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There was a proof in my differential equations class which stated that this function existed:

Find $f\in C^{\infty}(\mathbb{R})$ such that $f(x)=0$ if $x\leq 1/2$ and $f(x)=1$ if $x\geq 1$

At least prove that this function exists

Using a sine function it is pretty easy to find $f\in C^{1}(\mathbb{R})$

ATEMPT

I have tried to transform $f(x)=e^{-1/x^2}$ as $f(x)=0, f^{n)}(0)=0$ but cannot think of how to get it to be smooth on $x=1$

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    See for example https://math.stackexchange.com/q/638860/42969 – Martin R Apr 04 '23 at 13:54
  • Or directly: $\psi(t):=\int_{-1}^te^{-\frac{1}{1-s^2}}ds$ for $|t|\le1$ (and constant on $t\le-1$ and on $t\ge1$), $g(t):=\psi(t)/\psi(1),$ $f(x)=g(4x-3).$ – Anne Bauval Apr 04 '23 at 14:02

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