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This question came up when I was doing some reading into convolution squares of singular measures. Recall a function $f$ on the torus $T = [-1/2,1/2]$ is said to be $\alpha$-Hölder (for $0 < \alpha < 1$) if $\sup_{t \in \mathbb{T}} \sup_{h \neq 0} |h|^{-\alpha}|f(t+h)-f(t)| < \infty$. In this case, define this value, $\omega_\alpha(f) = \sup_{t \in \mathbb{T}} \sup_{h \neq 0} |h|^{-\alpha}|f(t+h)-f(t)|$. This behaves much like a metric, except functions differing by a constant will differ in $\omega_\alpha$ value. My primary question is this:

1) Is it true that the smooth functions are "dense" in the space of continuous $\alpha$-Hölder functions, i.e., for a given continuous $\alpha$-Hölder $f$ and $\varepsilon > 0$, does there exists a smooth function $g$ with $\omega_\alpha(f-g) < \varepsilon$?

To be precise, where this came up was worded somewhat differently. Suppose $K_n$ are positive, smooth functions supported on $[-1/n,1/n]$ with $\int K_n = 1$.

2) Given a fixed continuous function $f$ which is $\alpha$-Hölder and $\varepsilon > 0$, does there exist $N$ such that $n \geq N$ ensures $\omega_\alpha(f-f*K_n) < \varepsilon$?

This second formulation is stronger than the first, but is not needed for the final result, I believe.

To generalize, fix $0 < \alpha < 1$ and suppose $\psi$ is a function defined on $[0,1/2]$ that is strictly increasing, $\psi(0) = 0$, and $\psi(t) \geq t^{\alpha}$. Say that a function $f$ is $\psi$-Hölder if $\sup_{t \in \mathbb{T}} \sup_{h \neq 0} \psi(|h|)^{-1}|f(t+h)-f(t)| < \infty$. In this case, define this value, $\omega_\psi(f) = \sup_{t \in \mathbb{T}} \sup_{h \neq 0} \psi(|h|)^{-1}|f(t+h)-f(t)|$. Then we can ask 1) and 2) again with $\alpha$ replaced by $\psi$.

I suppose the motivation would be that the smooth functions are dense in the space of continuous functions under the usual metrics on function spaces, and this "Hölder metric" seems to be a natural way of defining a metric of the equivalence classes of functions (where $f$ and $g$ are equivalent if $f = g+c$ for a constant $c$). Any insight would be appreciated.

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I thnk the answer to 2) should be "yes". Certainly the trigonometric polynomials are dense in our space of $\alpha$-Holder functions -- this is usually known as a $Lip_\alpha$-space, by the way, see Katznelson's Intro to HA for instance for the terminology and the proof. If you haven't looked things up there, and can get hold of a copy of the book from your library or similar, that would be my first suggestion. –  Yemon Choi Jun 29 '10 at 7:28
    
I dn't have my copy of the book to hand, but some foraging on Google Books suggests that the kind of approximation results you want can be extracted from the general machinery in Chapter I, Section 2 of Katznelson's book. –  Yemon Choi Jun 29 '10 at 7:51
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Thank you for the response Yemon! I have a copy of Katznelson in fact, and I think you are referring to Exercise 6? I didn't notice this earlier, but I believe this is slightly different from what I have. The exercise gives that the trigonometric polynomials are dense in the space $lip_\alpha$, which is in general a subspace of $Lip_\alpha$, the difference being that we require in addition that $\lim_{h \to 0} \sup_t |h|^{-\alpha}|f(t+h)-f(t)| = 0$. –  Vince Jun 29 '10 at 18:02
    
This space is equivalent to what Katznelson calls $B_c$, the set of functions $f$ in the Banach space $B = Lip_\alpha$ such that $\tau \mapsto f_\tau$ is a continuous $B$-valued function. In light of this new information, it seems to me that $B_c \neq B$ in general, and the original question was about density of smooth functions in $B$. –  Vince Jun 29 '10 at 18:02
    
My intuition is still that smooth functions should be dense in Lip_\alpha, for essentially the same reason they're dense in $C^1(T)$. But perhaps I'm missing some subtlety. –  Yemon Choi Jun 29 '10 at 23:48
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2 Answers

up vote 9 down vote accepted

In our PDE seminar, we met the same kinds of questions, and we think the answer is "WRONG". The smooth functions is NOT dense in H{\"o}lder spaces.

An example is, $$f(x) = |x|^{1/2} \quad x \in (-1,1)$$ it is easy to check that $f$ is $1/2$-H{\"o}lder continuous.

