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An elementary proof was shown to me by Jan Malý. It has never been published and with his permission we included it in [1] (Proposition 28). The argument is very elementary (1.5 pages with all details), but still too long to be included here.

Theorem. For $n\geq 1$ and $p>1$, there is a bounded linear extension operator $$ E:W^{1,p}(\mathbb{R}^{n})\to W^{1,q}\cap C^\infty(\mathbb{R}^{n+1}_+), \quad \text{where $q=\frac{(n+1)p}{n}$.} $$ In other words, $W^{1,p}(\mathbb{R}^{n})$ continuously embeds into the trace space $W^{1-\frac{1}{q},q}(\mathbb{R}^{n})$ of $W^{1,q}(\mathbb{R}^{n+1}_+)$.

From this result the following corollary follows right away:

Corollary. If $\Omega\subset\mathbb{R}^n$, $n\geq 2$, is a bounded and smooth domain, then there is a bounded extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}\cap C^\infty(\Omega), \quad \text{where $1<p<\infty$ and $q=\frac{np}{n-1}$.} $$

Taking $p=n-1$ and $n>2$ yields the theorem asked in the question. We need to take $n>2$ since for $n=2$ we have $p=n-1=1$ and the result is false in that case (there are counterexamples, see [1]).

[1] P. Goldstein, P. Hajłasz, Jacobians of $W^{1,p}$ homeomorphisms, case $p=[n/2]$. (Preprint 2018), arXiv:1812.11888.

An elementary proof was shown to me by Jan Malý. It has never been published and with his permission we included it in [1] (Proposition 28). The argument is very elementary (1.5 pages with all details), but still too long to be included here.

Theorem. For $n\geq 1$ and $p>1$, there is a bounded linear extension operator $$ E:W^{1,p}(\mathbb{R}^{n})\to W^{1,q}\cap C^\infty(\mathbb{R}^{n+1}_+), \quad \text{where $q=\frac{(n+1)p}{n}$.} $$ In other words, $W^{1,p}(\mathbb{R}^{n})$ continuously embeds into the trace space $W^{1-\frac{1}{q},q}(\mathbb{R}^{n})$ of $W^{1,q}(\mathbb{R}^{n+1}_+)$.

From this result the following corollary follows right away:

Corollary. If $\Omega\subset\mathbb{R}^n$, $n\geq 2$, is a bounded and smooth domain, then there is a bounded extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}\cap C^\infty(\Omega), \quad \text{where $1<p<\infty$ and $q=\frac{np}{n-1}$.} $$

Taking $p=n-1$ and $n>2$ yields the theorem asked in the question. We need to take $n>2$ since for $n=2$ we have $p=n-1=1$ and the result is false in that case (there are counterexamples, see [1]).

[1] P. Goldstein, P. Hajłasz, Jacobians of $W^{1,p}$ homeomorphisms, case $p=[n/2]$. Calc. Var. Partial Differential Equations 58 (2019), no. 4, Art. 122, 28 pp.

An elementary proof was shown to me by Jan Malý. It has never been published and with his permission we included it in [1] (Proposition 28). The argument is very elementary (1.5 pages with all details), but still too long to be included here.

Theorem. For $n\geq 1$ and $p>1$, there is a bounded linear extension operator $$ E:W^{1,p}(\mathbb{R}^{n})\to W^{1,q}\cap C^\infty(\mathbb{R}^{n+1}_+), \quad \text{where $q=\frac{(n+1)p}{n}$.} $$ In other words, $W^{1,p}(\mathbb{R}^{n})$ continuously embeds into the trace space $W^{1-\frac{1}{q},q}(\mathbb{R}^{n})$ of $W^{1,q}(\mathbb{R}^{n+1}_+)$.

From this result the following corollary follows right away:

Corollary. If $\Omega\subset\mathbb{R}^n$, $n\geq 2$, is a bounded and smooth domain, then there is a bounded extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}\cap C^\infty(\Omega), \quad \text{where $1<p<\infty$ and $q=\frac{np}{n-1}$.} $$

Taking $p=n-1$ and $n>2$ yields the theorem asked in the question. We need to take $n>2$ since for $n=2$ we have $p=n-1=1$ and the result is false in that case (there are counterexamples, see [1]).

[1] P. Goldstein, P. Hajłasz, Jacobians of $W^{1,p}$ homeomorphisms, case $p=[n/2]$. (Preprint 2018).

