Return to Answer

The answer is no. Namely, for every nonempty open set $\Omega\subset\mathbb R\times\mathbb R$ there is a continuous function $u$ in $W^{1,2}_{\boldsymbol.}(\Omega)$ that has no continuous extension defined on the closure of $\Omega\,$. The (counter)example is constructed as follows.

Let $\langle\,x_i:i\in\mathbb N_0\,\rangle$ be any injective sequence in $\Omega$ with at least one accumulation point, and all accumulation points on the boundary of $\Omega\,$. Then one can choose a stictly positive sequence $\langle\,R_i:i\in\mathbb N_0\,\rangle$ in $\ell^{\,2}(\mathbb N_0)$ such that $\langle\,{\rm B\,}(x_i,R_i):i\in\mathbb N_0\,\rangle$ is a sequence of disjoint balls in $\Omega\,$. For $s\le 0$ letting $\ln\kern.3mm s=-\infty=-\ln\kern.3mm(+\infty)$ and defining $u_i:\Omega\to[\,0\,,1\,]$ by $$ x\mapsto\inf\,\{\,\sup\,\{\,R_i\ln\kern.5mm(1- \ln\frac{|\,x-x_i\,|}{R_i}\Bigg)\,,0\,\}\,,1\,\} $$ then the function $u=\sum_{\,i\,\in\,\mathbb N_0\,}u_i$ has the claimed property.

If there are points that you do not understand, leave a comment specifying the unclear parts.

The answer is no. Namely, for every nonempty open set $\Omega\subset\mathbb R\times\mathbb R$ there is a continuous function $u$ in $W^{1,2}_{\boldsymbol.}(\Omega)$ that has no continuous extension defined on the closure of $\Omega\,$. The (counter)example is constructed as follows.

Let $\langle\,x_i:i\in\mathbb N_0\,\rangle$ be any injective sequence in $\Omega$ with at least one accumulation point, and such that all accumulation points are on the boundary of $\Omega\,$. Then one can choose a stictly positive sequence $\langle\,R_i:i\in\mathbb N_0\,\rangle$ in $\ell^{\,2}(\mathbb N_0)$ such that $\langle\,{\rm B\,}(x_i,R_i):i\in\mathbb N_0\,\rangle$ is a sequence of disjoint balls in $\Omega\,$. For $s\le 0$ letting $\ln\kern.3mm s=-\infty=-\ln\kern.3mm(+\infty)$ and defining $u_i:\Omega\to[\,0\,,1\,]$ by $$ x\mapsto\inf\,\{\,\sup\,\{\,R_i\ln\kern.5mm(1- \ln\frac{|\,x-x_i\,|}{R_i}\Bigg)\,,0\,\}\,,1\,\} $$ then the function $u=\sum_{\,i\,\in\,\mathbb N_0\,}u_i$ has the claimed property.

If there are points that you do not understand, leave a comment specifying the unclear parts.

The answer is no. Namely, for every nonempty open set $\Omega\subset\mathbb R\times\mathbb R$ there is a continuous function $u$ in $W^{1,2}_{\boldsymbol.}(\Omega)$ that has no continuous extension defined on the closure of $\Omega\,$. The (counter)example is constructed as follows.

Let $\{x_i\}_{i\in\mathbb N_0}$ be any injective sequence in $\Omega$ with at least one accumulation point, and all accumulation points on the boundary of $\Omega\,$. Then one can choose a stictly positive sequence $\{R_i\}_{i\in\mathbb N_0}$ in $\ell^{\,2}(\mathbb N_0)$ such that $\big\{\mathrm B(x_i,R_i)\big\}_{i\in\mathbb N_0}$ is a sequence of disjoint balls in $\Omega\,$. For $s\le 0$ letting $\ln\kern.3mm s=-\infty=-\ln\kern.3mm(+\infty)$ and defining $u_i:\Omega\to[\,0\,,1\,]$ by $$ x\mapsto\inf\left\{\sup\left\{\,R_i\ln\left(1- \ln\tfrac{|\,x-x_i\,|}{R_i}\right),0\right\}\,,1\right\} $$ then the function $u=\sum_{\,i\,\in\,\mathbb N_0\,}u_i$ has the claimed property.

If there are points that you do not understand, leave a comment specifying the unclear parts.

