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Is it possible to deduce the general Sobolev embedding into Holder spaces,

$W^{k,p}(\Omega) \hookrightarrow C^{\ell, k-\frac{n}{p} - \ell}(\Omega)$

for $\ell, $p > \frac{n}{k-\ell}$ from the $k=1$ case,

$W^{k,p} \hookrightarrow C^{0, 1 - \frac{n}{p}}(\Omega)$

for $p > n$? If so, how?

I know this can be done for the embeddings into lower $W^{k,p}$ spaces in the case $p < \frac{n}{k-\ell}$, but I cannot find a way to do it in the $p > \frac{n}{k-\ell}$ case.

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Yes, this is very much possible. Let $u\in C^\infty_c(\Omega)$.By the Sobolev-Gagliardo-Nirenberg inequality, we can successively embed $W^{k,p}_0(\Omega) \hookrightarrow W^{k-1,\frac{np}{n-p}}_0(\Omega) \hookrightarrow W^{k-2,\frac{np}{n-2p}}_0(\Omega) \hookrightarrow \dots \hookrightarrow W^{k-r,\frac{np}{n-r p}}_0(\Omega)$ where $r< k$ is chosen minimal with $\frac{np}{n-rp}>n$, i.e. such that $(r+1)p > n$. So $r = \lfloor \frac np\rfloor$.

Then the special case for $p>n$ you mentioned $W_0^{1,p}(\Omega)\hookrightarrow C^{0,1-\frac np}(\Omega)$ gives us an embedding $W_0^{1,\frac{np}{n-rp}}(\Omega) \hookrightarrow C^{0,1-\frac n {np/(n-rp)}}(\Omega) = C^{0,1+\lfloor \frac np \rfloor -\frac np}(\Omega)$ Let $\gamma = 1+\lfloor \frac np \rfloor -\frac np$. Since $u\in W^{k-r,q}(\Omega)$ trivially implies $D^\alpha u\in W^{1,q}(\Omega)$ for $|\alpha|< k-r$, we can use all of the above to see that we have an estimate of the form $\vert D^\alpha u \vert_{C^{0,\gamma}(\Omega)} \le \tilde C \Vert D^\alpha u \Vert_{W_0^{1,\frac{np}{n-rp}}(\Omega)} \le C \Vert u \Vert_{W_0^{k,p}(\Omega)}, \quad \forall |\alpha|< k-r$ This shows that $|u|_{C^{k-r-1, \gamma}}\le C \Vert u \Vert_{W_0^{k,p}(\Omega)} ,\quad\forall u\in C_c^\infty(\Omega)$ for some constant $C>0$. By density of $C_c^\infty(\Omega)$ in $W_0^{k,p}(\Omega)$, we can extend this to arbitrary $u\in W_0^{k,p}(\Omega)$ and get an embedding $W_0^{k,p}(\Omega)\hookrightarrow C^{k-\lfloor \frac np \rfloor-1, \gamma}(\Omega)$ where $\gamma = 1+\lfloor \frac np \rfloor -\frac np$.

If you want to extend this result to $W^{k,p}(\Omega)$, then you will need a sufficiently smooth boundary so you can make use of an extension operator $T: W^{k,p}(\Omega) \hookrightarrow W^{k,p}_0(\mathbb R^n)$.