# Construction of a special continuous function

Let $\alpha\in\mathbb{R}^{d}$. Construct a continuous function $f:\mathbb{R}^{d}\rightarrow\mathbb{R}$ such that $0\leq f(x)\leq1$ for all $x\in\mathbb{R}^{d}$; $f(\alpha)=0$ and $f(x)=1$ if $|x-\alpha|\geq\epsilon$ for a given $\epsilon>0$.

My initial reaction is that $f$ is not continuous. There seems to be a jump at $\alpha$. I use the tag probability theory'' because this is from a probability class. Given this context, I wonder whether $f$ is a cumulative distribution function. I also considered constructing this directly by $f(x)={\displaystyle \frac{|x-\alpha|}{1+|x-\alpha|}}$. But it does not satisfy $f(x)=1$ if $|x-\alpha|\geq\epsilon$ for a given $\epsilon>0$.

Thank you!

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Thank you! I copied the original text. I guess it should mean for all $\epsilon>0$. I don't know how to modify the bump function to make it continuous. –  user16859 Apr 9 '12 at 3:24

Because $\epsilon$ is fixed ahead of time, you can make the function continuous. Start by letting $$f(x)=\frac1{\epsilon}\|x-\alpha\|$$ for $0\le\|x-\alpha\|\le\epsilon$; that makes $f$ increase linearly from $0$ to $1$ as you move from $\alpha$ out to the $d$-sphere of radius $\epsilon$ around $\alpha$. Now you need to make sure that $f(x)=1$ if $\|x-\alpha\|\ge\epsilon$. Just do it: use a two part definition. The problem asked for a continuous function, not one without ‘corners’.

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True, no need for bump functions –  you Apr 9 '12 at 3:26
Just one simple question: here the norm means the Euclidean norm. Right? en.wikipedia.org/wiki/Euclidean_norm#Euclidean_norm –  user16859 Apr 9 '12 at 3:54
@user16859: That’s what I intended, yes. –  Brian M. Scott Apr 9 '12 at 3:56

I don't understand how this relates to probability theory, because $f(x) = 1$ when $|x - \alpha| \ge \varepsilon$ means $f$ is not integrable over $\mathbb R^d$, but whatever. Let's say we don't want that.

One such function would be the following, for $p > 0$ :

$$f_p(x) = \begin{cases} 1 & \text{ if } |x - \alpha| \ge \varepsilon \\ \frac{|x- \alpha|^p}{\varepsilon} & \text{ if } |x - \alpha| < \varepsilon. \end{cases}$$ You can see that $f(\alpha) = 0$, $f$ is continuous, and $f(x) = 1$ if $|x-\alpha| \ge \varepsilon$. The best way to see continuity is that the function is defined on two sections with disjoint interiors, and those two sections coincide at the frontier and are continuous on the interior. Therefore $f$ is continuous.

Hope that helps,

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Yes. It does help. Two answers are basically the same and are provided basically at the same time. The first answer is 1 min earlier. Thank you. I mark this answer as helpful. –  user16859 Apr 9 '12 at 4:12