Let $G=(V,E)$ be a simple graph with $n$ vertices. The isoperimetric constant of $G$ is defined as 


$$
i(G) := \min\_{A \subset V,|A| \leq \frac n2} \frac{|\partial A|}{|A|}
$$


where $\partial A$ is the set edges $uv \in E$, with $v \in A$ and $u \in A^c$. 



> 
> Is there a graph $G$ such that $$i(G) + i(G^c) < \frac 12$$
>  where $G^c$ denotes the complement of $G$?
> 
> 
> 



Is it possible that both a graph and its complement have small connectivity?
It is *almost* obvious. Let $i(A)=\frac{|\partial\_G A|}{|A|}$ and $j(B)$ be the same ratio for a vertex set $B$ and the complement of $G$. Let $x=|A\cap B|, y=|A\cap B^c|, z=|A^c\cap B|, t=|A^c\cap B^c|$. The edges going between $A\cap B$ and $A^c\cap B^c$ are boundary edges for both $A$ and $B$ and so are the edges between $A^c\cap B$ and $A\cap B^c$. Thus, we have
$$
i(A)(x+y)+j(B)(x+z)\ge xt+yz
$$
If $i(A)+j(B)<1$, the left hand side is strictly less than $x+\max(y,z)$, which is at most $xt+yz$ if $t,y,z>0$. Note that if $t=0$, then $x=0$ because of the condition $|A|,|B|\le \frac n2$, so if $y,z>0$ in this case, we still have a contradiction. Finally, if, say, $y=0$, then $x>0$ (otherwise $A=\varnothing)$ and $t\ge \frac n2$ (because $|B|\le\frac n2$), so the RHS is at least $\frac n2$ while the LHS is less than $\max(|A|,|B|)\le \frac n2$ and we get a contradiction again. 


