(a) Let $u \in L_{\text {loc }}^{1}(\Omega)$. Given $1<p \leq \infty$, let $1 \leq q<\infty$ such that $\frac{1}{p}+\frac{1}{q}=1$. Suppose $D^{\alpha} u$ exists as weak derivative in $L^{p}(\Omega)$. Let $\varphi \in C_{c}^{\infty}(\Omega)$ be arbitrary. Then,
$$
\left|\int_{\Omega} u D^{\alpha} \varphi d x\right|=\left|(-1)^{|\alpha|} \int_{\Omega}\left(D^{\alpha} u\right) \varphi d x\right| \leq\left|D^{\alpha} u\right|_{L^{p}(\Omega)}|\varphi|_{L^{q}(\Omega)}
$$
by Hölder’s inequality which proves the first claim with constant $C=\left|D^{\alpha} u\right|_{L^{p}(\Omega)}$. Conversely, suppose
$$
\forall \varphi \in C_{c}^{\infty}(\Omega): \quad\left|\int_{\Omega} u D^{\alpha} \varphi d x\right| \leq C|\varphi|_{L^{q}(\Omega)}
$$
Then, since $C_{c}^{\infty}(\Omega)$ is dense in $L^{q}(\Omega)$ for $q<\infty$, the map
$$
f: \varphi \mapsto(-1)^{|\alpha|} \int_{\Omega} u D^{\alpha} \varphi d x
$$
defines a continuous linear functional $f \in\left(L^{q}(\Omega)\right)^{}$. Since $\left(L^{q}(\Omega)\right)^{}$ for $1 \leq q<\infty$ is isometrically isomorphic to $L^{p}(\Omega)$, there exists $g \in L^{p}(\Omega)$ such that
$$
\forall \varphi \in L^{q}(\Omega): \quad f(\varphi)=\int_{\Omega} g \varphi d x
$$
By definition of $f$ it follows that $g \in L^{p}(\Omega)$ is the weak derivative $D^{\alpha} u$ of $u$.
(b) Let $u=\chi_{j 0,[1}$ and $\varphi \in C_{c}^{\infty}(\mathbb{R})$. Then
$$
\left|\int_{\mathbb{R}} u \varphi^{\prime} d x\right|=\left|\int_{0}^{1} \varphi^{\prime} d x\right|=|\varphi(1)-\varphi(0)| \leq 2|\varphi|_{L^{\infty}(\mathbb{R})} .
$$
The function $u$ restricted to $\mathbb{R} \backslash{0,1}$ is differentiable with vanishing derivative. In particular, if $u$ had a weak derivative $u^{\prime} \in L_{\text {loc }}^{1}(\mathbb{R})$, then $u^{\prime}=0$ almost everywhere. A contradiction arises for test functions $\varphi \in C_{c}^{\infty}(\mathbb{R})$ with $\varphi(0) \neq \varphi(1)$ via
$$
0=\int_{\mathbb{R}} u^{\prime} \varphi d x=-\int_{\mathbb{R}} u \varphi^{\prime} d x=-\int_{0}^{1} \varphi^{\prime} d x=\varphi(0)-\varphi(1) .
$$

