Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be Banach and let $B(x,\varepsilon)$ be the closed ball of radius $\varepsilon>0$ around $x\in X$ and consider the sequence $$f_{n;x}(y)= \begin{cases} 1-n\cdot d(yB(x,\varepsilon)), & (1-n\cdot d(y,B(x,\varepsilon)))\ge0\\ 0, &\text{elsewhere}. \end{cases} $$ for $y\in X$ and $n=1,2,\ldots$ and where $d(y,B(x,\varepsilon))=\inf\{\parallel y-z\parallel\colon z\in B(x,\varepsilon)\}$. The pointwise limit of this sequence is the characteristic function on $B(x,\varepsilon)$.

Let $BC(X)$ denote the space of closed and bounded subset of $X$, then the map $$f\colon X\to BC(X): x\mapsto B(x,\varepsilon)$$ is continuous for the Hausdorff metric $d_H$ . And for $A\in BC(X)$ fixed, the map $$g_A\colon BC(X)\to\mathbb{R}: B\mapsto\delta(A,B),$$ is 1-Lipschitz, where $\delta(A,B)=\sup_{a\in A}\inf_{b\in B}\parallel a-b \parallel$.

Is there somebody who knows how to prove the following:

Let $\{x_n: n=1,2,\ldots\}$ be a convergent sequence in $X$ with limit $x_0$, then $f_{n,x_j}\to f_{n,x_0}$ pointwise as $j\to\infty$.

My main goal is to show that the map $p\colon X\to\mathbb{R}:x\mapsto\mu(B(x,\varepsilon))$ is Borel measurable for any (not necessarily finite or positive) Borel measure $\mu$ on $X$. If I would be able to prove $f_{n,x_j}\to f_{n,x_0}$, then I'm done.

share|cite|improve this question

Assume first that $y\in B(x_0,\varepsilon)$. Then $f_{n,x_0}(y)=1$. If $y\in B(x_j,\varepsilon)$, then $f_{n,x_j}(y)=1$. Otherwise, let $t=\varepsilon/d(y,x_j)<1$. Then $y_j=x_j+t(y-x_j)\in B(x_j,\varepsilon)$, since \begin{eqnarray} d(y_j,x_j)=d(x_j+t(y-x_j), x_j)&=&\|t(y-x_j)\|=t\|y-x_j\|=\varepsilon. \end{eqnarray} Then (note that below we use in a critical way that the distance is given by a norm) \begin{eqnarray} d(y,B(x_j,\varepsilon))+\varepsilon&\leq& d(y,y_j)+\varepsilon=d(y,y_j)+d(y_j,x_j)=\|(1-t)(y-x_j)\|+\|t(y-x_j)\|\\ &=&(1-t)\|y-x_j\|+t\|y-x_j\|=\|y-x_j\|\leq\|y-x_0\|+\|x_0-x_j\|\\ &\leq&\varepsilon+\|x_0-x_j\|, \end{eqnarray} so $$ d(y,B(x_j,\varepsilon))\leq\|x_0-x_j\|. $$ For $j$ big enough, $\|x_0-x_j\|\leq 1/n$, and so $$ f_{n,x_j}(y)=1-n\,d(y,B(x_j,\varepsilon))\geq1-n\|x_0-x_j\|\xrightarrow{j\to\infty}1. $$

The other case to consider is when $y\not\in B(x_0,\varepsilon)$. Then $f_{n,x_0}(y)=0$, and $d(y,x_0)>\varepsilon$. Then there exists $\delta>0$ such that $d(y,x_0)>\varepsilon+\delta$, and $$ d(y,x_j)\geq d(y,x_0)-d(x_0,x_j)>\varepsilon+\delta-d(x_0,x_j). $$ So, for $j$ big enough, we get $d(y,x_j)>\varepsilon$, i.e. $y\not\in B(x_j,\varepsilon)$ and $f_{n,x_j}(y)=0$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.