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Let $\mu$ be a probability measure on the circle $S^1=\mathbb{R}/\mathbb{Z}$ which is singular with respect to the Lebesgue measure $\lambda$. Consider the functions spaces $L^2(\mu)$ on the one hand, and $\mathrm{Hol}_\beta(S^1)$ on the other hand (space of Hölder functions of some exponent $\beta<1$ with the usual Hölder norm).

Given a pair of functions $(f,g)\in L^2(\mu)\times \mathrm{Hol}_\beta(S^1)$, consider the following property: $$(*) \qquad \exists (\varphi_n) \mbox{ a sequence of smooth functions on the circle such that }\\ \varphi_n' \to f \mbox{ in } L^2(\mu) \quad\mbox{and}\quad \varphi_n\to g \mbox{ in } \mathrm{Hol}_\beta(S^1).$$ Edit: the first convergence is about the derivative of $\varphi_n$, the ' was forgotten in the first version of the question.

My question is about the following general question:

Which pairs $(f,g)$ satisfy property $(*)$?

Under the above assumptions ($\mu$ singular and $\beta<1$), I think that many pairs satisfy $(*)$ (e.g. all pairs $(f,g)$ where $g$ is $\alpha$-Hölder for some $\alpha>\beta$) but the way I expect to be able to prove it would be somewhat tedious. It happens that I need such a result for some pairs, e.g. restricting to $g$ constant would be enough for my purpose.

My question: do you know a reference for this kind of result, or a way to prove such result without cutting in four too many epsilons, either in the general case or for some particular pairs?

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  • $\begingroup$ @ChristianRemling: sorry, the question had a mistake: the first convergence is about the derivative $\varphi_n'$. $\endgroup$ – Benoît Kloeckner May 25 '15 at 19:10
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    $\begingroup$ Oh, I see. But now I'm worried about something like $g=x^{\beta}$. How do you approximate this by smooth functions in Holder norm? $\endgroup$ – Christian Remling May 25 '15 at 19:23
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    $\begingroup$ Actually, this was discussed on MO before: mathoverflow.net/questions/29869/… $\endgroup$ – Christian Remling May 25 '15 at 19:26
  • $\begingroup$ @ChristianRemling: you are of course right. I will edit my guess. $\endgroup$ – Benoît Kloeckner May 25 '15 at 19:34
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Here is the solution I came up with for a restricted version of the question, which is sufficient for my main purpose. I will not accept my own answer, because I am still interested in a reference, a more complete solution, or a simpler proof.

I will prove the following.

Lemma. - Let $\mu$ be a measure on $\mathbb{S}^1$ which is singular with respect to the Lebesgue measure and without atoms; then for all $v\in L^2(\mu)$, and all $\beta<1$ there is a sequence of smooth functions $\varphi_n:\mathbb{S}^1\to \mathbb{R}$ such that $\varphi_n' \to v$ in $L^2(\mu)$ and $\varphi_n \to 0$ in $\mathrm{Hol}_\beta(\mathbb{S}^1)$.

Proof. We first claim that when $I\subset [0,1]$ is an interval of length $\ell$, $w:I\to \mathbb{R}$ is measurable and $\mu$-essentially bounded by some number $M$, and $\varepsilon>0$, there is a smooth function $\varphi:I\to\mathbb{R}$ such that $\varphi$ and all its derivatives vanish at the endpoints of $I$, $\lVert \varphi \rVert_\alpha \le 2\ell^{1-\beta} M$ and $\int_I (\varphi'-w)^2 \,d\mu \le \varepsilon^2\ell^2$

Let $\eta>0$ be arbitrary, to be chosen later on. Since $\mu$ is concentrated on a $\lambda$-negligible set, there is a finite set of intervals $I_1,\dots,I_k\subset I$ with disjoint interiors whose total length is less than $\eta$ and whose complement $J = I\setminus (I_1\cup\dots\cup I_k)$ is given by $\mu$ a mass less than $\eta$. Let $\bar w$ be the function which:

  • is constant on each $I_i$, with value the $\mu$-average of $w$ on $I_i$,

  • is constant on $J$, with value such that $\int_I \bar w \,d\lambda =0$.

By taking $\eta$ small enough and by dividing the intervals $I_i$ into smaller intervals, we can ensure that $\int (w-\bar w)^2 \,d\mu$ is arbitrarily small.

Define a Lipschitz function $\bar\varphi$ by $$\bar \varphi(x) := \int_a^x \bar w \,d\lambda$$ where $a=\min I$ is the starting point of $I$.

Let now $\varphi$ be a smooth approximation of $\bar\varphi$ in uniform norm, such that $\varphi'$ stays close to $\bar w$ in $L^2(\mu,I)$ norm (one might need to make $\varphi''$ very large, but this has no incidence on the claim). Since $\mu$ has no atom, we can further flatten down $\varphi$ at the endpoints of $I$ in order to have its derivatives vanish while keeping $\varphi'$ close to $\bar w$ in $L^2$ norm. This approximation can easily be performed in a way further ensuring that $|\varphi'|$ is bounded by $M$.

The uniform norm of $\varphi$ is then bounded by $M\eta$, and can thus be made arbitrarily small. Moreover for all $x,y\in I$: $$\frac{|\varphi(x)-\varphi(y)|}{|x-y|^\beta} \le \lVert\varphi'\rVert |x-y|^{1-\beta} \le M \ell^{1-\beta}$$ so that taking $\eta$ small enough, we get the desired control $\lVert \varphi \rVert_\alpha \le 2\ell^{1-\beta} M$, and the claim is proved.

Now, given $v$ and $\varepsilon>0$, choose $\bar v$ a $\mu$-essentially bounded function that is $\varepsilon$-close to $v$ in $L^2(\mu)$, call $M$ its essential bound, then choose $\ell$ small enough to ensure that $2\ell^{1-\beta} M<\varepsilon$. Divide $\mathbb{S}^1$ into intervals of length $\ell$ and apply the claim to each of them. The boundary conditions enable us to glue the smooth functions defined on each interval into a smooth function $\varphi$ defined on $\mathbb{S}^1$, and we get that $\varphi'$ is $\varepsilon$-close to $\bar v$ in $L^2(\mu)$. Moreover the variations of $\varphi$ on $\mathbb{S}^1$ are not greater than its variations on each interval, so that $\lVert \varphi\rVert_\beta <\varepsilon$. This proves the Lemma.

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