An elementary proof was shown to me by Jan Malý. It has never been published and with his permission we included it in [1] (Proposition 28). The argument is very elementary (1.5 pages with all details), but still too long to be included here.

Theorem. For $n\geq 1$ and $p>1$, there is a bounded linear extension operator $$ E:W^{1,p}(\mathbb{R}^{n})\to W^{1,q}\cap C^\infty(\mathbb{R}^{n+1}_+), \quad \text{where $q=\frac{(n+1)p}{n}$.} $$ In other words, $W^{1,p}(\mathbb{R}^{n})$ continuously embeds into the trace space $W^{1-\frac{1}{q},q}(\mathbb{R}^{n})$ of $W^{1,q}(\mathbb{R}^{n+1}_+)$.

From this result the following corollary follows right away:

Corollary. If $\Omega\subset\mathbb{R}^n$, $n\geq 2$, is a bounded and smooth domain, then there is a bounded extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}\cap C^\infty(\Omega), \quad \text{where $1<p<\infty$ and $q=\frac{np}{n-1}$.} $$

Taking $p=n-1$ and $n>2$ yields the theorem asked in the question. We need to take $n>2$ since for $n=2$ we have $p=n-1=1$ and the result is false in that case (there are counterexamples, see [1]).

[1] P. Goldstein, P. Hajłasz, Jacobians of $W^{1,p}$ homeomorphisms, case $p=[n/2]$. (Preprint 2018), arXiv:1812.11888.

An elementary was shown to me by Jan Malý. It has never been published and with his permission we included it in [1] (Proposition 28). The argument is very elementary (1.5 pages with all details), but still too long to be included here.

Theorem. For $n\geq 1$ and $p>1$, there is a bounded linear extension operator $$ E:W^{1,p}(\mathbb{R}^{n})\to W^{1,q}\cap C^\infty(\mathbb{R}^{n+1}_+), \quad \text{where $q=\frac{(n+1)p}{n}$.} $$ In other words, $W^{1,p}(\mathbb{R}^{n})$ continuously embeds into the trace space $W^{1-\frac{1}{q},q}(\mathbb{R}^{n})$ of $W^{1,q}(\mathbb{R}^{n+1}_+)$.

From this result the following corollary follows right away:

Corollary. If $\Omega\subset\mathbb{R}^n$, $n\geq 2$, is a bounded and smooth domain, then there is a bounded extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}\cap C^\infty(\Omega), \quad \text{where $1<p<\infty$ and $q=\frac{np}{n-1}$.} $$

Taking $p=n-1$ and $n>2$ yields the theorem asked in the question. We need to take $n>2$ since for $n=2$ we have $p=n-1=1$ and the result is false in that case (there are counterexamples, see [1]).

[1] P. Goldstein, P. Hajłasz, Jacobians of $W^{1,p}$ homeomorphisms, case $p=[n/2]$. (Preprint 2018).

An elementary proof was shown to me by Jan Malý. It has never been published and with his permission we included it in [1] (Proposition 28). The argument is very elementary (1.5 pages with all details), but still too long to be included here.

Theorem. For $n\geq 1$ and $p>1$, there is a bounded linear extension operator $$ E:W^{1,p}(\mathbb{R}^{n})\to W^{1,q}\cap C^\infty(\mathbb{R}^{n+1}_+), \quad \text{where $q=\frac{(n+1)p}{n}$.} $$ In other words, $W^{1,p}(\mathbb{R}^{n})$ continuously embeds into the trace space $W^{1-\frac{1}{q},q}(\mathbb{R}^{n})$ of $W^{1,q}(\mathbb{R}^{n+1}_+)$.

From this result the following corollary follows right away:

Corollary. If $\Omega\subset\mathbb{R}^n$, $n\geq 2$, is a bounded and smooth domain, then there is a bounded extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}\cap C^\infty(\Omega), \quad \text{where $1<p<\infty$ and $q=\frac{np}{n-1}$.} $$

Taking $p=n-1$ and $n>2$ yields the theorem asked in the question. We need to take $n>2$ since for $n=2$ we have $p=n-1=1$ and the result is false in that case (there are counterexamples, see [1]).

[1] P. Goldstein, P. Hajłasz, Jacobians of $W^{1,p}$ homeomorphisms, case $p=[n/2]$. (Preprint 2018).