# Convergence of a sequence of functions on closed balls

Let $X$ be a Banach space and $d$ be the induced metric. Let $S(x;r)$ denote the closed ball with radius $r$ at $x\in X$, that is,$$S(x;r)=\lbrace y\in X\colon d(x,y)\le r\rbrace.$$

Let $x,y\in X$ and define the sequence $\{f_n^x(y)\colon n\in\mathbf{N}\}$ where $$f_n^x(y)=\Big[1-n\cdot\mathrm{dist}(y,S(x;r))\Big]^+.$$ The plus sign denotes the positive part (see http://en.wikipedia.org/wiki/Positive_and_negative_parts) and $\mathrm{dist}(y,S(x;r))=\inf\{d(y,z)\colon z\in S(x;r)\}$

Does the sequence $\{f_n^x(y)\colon\in\mathbf{N}\}$ converge for $n\to\infty$ and what is the limit? My best guess is that this sequence converges to the characteristic function on the closed ball, but I don't know how to prove this.

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Thank you very much Davide Giraudo, you really helped me out. – bbold Mar 22 '12 at 20:07

## 1 Answer

If $y\notin S(x,r)$, since $S(x,r)$ is closed $\operatorname{dist}(y,S(x,r))>0$, and for $n\geq \frac 1{\operatorname{dist}(y,S(x,r))}+1$ we get $f_n^x(y)=0$. If $y\in S(x,r)$ then $f_n^x(y)=1$ for all $n$, so $f_n^x$ converges pointwise to the characteristic function of $S(x,r)$.

(the convergence is not uniform since $f_n^x$ is continuous and the limit is not)

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