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Can you give an example of continuously differentiable function that satisfy this properties:

Let $\{(a_i, b_i)\}_{i=1}^{\infty}$ be the family of open subintervals of $[0,1]$ with rationals endpoints. For each $i\geq 1$, let

$$g_i:[a_i,b_i]\to [0,1]$$

be a continuously differentiable function such that

  1. $g_i(a_i)=1$, $g_i(b_i)=0$,
  2. $\lim_{x\to a_i^+} g_i^{\prime}(x)=0$,
  3. $\lim_{x\to b_i^-} g_i^{\prime}(x)=0$.

I have tried to make some examples, this is my attempt:

  1. $g_i(x)=\frac{b_i-x}{b_i-a_i}, a_i\leq x \leq b_i$. But, this function not satisfy properties poin (3) and (4).
  2. \begin{align*} g_i(x)=\begin{cases} 1,& a_i\leq x < \frac{a_i+b_i}{2}\\ \frac{16x-4a_i-12b_i}{a_i-b_i},& \frac{a_i+b_i}{2}\leq x < \frac{a_i+3b_i}{4}\\ 0, & \frac{a_i+3b_i}{4}\leq x \leq b_i \end{cases} \end{align*} This function satisfies properties poin (1)-(4), but this is not continuously differentiable.

Finally, I got stuck on finding an example of that function. Any helps will be appreciated.

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  • $\begingroup$ Are your intervals disjoint? $\endgroup$ Apr 28, 2021 at 5:01
  • $\begingroup$ @Kavi Rama Murthy I think yes, my intervals are disjoint.. $\endgroup$
    – user136524
    Apr 28, 2021 at 6:10
  • $\begingroup$ The functions $$g_i(x):=\cos^2{\pi(x-a_i)\over2(b_i-a_i)}\qquad(i\geq1)$$ would do. – I think you have not properly described your problem, since you also talk about one function that should be constructed. Are you planning to concatenate the $g_i$ somehow? $\endgroup$ Apr 28, 2021 at 7:45
  • $\begingroup$ @Christian Blatter Yes, I will concatenate the $g_i$ with $f_i$, where $f_i$ is defined by $$f_i(x)=\begin{cases} 1, x<a_i\\ g_i(x), a_i≤x≤b_i\\ 0, x>b_i \end{cases}$$, then I will show that $f_i$ is continuously differentiable on $[0,1]$. Okay, I'll try the function that you give. $\endgroup$
    – user136524
    Apr 28, 2021 at 8:41

1 Answer 1

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Hint. There are 4 conditions so try a polynomial with 4 c0-efficients, i.e. a cubic $$g(x)=\frac {x-b}{a-b}\cdot (\,A(x-a)^2+B(x-a)+1\,).$$

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  • $\begingroup$ Are A and B arbitrary constant? Or can I take for example : A=1, B=2? $\endgroup$
    – user136524
    Apr 28, 2021 at 8:45
  • $\begingroup$ @MathLearner. Compute what g(x) and g'(x) are when x=a and when x=b, and compare those to Conditions 1.,2.,3. to find what A and B must be. $\endgroup$ Apr 29, 2021 at 10:34
  • $\begingroup$ oh I see.. thank you! I will try. $\endgroup$
    – user136524
    May 1, 2021 at 20:39

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