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Is there a general way to extend a smooth function on a closed interval $[a,b]$ to one that is defined on the entire $\mathbb R$?

It is not OK to reflect this function in points $a$ and $b$, and then in points $a -(b-a) $ and $b +(b-a) $ etc symmetrically because the result wouldn't be differentable in this points

hmakholm left over Monica
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user228494
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A very good question. As an observation, the derivatives at the end points are one-sided.

Indeed, reflection does not work if the function does not have the derivatives of odd order vanishing at the point. You can extend to the left of $a$ as a function of class $C^m$ if you can find a polynomial that has the prescribed derivatives at the point, up to order $m$. A $C^{\infty}$ is possible if you can find a $C^{\infty}$ function that has at $a$ matching derivatives with $f$ ( extend past $a$ with that function). This is possible indeed, due to the theorem of Borel on power series.

$\bf{Added:}$

The ideea : if $f$ is $C^{\infty}$ on $[a,b]$, $g$ is $C^{\infty}$ on $(-\infty, a]$, $h$ is $C^{\infty}$ on $[b, \infty)$, and $f$ and $g$ have all the derivative matching at $a$, $f$ and $h$ have all the derivatives matching at $b$, then $(f,g,h)$ provide an extension of $f$ to whole $\mathbb{R}$.

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orangeskid
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    (+0) This doesn't really answer the question. It seems to be assuring the OP that a continuation is possible, but not explaining how it can be done. – Cameron Buie Jun 19 '15 at 12:37
  • @Cameron Buie: Extend past the edge with that function. The OP had this idea right, but he needed a piece that matches, the symmetric one won't do. It's useful to start with the $C^m$ case to get the idea. – orangeskid Jun 19 '15 at 18:14
  • @Cameron Buie: I added some extra details, thanks for the feedback. – orangeskid Jun 19 '15 at 18:24
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    The answer still doesn't explain how to find $g$ and $h$, other than reference a theorem stating that they have to exist, somewhere. – hmakholm left over Monica Jun 19 '15 at 19:59
  • @Henning Makholm: $h$ is the restriction to $(-\infty, a]$ of a $C^{\infty}$ function on $\mathbb{R}$ that has matching derivatives at $a$ with $f$ ; the latter exists by Borel's theorem. – orangeskid Jun 19 '15 at 21:03
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    @orangeskid: Exactly: You have pointed to an existence theorem, but not explained how to actually find that function. – hmakholm left over Monica Jun 19 '15 at 21:19