In my study I have faced the problem showing the property below: regularity of mollification on boundary. (However, I don't know whether this is true or not although I hope.)
Let $f\in C^{1}((0,2]^{2})$ be a function such that it attains a strict maximum at a point $(2,1)$. Define a function $g\in C([0,2]^{2})$ by $$g:=f\quad\text{in $B_{1/2}(2,1)$}\quad\text{and}\quad g:=const.\quad\text{in $[0,2]^{2}\setminus B_{1/2}(2,1)$}.$$ Then $g\ast\eta_{\varepsilon}$ belongs to $C^{\infty}([0,2]^{2})$.
Here, $B_{r}(a,b)$ is a open ball of a radius $r$ with centered at $(a,b)$ and $\eta_{\varepsilon}:=\varepsilon^{-2}\eta(x/\varepsilon)$ for the Friedrichs mollifier. Please be careful that the function $g$ is continuous.
I think this probelm is very delicate, but I guess that it is true if the following holds; this is my question.
(Question:) $C^{1}[a,b]$ functions can be extended to $C^{1}(a-\varepsilon,b+\varepsilon)$ functions for $\varepsilon>0$. Is this true?
Sketch of my proof: Assume that the statement of my question holds for $\varepsilon\in(1/2,1)$. Denote by $\tilde{f}$ a extended function of $f$ and define $\tilde{g}\in C([0,2+\varepsilon]^{2})$ similarly for $\tilde{f}$. Then the standard argument implies that $\tilde{g}\ast\eta_{\varepsilon}\in C^{\infty}([0,2+\varepsilon])$. This particularly means that $g\ast\eta_{\varepsilon}=\tilde{g}\ast\eta_{\varepsilon}\in C^{\infty}([0,2]^{2})$.
I'm glad if you tell me an answer for my question or some comments for my proof.