Let $C^{m,\alpha}_M([0,1])$ be a Holder ball consisting of real-valued functions $g$ on $[0,1]$ such that $$ \|g\|_{C^{m,\alpha}} := \max_{0\leq j \leq m } \sup_{x\in [0,1]} |g^{(j)}(x)| + \sup_{x,y\in [0,1], x\neq y} \frac{ |g^{(m)}(x) -g^{(m)}(y)|}{|x-y|^\alpha} \leq M. $$ Let $f \in C^{m,\alpha}_M([0,1])$ be fixed.

By Weierstrass approximation theorem, there exists a sequence of polynomials $$p_n(x) = \sum_{j=0}^{J_n} \alpha_{n,j} x^j$$ such that $\|f-p_n\|_{\infty} \to 0$ as $n \to \infty$, where $\|\cdot\|_\infty$ denotes the uniform norm. Here $J_n \to \infty$.


Does there exist approximating polynomials $p_n \in C^{m,\alpha}_{M'}([0,1])$ for some $M'>0$, such that $\|f-p_n\|_{\infty} \to 0$? If this is not true in general, for what kind of functions $f$ it may hold?

Motivation of the question:

I am reading some papers in nonparametric statistics, Chen (2007) and Chen and Ai (2003). They rely on approximating an unknown function in some space, i.e. $f\in C_M^{m,\alpha}$ here, using functions in some approximating spaces increasing with $n$, i.e. space of $J_n$-th order polynomials here.

To have desirable statistical properties, a key assumption they used is that, the approximating spaces are subsets of the original space $C^{m,\alpha}_M([0,1])$. Their practice corresponds to the question. I am not sure whether adding such further restriction on the approximating spaces will cause serious issues on the approximating ability. This issue is not discussed in the listed papers, nor in other related papers I have checked.

Any reference or discussions is greatly appreciated!

  • 1
    $\begingroup$ Asking multiple questions in one post is not encouraged on MathOverflow. Also, this strategy is not beneficial for askers, in most cases. $\endgroup$ Jul 14, 2022 at 17:13
  • $\begingroup$ @IosifPinelis Sure! I have revised to keep the most essential question. $\endgroup$
    – masala
    Jul 14, 2022 at 17:28
  • $\begingroup$ This is still two questions. Also, $\alpha$ is missing in the definition of $\|g\|_{C^{m,\alpha}}$. Also, you probably want to relax the condition $p_n \in C^{m,\alpha}_M([0,1])$ to something like $p_n \in C^{m,\alpha}_{2M}([0,1])$. $\endgroup$ Jul 14, 2022 at 17:57
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    $\begingroup$ Do you mean trying to approximate in Holder norm? The answer is no. Even $C^\infty$ is not dense in $C^{k,\alpha}$. See e.g. mathoverflow.net/questions/29869/… But you can get approximation in slightly worse space (replace $\alpha$ by $\beta\in (0,\alpha)$. (At least by smooth functions, I think (but may be wrong) should also work for polynomials. See mathoverflow.net/questions/244377/… ) $\endgroup$ Jul 14, 2022 at 17:57
  • $\begingroup$ Reading @IosifPinelis's comment, I suppose another possibility is that the unspecified norm in your Question is the uniform norm, and you are interested in whether you can require the approximation sequence be drawn from the same Holder ball. Is that the correct interpretation? $\endgroup$ Jul 14, 2022 at 18:03

1 Answer 1


The revised question, where you just want uniformly bounded Holder norm on the approximating polynomials, the answer is yes.

In fact, you have more.

Theorem 1 If $f\in C^{m,\alpha}_M([0,1])$ is such that the function $\tilde{f}(x) = \begin{cases} f(x) & x\in [0,1]\\ 0 & x\not\in [0,1]\end{cases}$ is in $C^{m,\alpha}(\mathbb{R})$, then there exists a polynomial approximation sequence in $C^{m,\alpha}_M([0,1])$.

This theorem is based on a proof of the Stone-Weierstrass Theorem using a polynomial approximation to identity. See e.g. Theorem 11.7.1 of Lebl's textbook (volume 2).

A nice property of the convolution argument is that:

If $\phi$ is a non-negative function with total integral 1, then $\|\phi*f\|_{\infty} \leq \|f\|_\infty$.

Since $(\phi*g)^{(j)} = \phi* (g^{(j)})$, for any function $g$, applying this to $\tilde{f}$ you see that the convolution by the polynomial approximation to identity means that the polynomials constructed have $C^m$ norms bounded by that of $\tilde{f}$.

Similarly, letting $g = \tilde{f}^{(m)}$, observe for each fixed $t\neq 0$, $x\mapsto \frac{1}{|t|^\alpha}({g}(x+t) - {g}(x))$ is a $C^{0,\alpha}$ function, we have that

$$ \frac{1}{|t|^\alpha} \| \phi*{g}(x+t) - \phi*{g}(x)\|_{L^\infty_x} \leq \frac{1}{|t|^\alpha} \|{g}(x+t) - {g}(x)\|_{L^\infty_x} $$

Taking the sup of both sides over $t\neq 0$ this shows that the Holder semi-norm of $\phi*\tilde{f}$ is bounded by that of $\tilde{f}$. And this proves the Theorem.

Now, given $f\in C^{m,\alpha}_M([0,1])$, there exists a polynomial $q$ such that $f-q$ satisfies the hypotheses of Theorem 1. The polynomial $q$ is determined entirely by $f(0), f'(0), \ldots, f^{(m)}(0), f(1), f'(1), \ldots, f^{(m)}(1)$.) You can select $q$ to be of degree $2m+1$, in which case the $q$ is uniquely determined by those values. The relation between those values and the coefficients of $q$ are given by a linear operator, depending only on $m$. This means that there exists $\tilde{M}$ that uniformly bounds the coefficients of $q$, where $\tilde{M}$ depends only on $M$ and on $m$ (since by assumption $f(0), \ldots f^{(m)}(1)$ are all bounded by $M$). (Note that in the context of Theorem 1, this unique $q$ is the 0 polynomial, and hence $\tilde{M} = 0$.) The mapping from $\mathbb{R}^{2m+2}\times \mathbb{R} \to \mathbb{R}$ where the first factor represents the coefficients of $q$ and the last factor represents $x$ mapping $(q,x)\mapsto q(x)$ is continuous (in fact a polynomial), and hence achieves a maximum on compact regions. Therefore knowing that the coefficients are bounded by $\tilde{M}$ and $x\in [0,1]$ means that there exists $\hat{M}$ depending only on $M$ and $m$ that bounds the $C^{m+1}$ and hence $C^{m,\alpha}$ norm of $q$. Finally set $M' = M + 2 \hat{M}$.

So you conclude:

Theorem 2 Given $m, \alpha,M$, there exists $M' = M'(m,M,\alpha)$ such that whenever $f\in C^{m,\alpha}_M([0,1])$, you can approximate $f$ by polynomials $p_j$ in the uniform norm, with $p_j$ selected from $C^{m,\alpha}_{M'}([0,1])$.

  • $\begingroup$ Thank you so much!!! You saved my day! $\endgroup$
    – masala
    Jul 14, 2022 at 22:00

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