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Do we know any function spaces dense in Zygmund space $C_*^s$(a special case of Besov space, i.e. $C_*^s = B^s_{\infty,\infty}$) or Hölder space$C^{k,r}$, with underlying field $\mathbb{R}^d$?

Will $C^\infty$ do the job?

newbie
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1 Answers1

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If a space is not separable, you should not expect it to have any nice dense subspaces, such as $C^\infty$. Indeed, $C^\infty$ is separable with respect to any Zygmund- or Hölder-norm; a countable dense subset can be constructed from a countable partition of unity multiplied by polynomials with rational coefficients.

For example, there is no (norm) dense subspace of $L^\infty$ that is worth mentioning.

None of the spaces you mention are separable. For example, the functions $\max(x-s,0)^{k+\alpha}$ for $s\in\mathbb R$ form an uncountable set in which any two points are at distance at least $1$ in the $C^{k,\alpha}(\mathbb R)$ norm. Same example (with $\alpha=1$) works for Zygmund spaces.

For $0<\alpha<1$, the closure of $C^\infty$ functions in the Hölder space $C^{k,\alpha}$ is known as the little Hölder space $c^{k,\alpha}$. Its elements are described by the $k$th derivative having the modulus of continuity $\omega(\delta)$ such that $$\lim_{\delta\to 0}\frac{\omega(\delta)}{\delta^\alpha}= 0 \tag1$$ Indeed, (1) holds for $C^\infty$ functions and is preserved under convergence in the norm.

When $\alpha=1$, the closure of $C^\infty$ functions in $C^{k,1}$ is simply $C^{k+1}$. The reason is that convergence of $C^\infty$ functions in $C^{k,1}$ norm implies uniform convergence of their derivatives of order $k+1$. Thus, the continuity of $(k+1)$th derivative is preserved.

The situation in the Zygmund spaces is similar to what happens what $\alpha$ is fractional: the closure of $C^\infty$ is the space of functions whose second modulus of continuity $\omega_2$ satisfies a vanishing condition instead of the boundedness condition. This is also called a little Zygmund space. For example, functions in $c^1_*$ satisfy $$\lim_{\delta\to 0}\sup_x \sup_{|h|<\delta}\left| \frac{f(x+h)-2f(x)+f(x-h)}{h} \right|=0 \tag2$$

Zygmund himself observed the difference between the big and little spaces on the first page of his paper Smooth functions which you can read without subscription.

ˈjuː.zɚ79365
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  • So in conclusion, for these two spaces we don't have nice dense subsets. Do we have the following inclusion $C^{\lfloor s \rfloor, s-\lfloor s \rfloor} \subset c_*^s$? And what's the relation between small Hölder space and small Zygmund space? – newbie Jun 30 '13 at 10:17
  • @newbie I have no experience with fractional order Zygmund spaces, but I think the answer is no. The function $\max(x,0)^{1/2}$ is in $C^{0,1/2}$, and I think it's not in $c_*^{1/2}$. – ˈjuː.zɚ79365 Jun 30 '13 at 10:21
  • Thanks for the quick reply. My question orientates from this paper. On the top of page 8, they only show the case for smooth functions and then generalize by approximation(which i guess should be density argument). The $C^\alpha$ in the paper is defined as $B_{\infty,\infty}^\alpha$, which should be Zygmund space. But I think if it works for Hölder continuous is already good enough. So, I am wondering if there is a flaw in their proof. – newbie Jun 30 '13 at 10:28
  • @newbie For fractional $\alpha$ the space $B^\alpha_{\infty,\infty}$ is $C^{\lfloor \alpha \rfloor, \alpha-\lfloor \alpha \rfloor}$, as they say on page 4. Since smooth functions are not dense in this space in the norm topology, I do not know what kind of approximation they had in mind. – ˈjuː.zɚ79365 Jun 30 '13 at 10:42