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Is there a smooth (infinitely times continuously differentiable) function $f:\mathbb R\to\mathbb R$ with $f(x)=\frac1x$ for all $x\ge1$? Intuitively, it should be possible to construct such a function.

0xbadf00d
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  • Is your function not an example of a smooth function? – Jacob Mar 23 '19 at 15:40
  • @JacobJones The OP wants a smooth function over $\mathbb{R}$ that extends $1/x$. – egreg Mar 23 '19 at 15:42
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    See https://en.wikipedia.org/wiki/Whitney_extension_theorem#Extension_in_a_half_space – egreg Mar 23 '19 at 15:45
  • Note that such a function cannot be analytic. So any example will be a $C^\infty$ but not analytic function, typically constructed using the example $f(x) = e^{-1/x^2}$, $f(x) = 0$ for $x \leq 0$ as a starting point. This function is nice and smooth on $\mathbb{R}$ but runs into trouble near the origin in the complex plane. – Jair Taylor Mar 23 '19 at 17:14

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Start with the smooth function $$ f(x)=\cases{e^{-1/x^2}& if $x>0$\\0& otherwise} $$ Now consider the function $f(x-0.5)f(1-x)$. It is smooth, on $(0.5,1)$ it is strictly positive, and it is $0$ everywhere outside $(0.5,1)$.

Now define $$ g(x)=\int_{-\infty}^xf(t-0.5)f(1-t)dt $$ It is smooth, it is constantly $0$ on $(-\infty, 0.5)$, and it is constantly equal to $g(1)>0$ on $(1,\infty)$.

Now $h(x)=\frac{g(x)}{g(1)x}$ is is the function you're after (with the specification that $h(0)=0$).

It is more common to see this done with $f(x)f(1-x)$. I chose not to do this in order to entirely avoid any discussion about whether the limiting behaviour of $\frac1x$ as $x\to0$ interferes.

Arthur
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This answer to a previous question gives the construction of a $C^\infty$ bump function, which can also be found in many analysis textbooks. If you read and understand it, you can most likely see how to generalize that construction to produce the function you desire, as well as many other similar functions.

Lee Mosher
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