Given $x=\left(x_{1}, \ldots, x_{n-1}, x_{n}\right) \in \mathbb{R}{+}^{n}$, let $\bar{x}=\left(x{1}, \ldots, x_{n-1},-x_{n}\right)$ denote its reflection in
the plane $\partial \mathbb{R}{+}^{n} .$ Let $\Phi: \mathbb{R}^{n} \backslash{0} \rightarrow \mathbb{R}$ be the fundamental solution of Laplace’s equation as given on the problem set. Then the function $$ \phi^{x}(y):=\Phi(y-\bar{x})=\Phi\left(y{1}-x_{1}, \ldots, y_{n-1}-x_{n-1}, y_{n}+x_{n}\right)
$$
satisfies
$$
\left{\begin{aligned}
\Delta \phi^{x} &=0 & & \text { in } \mathbb{R}{+}^{n} \ \phi^{x}(y) &=\Phi(y-x) & & \text { for } y \in \partial \mathbb{R}{+}^{n}
\end{aligned}\right.
$$
because $y-\bar{x} \neq 0$ for every $y \in \mathbb{R}{+}^{n}$ and since by symmetry of $\Phi$ $$ \forall y \in \partial \mathbb{R}{+}^{n}: \quad \Phi(y-x)=\Phi(\overline{y-x})=\Phi(\bar{y}-\bar{x})=\Phi(y-\bar{x})=\phi^{x}(y)
$$
Hence, Green’s function for the upper half-space is
$$
\begin{aligned}
G(x, y) &=\Phi(y-x)-\phi^{x}(y)=\Phi(y-x)-\Phi(y-\bar{x}) \
&=\left{\begin{array}{ll}
-\frac{1}{2 \pi}(\log |y-x|-\log |y-\bar{x}|), & (n=2) \
\frac{1}{n(n-2)\left|B_{1}\right|}\left(|y-x|^{2-n}-|y-\bar{x}|^{2-n}\right), & (n \neq 2)
\end{array}\right.
\end{aligned}
$$
Remark. Since the domain $\mathbb{R}{+}^{n}$ is unbounded, the representation formula (as given on the problem set) for solutions of the equation $-\Delta u=f$ in $\mathbb{R}{+}^{n}$ with boundary data $\left.u\right|{\partial \mathbb{R}{+}^{n}}=g$ has to be checked separately.

(a) Let $\Omega=]-1,1\left[^{2} \backslash([0,1[\times{0})\right.$ and let $u: \Omega \rightarrow \mathbb{R}$ be given by
$$
u\left(x_{1}, x_{2}\right):=\left{\begin{array}{ll}
x_{1} & \text { if } x_{1}>0 \text { and } x_{2}>0 \
0 & \text { otherwise. }
\end{array}\right.
$$
As shown in Problem $5.1, u \in W^{1, \infty}(\Omega) .$ Since $\Omega$ is bounded, $u \in W^{1, p}(\Omega)$ for any $1 \leq p \leq \infty$. Suppose, there exists an extension operator $E: W^{1, p}(\Omega) \rightarrow W^{1, p}\left(\mathbb{R}^{2}\right)$ such that $\left.(E u)\right|{\Omega}=u$ almost everywhere in $\Omega .$ Let $\left.Q:=\right]-1,1\left[^{2}\right.$ and $v:=\left.(E u)\right|{Q}$. Then $E u \in W^{1, p}\left(\mathbb{R}^{n}\right)$ implies $v \in W^{1, p}(Q)$. Consequently, as shown in Problem 5.5, $\left(x_{2} \mapsto v\left(x_{1}, x_{2}\right)\right) \in W^{1, p}(]-1,[)1$ for almost every $\left.x_{1} \in\right]-1,1[.$ Moreover, since $[0,1[\times{0}$
has measure zero, $v\left(x_{1}, x_{2}\right)=u\left(x_{1}, x_{2}\right)$ for almost every $\left(x_{1}, x_{2}\right) \in Q$.

Hence, there exists some fixed $\left.x_{1} \in\right] \frac{1}{2}, 1\left[\right.$ such that $\left(g: x_{2} \mapsto v\left(x_{1}, x_{2}\right)\right) \in W^{1, p}(]-1,[)1$ and such that $g\left(x_{2}\right)=u\left(x_{1}, x_{2}\right)$ for almost every $\left.x_{2} \in\right]-1,1[.$ By Sobolev’s embedding in dimension one, $g$ and hence $x_{2} \mapsto u\left(x_{1}, x_{2}\right)$ has a representative in $C^{0}(]-1,[)1$. However, since we chose $x_{1}>\frac{1}{2}$, this contradicts discontinuity of
$$
x_{2} \mapsto u\left(x_{1}, x_{2}\right)=\left{\begin{array}{ll}
x_{1} & \text { for } x_{2}>0 \
0 & \text { for } x_{2}<0
\end{array}\right.
$$
(b) The issue is that $\Omega$ is not a topological manifold with boundary. In particular, every point $x \in[0,1] \times{0}$ belongs to the topological boundary of $\Omega$ but doesn’t admit an open neighbourhood $U$ such that $U \cap \Omega$ is even only homeomorphic to $Q_{+}$ (compare with the definition of open set with $C^{k}$ boundary given in class).

