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.


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    $\begingroup$ See Fabio's question and answer "Extension of continuous and smooth functions" from MO. $\endgroup$ Commented Nov 2, 2016 at 11:37
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    $\begingroup$ Linked. $\endgroup$ Commented Nov 2, 2016 at 11:37
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    $\begingroup$ 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 $(0,0)$. $\endgroup$
    – Jochen
    Commented Aug 1, 2023 at 13:34

1 Answer 1


I recommend looking at https://www.ams.org/journals/proc/1964-015-04/S0002-9939-1964-0165392-8/S0002-9939-1964-0165392-8.pdf

for a short note on extending smooth functions from the halfspace to euclidean space. I originally got this reference from a stackexchange question which I am unable to find anymore.


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