Consider a sequence of open sets in $R^n$: $\Omega\_1 \supset \Omega\_2 \supset\cdots$. Consider that this sets are bounded, convex with the boundary piecewise smooth .When i say smooth i mean $C^{\infty}$


I believe that the set $\operatorname{int}\left(\,\overline{\bigcap \Omega\_i}\,\right)$ have the boundary piecewise smooth . but i have no idea to how to prove this ...
Someone can give me a help to prove or disprove this?


thanks in advance



The boundary of this set is piecewise smooth?
Since you can assume that $\Omega\_{n+1}$ is strictly contained in the interior of $\Omega\_n$ for all $n$, there is no reason to assume that the boundary of the limit has any interesting properties. Perhaps you are missing an hypothesis.


In any case, perhaps the following construction will give you food for thought.


Work in $\Bbb{R}^2$. Consider graphs of concave functions over $[0,1]$. The regions beneath these graphs are convex, and if $f\geqslant g$, then the region beneath $f$ is contained in the region beneath $g$. We define the sequence $(g\_n)$ of functions over $[0,1]$ inductively by:


1) $g\_0(t) = -t$ if $0\leqslant t<1/2$; and


$g\_0(t) = t-1$ if $1/2\leqslant t\leqslant 1$.


2) for all $n$, given $g\_0,...,g\_n$, $g\_{n+1}$ is defined by:


$g\_{n+1}(t) = g\_0(t) + g\_n(2t)/2$ if $0\leqslant t<1/2$; and


$g\_{n+1}(t) = g\_0(t) + g\_n(2t-1)/2$ if $1/2\leqslant t\leqslant 1$.


Observe that $-1\leqslant g\_n\leqslant 0$ for all $n$, and the sequence $(g\_n)$ converges to a continuous nowhere differentiable function. 


For all $n$, let $f\_n$ be such that $f\_n''=2g\_n$, $f\_n(0)=f\_n'(0)=0$. $f\_n$ is a decreasing family of non-positive concave functions bounded below by the function $f\_n(x)=-x^2$. The limit $f\_\infty$ is a concave function which is nowhere $C^3$. In particular, it is not smooth.


