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Let's consider the equation $$ \Delta u(x,y)=0\;\in\;\mathbb{R}\times\mathbb{R}^+,\;u(x,0)=g(x), $$ where $g$ is an integrable, smooth enough function. Let me write $P_y$ for the Poisson kernel for the half plane. The question: Is the bound

$$ \|P_y*g\|_{L^2(\mathbb{R}\times\mathbb{R}^+)}\leq C\|g\|_{H^1(\mathbb{R})} $$


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Of course it is not. Reformulated for Fourier images, your question turns rather trivial. Indeed, denote by $\hat{u}$ the Fourier image of $u$, i.e., $$ \hat{u}(\xi,y)=F[u]=\int\limits_{-\infty}^{+\infty}u(x,y)e^{-i\xi x}dx, $$ and notice that $\hat{u}(\xi,y)=\hat{g}(\xi)e^{-|\xi|y}=F[P_y\ast g]$, while $$ \|\hat{u}\|^2_{L^2(\mathbb{R}^2_{+})}=\int\limits_{0}^{+\infty}\frac{|\hat{g}(\xi)|^2}{2|\xi|}d\xi=2\pi\|P_y\ast g\|^2_{L^2(\mathbb{R}^2_{+})}\,, $$ where $\mathbb{R}^2_{+}=\{(x,y)\in \mathbb{R}^2\,\colon\,y>0\}$. Thus, reformulated for Fourier images, your question is: does there exist a positive $C$ such that $$ \int\limits_{-\infty}^{+\infty}\frac{|\hat{g}(\xi)|^2}{2|\xi|}d\xi\leqslant C\int\limits_{-\infty}^{+\infty}|\hat{g}(\xi)|^2(1+\xi^2)\,d\xi\quad \forall\, g\in H^1(\mathbb{R}) \tag{$\ast$} $$ and hence for all $g\in S(\mathbb{R})$? Here, as usual, $S(\mathbb{R})$ stands for the Schwartz space of rapidly decreasing functions on $\mathbb{R}$. To disprove $(\ast)$, routinely choose $\hat{g}(\xi)= \varphi(\lambda\xi)$ with some nonzero function $\varphi\in S(\mathbb{R})$ and arbitrary parameter $\lambda>0$. This yields the inequality $$ \int\limits_{-\infty}^{+\infty}\frac{|\varphi(\lambda\xi)|^2}{2|\xi|}d\xi\leqslant C\int\limits_{-\infty}^{+\infty}|\varphi(\lambda\xi)|^2(1+\xi^2)\,d\xi\quad \forall\, \lambda>0, $$ that can be rewritten in the form $$ \int\limits_{-\infty}^{+\infty}\frac{|\varphi(\eta)|^2}{2|\eta|}d\eta\leqslant \frac{C}{\lambda}\int\limits_{-\infty}^{+\infty}|\varphi(\eta)|^2\biggl(1+\frac{\eta^2}{\lambda^2}\biggr)\,d\eta\quad \forall\, \lambda>0\tag{$\ast\ast$} $$ Passing to the limit in $(\ast\ast)$ as $\lambda\to\infty$ results in contradiction $$ \int\limits_{-\infty}^{+\infty}\frac{|\varphi(\eta)|^2}{2|\eta|}d\eta=0 $$ for a nonzero function $\varphi\in S(\mathbb{R})$, which disproves $(\ast)$.

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