Comparison of Bessel Capacities The Bessel kernels $G_{\alpha},\, \alpha>0$ are defined by their Fourier transform
$
\hat G_{\alpha}(\xi):= \frac 1 { (1+4\pi ^{2}\vert \xi \vert ^{2})^{\alpha/2}}.
$
Bessel $(\alpha, p)$-capacity of a set $E\subset\mathbb{R}^n$ is defined by
$$
B_{\alpha, p}(E):=\inf\{\|h\|_{L^p}^p\colon\, g_{\alpha}\star h\geq  1\quad on\quad E,\, h\geq 0 \}.
$$
I am interested in the conditions on exponents $\alpha, \beta, p, q$ for which one has the implication  $$B_{\alpha,\, p}(E) =0\implies B_{\beta,\, q}(E) = 0.$$
The Sobolev embedding theorem suggest the condition
${\frac {1}{p}}-{\frac {\alpha}{n}}<{\frac {1}{q}}-{\frac {\beta }{n}}.$
I would like to know if this is optimal. In particular I would like to know what is the maximum one can obtain form $B_{2,\, 1+\epsilon}(E) =0$ about the capacities $B_{1,\, q}(E).$
 A: 
Theorem. If $\beta q<\alpha p\leq n$, then  $$ B_{\alpha,p}(E)=0 \quad \Rightarrow B_{\beta,q}(E)=0. $$

Remark. If $\alpha p>n$, then $B_{\alpha,p}(\{ x\})>0$ (Remark 2.6.15 in [1]) so the only set with zero capacity is the exmpty set making the problem trivial. This is why in the statement of the theorem we assume that $\alpha p<n$.
In the proof we will need the following result (Theorem 2.6.16 in [1]). Here $\mathcal H^s$ stands for the $s$-dimensional Hausdorff measure.

Lemma. If $p>1$ and $\alpha p\leq n$, then $$ \mathcal{H}^{n-\alpha p}(E)<\infty \quad \Longrightarrow \quad
 B_{\alpha, p}(E)=0, $$ $$ B_{\alpha, p}(E)=0 \quad \Longrightarrow \quad
 \forall \epsilon>0\ \ \ \ \mathcal{H}^{n-\alpha p+\epsilon}(E)=0 $$

Proof of the theorem.
If $\beta q<\alpha p$, then $\beta q=\alpha p-\epsilon$ so
the lemma yields
$$
B_{\alpha, p}(E)=0
\Rightarrow 
\mathcal{H}^{n-\beta q}(E)=\mathcal{H}^{n-\alpha p+\epsilon}(E)=0
\Rightarrow B_{\beta,q}(E)=0.
$$
The proof is complete.
[1] W. P. Ziemer, Weakly differentiable functions. Sobolev spaces and functions of bounded variation. Graduate Texts in Mathematics, 120. Springer-Verlag, New York, 1989. 
Remark. This is not the simplest proof, because the lemma is not easy, but it is short. I am pretty sure that if you search books that deal with Bessel capacity you will find other arguments too.
