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We assume that we have a $\alpha$-Hölder continuous function $f$ on an interval $[0,1]$ with $f(0)=0$.

I am wondering if there exists an explicit construction of a sequence $f_{n} \in C_c^{\infty}(\mathbb R)$ such that

$$\lVert f-f_n\rVert_{C^{\alpha}([0,1])} \le \frac{1}{n}$$

and $\lvert f_n(x)\rvert \le \lvert f(x)\rvert$ on $[0,1]$. The usual convolution idea does not respect the last condition.

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    $\begingroup$ Please revert this edit, accept Alexandre's answer to your original question, and post your modified version as a new question. $\endgroup$
    – Nik Weaver
    Commented Apr 11, 2023 at 11:46

2 Answers 2

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It is not possible: your condition $|f_n|<|f|$ implies that the zero set of $f_n$ is contained in the zero set of $f$. So $f(x)=|x|^{\alpha}$ cannot be approximated by a smooth function $f_n$, since $f_n(x)=(c+o(1))x$, and $\sup|f(x)-f_n(x)|\geq (1+o(x))|x|^\alpha$ for some small $|x|$.

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    $\begingroup$ You are right but this does not affect the answer. I corrected. $\endgroup$ Commented Apr 10, 2023 at 13:03
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    $\begingroup$ I think that actually the zero set of $f_n$ must contain, not necessarily be contained in, the zero set of $f$ (and it is a non-strict inequality). $\endgroup$
    – LSpice
    Commented Apr 10, 2023 at 13:05
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Alexandre has answered correctly, but I wanted to add that the norm closure of $C^\infty[0,1]$ in the $\alpha$-Hölder space is the "little" $\alpha$-Hölder space (a.k.a. little Lipschitz space) consisting of those $\alpha$-Hölder functions that are locally flat in the sense that $\lim_{y \to x}\frac{|f(x) - f(y)|}{|x - y|^\alpha} = 0$ for all $x \in [0,1]$. It's easy enough to see that every $C^\infty$ function satisfies this condition and that the set of locally flat $\alpha$-Holder functions is norm closed. (Alexandre's function $|x|^\alpha$ fails to be locally flat at $x = 0$.) To get density we need to use "uniform separation of points"; see Corollary 8.30 of my book Lipschitz Algebras (second edition).

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  • $\begingroup$ thank you, does this actually mean the approximation I asked for is possible in every $C^{\beta}$ norm with $\beta< \alpha$? $\endgroup$ Commented Apr 10, 2023 at 21:46
  • $\begingroup$ Yes, if $\alpha < \beta$ then every $\alpha$-Holder function is not only $\beta$-Holder, but even $\beta$-locally flat. So you are right, anything in $C^\alpha$ can be approximated in $\beta$-Holder norm by $C^\infty$ functions. (That's not to say approximable from below, though.) $\endgroup$
    – Nik Weaver
    Commented Apr 10, 2023 at 22:26

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