Sobolev’s embedding (in the case $n=2=p$ ) states that the space $H^{1}\left(\mathbb{R}^{2}\right)$ embeds into $L^{q}\left(\mathbb{R}^{2}\right)$ for any $2 \leq q<\infty$, in particular for $q=4$. The Sobolev inequality states
$$
\exists C<\infty \quad \forall u \in H^{1}\left(\mathbb{R}^{2}\right): \quad|u|_{L^{4}\left(\mathbb{R}^{2}\right)} \leq C|u|_{H^{1}\left(\mathbb{R}^{n}\right)}
$$
In this special case, we claim that the following inequality also holds.
$$
\forall u \in H^{1}\left(\mathbb{R}^{2}\right): \quad|u|_{L^{4}\left(\mathbb{R}^{2}\right)}^{4} \leq 4|u|_{L^{2}\left(\mathbb{R}^{2}\right)}^{2}|\nabla u|_{L^{2}\left(\mathbb{R}^{2}\right)}^{2}
$$
Since $C_{c}^{\infty}\left(\mathbb{R}^{2}\right)$ is dense in $H^{1}\left(\mathbb{R}^{2}\right)$, it suffices to prove the inequality for $u \in C_{c}^{\infty}\left(\mathbb{R}^{2}\right)$. Let $u \in C_{c}^{\infty}\left(\mathbb{R}^{2}\right)$ and $\left(x_{1}, x_{2}\right) \in \mathbb{R}^{2}$. Then,
$$
\begin{aligned}
\left|u^{2}\left(x_{1}, x_{2}\right)\right| &=\left|\int_{-\infty}^{x_{1}} \frac{\partial u^{2}}{\partial x_{1}}\left(s, x_{2}\right) d s\right|=\left|\int_{-\infty}^{x_{1}} 2 u\left(s, x_{2}\right) \frac{\partial u}{\partial x_{1}}\left(s, x_{2}\right) d s\right| \
& \leq 2 \int_{\mathbb{R}}\left|u\left(s, x_{2}\right) | \nabla u\left(s, x_{2}\right)\right| d s
\end{aligned}
$$

Analogously,
$$
\left|u^{2}\left(x_{1}, x_{2}\right)\right| \leq 2 \int_{\mathbb{R}}\left|u\left(x_{1}, t\right) | \nabla u\left(x_{1}, t\right)\right| d t
$$
Hence, by Fubini’s theorem and the Cauchy-Schwarz inequality
$$
\begin{aligned}
|u|_{L^{4}\left(\mathbb{R}^{2}\right)}^{4} &=\int_{\mathbb{R}} \int_{\mathbb{R}}\left|u\left(x_{1}, x_{2}\right)\right|^{4} d x_{1} d x_{2}=\int_{\mathbb{R}} \int_{\mathbb{R}}\left|u^{2}\left(x_{1}, x_{2}\right) | u^{2}\left(x_{1}, x_{2}\right)\right| d x_{1} d x_{2} \
& \leq 2 \int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left|u\left(s, x_{2}\right) | \nabla u\left(s, x_{2}\right)\right| d s\right) \int_{\mathbb{R}}\left|u^{2}\left(x_{1}, x_{2}\right)\right| d x_{1} d x_{2} \
& \leq 4 \int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left|u\left(s, x_{2}\right) | \nabla u\left(s, x_{2}\right)\right| d s\right) d x_{2} \int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left|u\left(x_{1}, t\right) | \nabla u\left(x_{1}, t\right)\right| d t\right) d x_{1} \
&=4\left(\int_{\mathbb{R}^{2}}|u||\nabla u| d x\right)^{2} \leq 4|u|_{L^{2}\left(\mathbb{R}^{2}\right)}^{2}|\nabla u|_{L^{2}\left(\mathbb{R}^{2}\right)}^{2}
\end{aligned}
$$