I want to find a visual proof of the following fact:



> 
> For any convex figure in the plane there is a sequence of Steiner's symmetrizations that makes it arbitrary close to a circular disc.
> 
> 
> 


All proofs I know require some integral estimates. I would prefer a more visual proof (even if it is more involved). 



Visual proof of convergence for Steiner's symmetrization
(1) Let
$$ C = {\rm conv}\ B\bigcup \{x\}$$ where $B$ is a closed ball of center $o$ and
$x$ is not in $B$.


Consider arc $A\subset B$, $$ \partial B \bigcap {\rm Int}\ C $$


Note that by any Steiner symmetrization $S\_L$, $A$ is still in the interior ${\rm Int}\ S\_L(
C)$.


If $x\_1$ is close to $x$ s.t. $\overrightarrow{ox}$ and
$\overrightarrow{ox\_1}$ are different, then $S\_L(C)$ where $L=(
\overrightarrow{ox\_1})^\perp$ has arc $A\_1$ which contains $A$.
After a finite process, $\partial B-A\_n$ is a one point set. Hence
through a some Steiner symmetrization, $A\_{n+1}=\partial B$. Hence
the resulting convex and compact set contains a ball $B'$ whose
radius is
strictly larger than that of $B$. 


(2) Fix a point $o\in \mathbb{E}^2$ which is the origin.
Here assume that
all hyperplane pass through $o$.


Assume that $C$ is a convex and compact subset. If $x$ is interior
point in $C$, then define $v=\overrightarrow{ox}$
so that $o$ is an interior point of $S\_L(C)$ for $v^\perp=L$.


In further, if $B$ is a largest closed ball in $C$ of center $o$,
then there is an open arc of non-zero length in $
\partial B \bigcap {\rm Int}\ C$.


Hence we have $B'$ in $(1)$. So repeat the process.


If there is no such arc in $\partial B \bigcap {\rm Int}\ C$, then $B=C$.


