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Dec 10, 2021 at 0:39 vote accept CommunityBot
Dec 9, 2021 at 21:40 history edited Iosif Pinelis CC BY-SA 4.0
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Dec 9, 2021 at 21:37 comment added Iosif Pinelis @Dal : Your original question has been fully answered, and I even answered all your extra questions. Also, multiple questions for one post are discouraged on MathOverflow, anyway. So, let us have a closure here.
Dec 9, 2021 at 21:37 comment added Iosif Pinelis @Dal : No to (3) and (4) either. Indeed, if for $F:=f_\epsilon$ we have $F''\le CF'$ and $F'>0$ on an interval $(a,b]$, then on $(a,b]$ we have $(\ln F')'\le C$ and hence $\ln F'(x)-\ln F'(a)\le C(x-a)$ and $F'(x)\le F'(a)e^ {C(x-a)}$ for $x\in[a,b]$, which implies $F'\ge0$ on $[a,b]$ if $F'(a)=0$. So, we get a contradiction with the condition $F'>0$ on an interval $(a,b]$. Now take $a:=\max\{z\colon F(z)=\bar x\}$.
Dec 9, 2021 at 21:19 comment added Iosif Pinelis @Dal : (1) I have added the explicit expression for $h$. (2) No, because a convex function which equals a constant $c$ on a nonzero-length interval must be $\ge c$ everywhere and therefore cannot approximate the identity function.
Dec 8, 2021 at 10:35 comment added user139844 Other two questions: (3) what is the dependence of $f'_\epsilon$ on $\epsilon$? Is it true that $f''_\epsilon \le C_\epsilon f'_\epsilon$? (4) Also, can the construction be generalized to get a $C^3$ function? If yes, on this $C^3$ function, do we have a better control on the second derivative, e.g. something like $f''_\epsilon \le C_\epsilon f'_\epsilon$?
Dec 8, 2021 at 7:38 comment added user139844 Thank you so much. I have a couple of questions: (1) how can one compute $h$ explicitly? (2) Is it possible for the function $f_\epsilon$ to be built also to be convex?
Dec 7, 2021 at 22:54 history edited Iosif Pinelis CC BY-SA 4.0
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Dec 7, 2021 at 22:42 history answered Iosif Pinelis CC BY-SA 4.0