous, we could (if $v$ was also continuous) conclude $u=v$ not only in the sense of distributions, but pointwise, by appealing to the uniqueness of the Dirichlet problem. I have proved in the past that weakly harmonic implies classically harmonic in the past using a mollification argument, and it can be found here
EDIT. Actually, it turns out the weak-classical equivalence is true even when $v$ is merely $L^{1}_{\text{loc}}.$ Since this space is embedded into $L^{\infty},$ the claim follows (assuming my proof accurate). Is there a source for this stronger result somewhere?
