# Smooth version of Tietze extension theorem

Thanks to the properties of mollifications, it is very easy to prove Urysohn's lemma in the euclidean space with the big plus that the constructed function is smooth.

I was wondering if something of the kind could be achieved with Tietze extension theorem in the euclidean setting (i.e, can we always find smooth extensions of smooth functions in a closed set of the euclidean space?)

The proof of Tietze's theorem runs by constructing iteratively a family of functions with the aid of Urysohn's lemma. This family extends the original function and converges uniformly.

This is very nice, but I if I try to replicate the proof, I have no way to prove the uniform convergence of the derivatives, so I guess that if the theorem is true it will not follow these lines.

Thanks

• See Fabio's question and answer "Extension of continuous and smooth functions" from MO. Commented Nov 2, 2016 at 11:37
• Commented Nov 2, 2016 at 11:37
• The answer depends on your definition of smoothness on compact sets. Whitney's extension theorem is about a notion of smoothness which allows smooth extensions of all smooth functions. On the other hand, there is the following classical example: Let $K=\{(x,y)\in\mathbb R^2: |y|\ge e^{-1/x}$ if $x>0\}$ and $f(x,y)=e^{-1/2x}$ if $x,y>0$ and $=0$ else. Then $f$ does not have an extension to a $C^1$-function on $\mathbb R^2$ because it is not Lipschitz continuous near $(0,0)$. Commented Aug 1, 2023 at 13:34