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Given a one-dimensional closed and bounded interval, say $[0,L]$ for some $L > 0$, and fix $\epsilon >0$ to be sufficiently small, i.e., $\epsilon \ll 1$. May I know does there exists a smooth non-negative function (let's call it $f$) on $\mathbb{R}$ such that $f(x) \equiv 0$ for $x \in (-\infty,0] \cup [L,\infty)$ and $f(x) \equiv 1$ for $x \in [\epsilon, L-\epsilon]$ ? I am more eager to see an explicit example of such a function if possible, I would appreciate any help or reference related to this question.

Edit: If possible, I hope such construction does not involve summation of a series.

Fei Cao
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  • @Thomas I of course have checked the wiki, this does not solve my problem. – Fei Cao Jan 13 '22 at 06:27
  • I want $f$ to be identically $1$ on a "large" interval (almost of size $L - 0 = L$), if it sounds too easy to you, you don't you give a sketch of the "scaling" and "translating" you are referring to? Your newly posted wiki is highly non-trivial and I just want to convince you that the details are not that easy – Fei Cao Jan 13 '22 at 06:30
  • combine the article on mollification with this one: https://en.wikipedia.org/wiki/Partition_of_unity - subdivide the interval in question in subintervals of length $ \varepsilon$, construct a partition of unity subordinate to that partition, and sum the functions (there are finitely many only) on all intevals with the exception of those at the boundary. (I do not claim it's easy. But it is well known). – Thomas Jan 13 '22 at 06:30
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    "Well-known" is a pretty bad phrase in my opinion, mathematicians all have different background, and as you said, you do not claim it's easy, so why do you claim it's well-known then? – Fei Cao Jan 13 '22 at 06:34
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    Because it is well known. These kind of functions are used extensively in analysis to get global results from local ones. This is a well established technique which is taught in every course on analysis, the latest when it comes to integration theory. – Thomas Jan 13 '22 at 06:38
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    @Thomas I am ready to offer bounties to this "well-known" fact. Please note that in my edited post, I want a explicit example which does not involving summation of a series. – Fei Cao Jan 13 '22 at 06:41
  • A theoretical answer is to take the convolution of the characteristic function of $[O,L]$ with a smooth function with support on $[0,\varepsilon]$. Besides @Thomas I don't see the need to consider a partition of unity. – Jean Marie Jan 13 '22 at 06:47
  • @JeanMarie May I know a explicitly written example? Nothing is more transparent than a explicit definition of $f$ – Fei Cao Jan 13 '22 at 06:51
  • I said that it is theoretical answer ! An example of a smooth $C^n$ function with support on [0,\varepsilon] : a "cardinal spline" with degree $n+1$ (have you heard about these functions which are defined as convolution powers of characteristic functions ?) – Jean Marie Jan 13 '22 at 06:55
  • @JeanMarie Actually I have no clue as to the "cardinal spline". I hope such example does not involve summation of a series. – Fei Cao Jan 13 '22 at 06:57
  • In fact, more practically, see this question and its answers: https://math.stackexchange.com/q/799567 – Jean Marie Jan 13 '22 at 07:06
  • Unless you receive an answer by the weekend consider pinging me again and I may find the time on Saturday or Sunday... – Thomas Jan 13 '22 at 10:06

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This is covered for example by my answer to Smooth function passing through a countable number of points.

The lemma proved there says that there exists a smooth ($C^\infty$) function $f_0 : \mathbb R \to \mathbb R$ such that $f_0 = 0$ on $(-\infty,0]$ and $f_0 = 1$ on $[\epsilon,\infty)$. The function $f_0$ is given explicitly in the proof of the lemma and you see that $f_0(x) \in [0,1]$ for all $x$.

Define $$f(x) = f_0(x) \cdot f_0(L - x). $$

This is clearly a smooth function. We have

  1. $f_0(x) \in [0,1]$ for all $x$

  2. $f(x) = 0$ for $x \in (-\infty,0]$ and $x \in [L,\infty)$.

  3. $f(x) = 1$ for $x \in [\epsilon,L-\epsilon]$.

Paul Frost
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