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I need to prove that exists a indefinitely differentiable function $F$ in $\mathbb{R}$ such that $F(x)=0$ if $x\leq a$, $F(x)=1$ if $x\geq b$ and $F$ is strictly increasing in $[a,b]$.

To simplify, I write $f:=\left.F\right|_{[a,b]}$.

My first try use the exponential function. The function $$f(x)=e^{\frac{x-b}{x-a}}$$ solves the problem for $x\rightarrow a^{+}$. $f(b)=e^{0}=1$, but $F$ will not be differentiable at $b$.

Any tips?

Mateus Rocha
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  • Very related: https://math.stackexchange.com/questions/1283783/an-example-of-an-infinitely-differentiable-function-with-compact-support – Tito Eliatron Nov 07 '18 at 19:35

1 Answers1

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Define

$$f(x)= \begin{cases} 0, \;\textrm{if}\;x\leq a\;\textrm{or}\; x\geq b, \\ e^{\frac{b-a}{(x-a)(x-b)}}\; \textrm{if}\; a<x<b. \end{cases} $$ Now, define the constant $$c=\int_{a}^{b}f(x)dx=\int_{a}^{b}e^{\frac{b-a}{(x-a)(x-b)}}dx.$$

So, we can take $$F(x)=\frac{1}{c}\int_{-\infty}^{x}f(x)dx.$$

For all $x\leq a$, $F(x)=0$ and, for all $x\geq b$, $F(x)=1.$ $f$ is non-negative, so $F$ is striclty increasing in $[a,b]$. To finish, $F$ is indefinitely differentiable.

Mateus Rocha
  • 2,646