For details, for any $g \in C^{1}((-1,1))$, then the derivative of $g$ is continuous at $0$, so we have $$ \lim_{x \to 0} \frac{|g(x)-g(0)|}{|x|^{1/2}} = \lim_{x \to 0} |x|^{1/2}\frac{|g(x)-g(0)|}{|x|} = 0 $$
and $$ \omega_{1/2}(g-f) \ge \frac{|(g(x)-f(x))-(g(0)-f(0))|}{|x|^{1/2}} \ge |\frac{|(g(x)-g(0)|}{|x|^{1/2}}-\frac{|f(x)-f(0)|}{|x|^{1/2}}| $$ but
$$ \frac{|f(x)-f(0)|}{|x|^{1/2}}=1 \quad x \in (-1,1) \quad x \neq 0 $$ let $x \to 0$, we obtain $\omega_{1/2}(g-f) \ge 1$.

Thus, for any $g \in C^{1}((-1,1))$, we have $\omega_{1/2}(g-f)\ge 1$.

For $0< \alpha <1$ we can make similar examples, but when $\alpha = 1$, the proof of the counter-example may be different.

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Right. When $\alpha=1$ take $f=|x|$ and write $$\omega_1(f-g) \ge |\frac{g(x)-g(0)}{|x|}-\frac{f(x)-f(0)}{|x|}|.$$ The second fraction is 1, the first one tends to $g'(0)$ if $x\to0^+$ and to $-g'(0)$ if $x\to0^-$. If $g'(0)\ge0$ taking the left limit gives $\omega_1(f-g)\ge 1 + |g'(0)|$, if $g'(0)<0$ use the right limit. –  Piero D'Ancona Aug 12 '10 at 21:03
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The answer is yes (Edit: my "proof" below seems incomplete; read the comment below). I will prove it in $\mathbb{R}^N$; the proof is easily adapted to the torus. Let $\phi\colon\mathbb{R}^N\to\mathbb{R}^N$ be a smooth function with compact support, non-negative, and such that $\int_{\mathbb{R}^N}\phi(x)dx=1$. Define $\phi_\varepsilon(x)=\varepsilon^{-N}\phi(\varepsilon^{-1}x)$. Given an $\alpha$-Hölder function $f\colon\mathbb{R}^N\to\mathbb{R}^N$, $\phi_\varepsilon*f$ is smooth and $\lim_{\varepsilon\to0}w_\alpha(f-\phi_\varepsilon*f)=0$. To prove it, let $g_\varepsilon=f-\phi_\varepsilon*f$. Then, for any $x,y\in\mathbb{R}^N$, $$ |g_\varepsilon(x)-g_\varepsilon(y)|\le\int_{\mathbb{R}^N}\phi(t)|(f(x-\varepsilon t)-f(x))-(f(y-\varepsilon t)-f(y))|dt. $$ Let $$ G_\varepsilon(t)=\sup_{x\ne y}\frac{|(f(x-\varepsilon t)-f(x))-(f(y-\varepsilon t)-f(y))|}{|x-y|^\alpha} $$ $$ \qquad\qquad\qquad\le\sup_{x\ne y}\frac{|f(x-\varepsilon t)-f(y-\varepsilon t)|+|f(x)-f(y)|}{|x-y|^\alpha}\le2w_\alpha(f). $$ Then, for any $x,y\in\mathbb{R}^N$, $x\ne y$ $$ \frac{|g_\varepsilon(x)-g_\varepsilon(y)|}{|x-y|^\alpha} \le\int_{\mathbb{R}^N}\phi(t)\frac{|(f(x-\varepsilon t)-f(x))-(f(y-\varepsilon t)-f(y))|}{|x-y|^\alpha}dt\le\int_{\mathbb{R}^N}\phi(t)G_\varepsilon(t)dt. $$ Taking the supremum over $x,y\in\mathbb{R}^N$ we get $$ w_\alpha(g_\varepsilon)\le\int_{\mathbb{R}^N}\phi(t)G_\varepsilon(t)dt. $$ As we have seen, $G_\varepsilon(t)$ is bounded and, since $f$ is continuous, $\lim_{\varepsilon\to0}G_\varepsilon(t)=0$ for all $t\in\mathbb{R}^N$. The Lebesgue dominated convergence theorem implies the result.

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Hi Julian, thanks for replying. I followed most of what you wrote, except I'm unclear about the last step. I am guessing you meant $g_\varepsilon$ in place of $g$? Could you please provide some insight as to how the inequality $\omega_\alpha(g_\varepsilon) \leq \int \phi(t)G_\varepsilon(t) dt$ holds? I ask because I thought I had a counterexample in mind in the boundary case of $\alpha = 1$, and it didn't seem like there was a dependence on the choice of $\alpha$, unless it appears in that particular inequality. Of course, I could be mistaken. –  Vince Jul 2 '10 at 5:23
    
I have included some more detail, but I have also realizaed that it is not clear at all that $\lim_{\varepsilon\to0}G_\varepsilon(t)=0$. –  Julián Aguirre Jul 3 '10 at 22:14
    
You want $\phi_{\varepsilon}(x)=\varepsilon^{-n}\phi(x/\varepsilon)$ to get a mollifier. –  Ben McKay Apr 6 '13 at 18:14
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