Lemma 3.3 cannot be true as stated.
The classical Holder spaces (which the authors denote by $\mathcal{H}^\eta$), for $\eta$ non-integral and positive, is exactly equal to the Besov space $B^{\eta}_{\infty,\infty}$, and so the embedding from Holder into Besov of the same regularity cannot be compact. (It is the identity, and so is continuous.)
For the first embedding, compactness is also false: let $w^{(i)}$ be an enumeration of points in $\mathbb{Z}^p$, and set $f_i = \psi_{0w^{(i)}}$. Then you can check that by their definition the $B^{s,b}_{\infty,\infty}$ norm of $(f_i - f_j)$ is exactly $2$ when $i\neq j$, for any $b$. So this is a bounded sequence in $B^{\alpha,1+\epsilon}_{\infty,\infty}$ that has no convergent subsequence in $B^{\alpha,0}_{\infty,\infty}$.
(More generally, any of the spaces in the $B^{\alpha,\epsilon}_{p,q}$ scale is translation invariant, and so on a non-compact domain you can generate bounded non-convergent sequences by translations, which will prevent compactness of embedding between them.)
As the OP mentioned, in the MSE version of the question, it was shown that $B^{k,1+\epsilon}_{\infty,\infty}$ embeds into $B^{k}_{\infty,1}$ which then embeds into $\mathcal{H}^k$ for $k$ a non-negative ineger. The original article in question also mentioned compactness, so I wish to add one more remark:
Observe that given the smoothness and compact support of the wavelet functions, you have that
$$ |\psi_{j\ell w}(x) - \psi_{j\ell w}(y)| \leq 2^{jp/2} M \min(1, 2^j|x-y|) $$
(The first uses triangle inequality, the second uses the fundamental theorem of calculus, and so $M$ depends on the uniform bound on the wavelet and its first derivative.)
Additionally, the compact support of the wavelet functions means that there is a universal constant $c$ such that at every point $x$ and at every scale $j$ there is at most $c$ of the points $w$ such that $\psi_{j\ell w}(x) \neq 0$.
So using the wavelet expansion, you can estimate
$$ |f(x) - f(y)| \leq 2c M |x-y| \Big[ \sup_w |\alpha_f(w)| + \sum_{j \leq j_*} \sum_\ell \sup_w |\alpha_f(j,\ell,w)| 2^{j(p/2+1)} \Big] \\+ 2c M \sum_{j > j_*} \sum_\ell \sup_w|\alpha_f(j,\ell,w)|2^{jp/2}$$
where $j_*$ is the largest value satisfying $2^{j_*}|x-y| \leq 1$.
The sum inside the brackets can be estimated by
$$ \sum_{j\leq j_*} \sum_\ell \sup_w |\alpha_f(j,\ell,w)| 2^{j(p/2) + j)} \leq \Big[ \sup_j \Big(2^{jp/2} (1+j)^b \sum_\ell \sup_w |\alpha_f(j,\ell,w)| \Big) \Big]\cdot \sum_{j \leq j_*} \frac{2^j}{(1+j)^b} $$
Similarly, the second sum outisde brackets can be estimated by
$$ \sum_{j> j_*} \sum_\ell \sup_w |\alpha(j,\ell,w)|2^{jp/2} \leq \sum_{j > j_*} \frac{1}{(1+j)^b} \cdot \Big[ \sup_j \Big(2^{jp/2} (1+j)^b \sum_\ell \sup_w |\alpha_f(j,\ell,w)| \Big) \Big] $$
So we find that
$$ |f(x) - f(y)| \leq 2cM \|f\|_{B^{0,b}_{\infty,\infty}} \cdot \Big(\sum_{j \leq j_*} \frac{2^{j-j_*}}{(1+j)^b} + \sum_{j > j_*} \frac{1}{(1+j)^b} \Big) $$
when $b > 1$, the final sum is absolutely summable, and is bounded by $(1+j_*)^{1-b}$. The first sum can be estimated in different ways, a very cheap and inefficient way is to split the sum into $j < j_*/2$ and $j \in [j_*/2,j_*]$. We get $$ \sum_{j \leq j_*/2} \frac{2^{j-j_*}}{(1+j)^b} \leq \sum_{j \leq j_*/2} \frac{2^{-j_*/2}}{1} \leq j_* 2^{-j_*/2 - 1} $$ and $$ \sum_{j = j_*/2}^{j_*} \frac{2^{j-j_*}}{(1+j)^b} \leq \sum_{j = j_*/2}^{j_*} \frac{1}{(1+j_*/2)^b} \leq \frac{j_*/2}{(1+j_*/2)^b} \leq (1+j_*/2)^{1-b} $$
To summarize, we find, when $b > 1$, that for $|x-y| < \frac12$ $$ |f(x) - f(y)| \leq \|f\|_{B^{0,b}_{\infty,\infty}} \cdot O( |\log(|x-y|)|^{1-b}) $$ This allows us to draw two conclusions:
- The continuous embedding of $B^{0,b}_{\infty,\infty}$ into $\mathcal{H}^0$.
- That the unit ball in $B^{0,b}_{\infty,\infty}$ is in fact uniformly equicontinuous, and so restricting to functions with support within a fixed compact set $K$, the embedding is in fact compact.
