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Let $X$ be a Banach space and endow the space $BC(X)$, the space of all bounded closed subsets of $X$, with the Hausdorff distance $d_H$. Fix $C_0\in BC(X)$. Is it true that $d_H(A,B)=d_H(A+C_0,B+C_0)$ for all $A,B\in BC(X)$?

Edit: Is it true when $A,B,C_0$ are convex sets?

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2 Answers 2

Counterexample in $\mathbb R$: $A=[-1,1]$, $B=\{-1,1\}$, $C_0=[-1,1]$. Here $d_H(A,B)=1$ but $A+C_0=B+C_0=[-2,2]$. Did you want the sets to be convex, by any chance?

Okay, let's consider the convex case. Let $X$ be a real Banach space; if it's complex, forget the complex structure. Let $S$ be the unit sphere of the dual space $X^*$. Given a bounded closed set $A\subset X$, define its support function $h_A\colon S\to\mathbb R$ by $h_A(\phi)=\sup_{A}\phi$. The definition implies $h_{A+C}=h_A+h_C$ for any bounded closed sets $A,C$. In particular, the quantity $\rho(A,B)=\sup_{S}|h_A-h_B|$ is invariant under addition/translation: $\rho(A+C,B+C)=\rho(A,B)$.

So far so good, but what does $\rho$ have to do with the Hausdorff metric? The inequality $\rho(A,B)\le d_H(A,B)$ holds for general $A,B\in BC(X)$: just use the definition of $d_H$ together with the fact that $\phi$ has norm $1$. The reverse inequality is where we need convexity... and it seems that we need $X$ to be a Hilbert space too... (?)

So, suppose $X$ is Hilbert and $A$ and $B$ are convex. We can find a point $p$ in one of these sets (say, in $A$) which is at distance $>d_H(A,B)-\epsilon$ from the other set (namely, $B$). There exists $q\in B$ that minimizes the distance to $p$ (a closed convex set in a Hilbert space has a unique nearest point to any point of its complement). Let $\phi$ be the norming functional of $p-q$, simply $\phi=(p-q)/\|p-q\|$. Then $h_A(\phi)-h_B(\phi)>d_H(A,B)-\epsilon$ as required.

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$\newcommand{\ep}{\varepsilon}$$\newcommand{\f}{\phi}$ NOTE: this is an extension of the answer given by Leonid Kovalev. I continu with the notation from his answer.

The Hausdorff distance is given by $$d_h(A,B)=\inf\{\ep>0\colon A\subset B+B_\ep(0)\text{ and }B\subset A+B_\ep(0)\},$$ where $B_\ep(0)$ is the open ball of radius $\ep$ centered at $0$ (see here).

The identity $d_h(A,B)=\sup_{\phi\in S}h_A(\phi)-h_B(\phi)|$ holds for any convex, bounded closed, nonvoid subset of a Banach space $X$ (It need not be Hilbert)

First note that if $A=B$, then the result is clear. Now assume $A\neq B$ and let $\ep>0$ such that $A\subset B+B_\ep(0)$.

Then $h_A(\f)\le h_B(\f)+\ep$ and $h_B(\f)\le h_A(\f)+\ep$ for all $\f\in S.$ Thus $|h_A(\f)-h_B(\f)|<\ep$, hence $\sup_{\f\in S}|h_A(\f)-h_B(\f)|\le d_h(A,B).$

Next, define $$\ep:=\sup_{\f\in S}|h_A(\f)-h_B(\f)|>0,$$then $A\subset B+B_\ep(0)$ and $B\subset A+B_\ep(0).$ Thus $d_h(A,B)\le\ep$.

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