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 $B\_{1,\, q}(E).$



Comparison of Bessel Capacities

> 
> **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.


