$P_{x}(T_{A}<\infty)<P_{x}(T_{B}<\infty)$ imply $Cap_{N}(A)<Cap_{N}(B)$, where $Cap_{N}$ is Newtonian capacity

Question

We start a Brownian motion at $x\in [B(0,r)]^{c}$, where $B(0,r)$ is a large enough ball that contains compact sets $A$ and $B$. In other words, the B.M. starts on the exterior of $A$ and $B$.

Then if the hitting probabilities for $A$, $B$ satisfy $P_{x}(T_{A}<\infty)<P_{x}(T_{B}<\infty)$ for all $x$ as described above, does this imply $Cap_{N}(A)<Cap_{N}(B)$, where $Cap_{N}$ is Newtonian capacity? (**)

For compact sets $K$, we have the following inequality

$$\frac{1}{2}Cap_{N}(K)\leq P_{x}(T_{K}<\infty)\leq Cap_{N}(K)$$

found in http://www.math.upenn.edu/~pemantle/papers/martin.pdf Proposition 1.1.

How can I go about answering (**)?

Thanks

Answer 1

No. Imagine $A=\{-1\}\times [-1,1]^2$ and $B=\{1\}\times[-1,1]^2$. If $x=(10,0,0)$, then $P_x(T_A<\infty) < P_x(T_B<\infty)$, whereas if $x=(-10,0,0)$, then the opposite inequality holds. Of course $A$ and $B$ have same capacity.