The answer is no. Namely, for every nonempty open set $\Omega\subset\mathbb R\times\mathbb R$ there is a continuous function $u$ in $W^{1,2}_{\boldsymbol.}(\Omega)$ that has no continuous extension defined on the closure of $\Omega\,$. The (counter)example is constructed as follows.

Let $\langle\,x_i:i\in\mathbb N_0\,\rangle$ be any injective sequence in $\Omega$ with at least one accumulation point, and all accumulation points on the boundary of $\Omega\,$. Then one can choose a stictly positive sequence $\langle\,R_i:i\in\mathbb N_0\,\rangle$ in $\ell^{\,2}(\mathbb N_0)$ such that $\langle\,{\rm B\,}(x_i,R_i):i\in\mathbb N_0\,\rangle$ is a sequence of disjoint balls in $\Omega\,$. For $s\le 0$ letting $\ln\kern.3mm s=-\infty=-\ln\kern.3mm(+\infty)$ and defining $u_i:\Omega\to[\,0\,,1\,]$ by $$ x\mapsto\inf\,\{\,\sup\,\{\,R_i\ln\kern.5mm(1- \ln\frac{|\,x-x_i\,|}{R_i}\Bigg)\,,0\,\}\,,1\,\} $$ then the function $u=\sum_{\,i\,\in\,\mathbb N_0\,}u_i$ has the claimed property.

If there are points that you do not understand, leave a comment specifying the unclear parts.

The answer is no. Namely, for every nonempty open set $\Omega\subset\mathbb R\times\mathbb R$ there is a continuous function $u$ in $W^{1,2}_{\boldsymbol.}(\Omega)$ that has no continuous extension defined on the closure of $\Omega\,$. The (counter)example is constructed as follows.

Let $\langle\,x_i:i\in\mathbb N_0\,\rangle$ be any injective sequence in $\Omega$ with at least one accumulation point, and all accumulation points on the boundary of $\Omega\,$. Then one can choose a stictly positive sequence $\langle\,R_i:i\in\mathbb N_0\,\rangle$ in $\ell^{\,2}(\mathbb N_0)$ such that $\langle\,{\rm B\,}(x_i,R_i):i\in\mathbb N_0\,\rangle$ is a sequence of disjoint balls in $\Omega\,$. For $s\le 0$ letting $\ln\kern.3mm s=-\infty=-\ln\kern.3mm(+\infty)$ and defining $u_i:\Omega\to[\,0\,,1\,]$ by $$ x\mapsto\inf\,\{\,\sup\,\{\,R_i\ln\kern.5mm(1- \ln\frac{|\,x-x_i\,|}{R_i}\Bigg)\,,0\,\}\,,1\,\} $$ then the function $u=\sum_{\,i\,\in\,\mathbb N_0\,}u_i$ has the claimed property.

If there are points that you do not understand, leave a comment specifying the unclear parts.

The answer is no. Namely, for every nonempty open set $\Omega\subset\mathbb R\times\mathbb R$ there is a continuous function $u$ in $W^{1,2}_{\boldsymbol.}(\Omega)$ that has no continuous extension defined on the closure of $\Omega\,$. The (counter)example is constructed as follows.

Let $\{x_i\}_{i\in\mathbb N_0}$ be any injective sequence in $\Omega$ with at least one accumulation point, and all accumulation points on the boundary of $\Omega\,$. Then one can choose a stictly positive sequence $\{R_i\}_{i\in\mathbb N_0}$ in $\ell^{\,2}(\mathbb N_0)$ such that $\big\{\mathrm B(x_i,R_i)\big\}_{i\in\mathbb N_0}$ is a sequence of disjoint balls in $\Omega\,$. For $s\le 0$ letting $\ln\kern.3mm s=-\infty=-\ln\kern.3mm(+\infty)$ and defining $u_i:\Omega\to[\,0\,,1\,]$ by $$ x\mapsto\inf\left\{\sup\left\{\,R_i\ln\left(1- \ln\tfrac{|\,x-x_i\,|}{R_i}\right),0\right\}\,,1\right\} $$ then the function $u=\sum_{\,i\,\in\,\mathbb N_0\,}u_i$ has the claimed property.

If there are points that you do not understand, leave a comment specifying the unclear parts.