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I'm struggling either proving or disproving the following statement:

Let $K\subset \mathbb{R}$ be compact, and $S = \mathrm{span}\{p_k, k = 0, 1, \ldots\}$, where $p_k$'s are polynomials over $K$. If $S$ is dense in $C^1(K)$ with respect to sup-norm, then for any $f\in C^1(K)$, there exists $\{g_n\} \subset S$ such that $g_n$ uniformly approximates $f$ with $f'$ uniformly approximated by $g_n'$.

This statement seems to be true, although $S$ is only a subset of all polynomials over $K$. If the statement appears false, is there any prior assumption that makes it true? Any reference or direct answer would greatly help me.

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    $\begingroup$ Is $C(K)$ supposed to be $\mathrm C^1(K)$? Otherwise, it's not even clear what $f'$ means. (Also, where does this come from? It looks like a homework question.) $\endgroup$
    – LSpice
    Commented Oct 7, 2020 at 18:58
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    $\begingroup$ @LSpice Yes. It should be $C^1(K)$. I've changed the original post. It's actually an intermediate step in a theorem that I would like to prove in my research work. I tried to look up real-analysis references. But I didn't find anything useful... $\endgroup$
    – mw19930312
    Commented Oct 7, 2020 at 19:07
  • $\begingroup$ Let's start with the case in which $K$ is an interval $[a,b]$. If $h$ is a polynomial that $\varepsilon$-approximates $f'$, then $H(x):=f(a)+\int_a^x h(z)\,\mathrm{d}z$ is a polynomial that $\delta$-approximates $f$, where $\delta:=\varepsilon |b-a|$. $\endgroup$
    – Algernon
    Commented Oct 7, 2020 at 19:31
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    $\begingroup$ @Algernon Thanks for the answer here. I'm more interested in the case where $S$ is a subset of all polynomials. The tricky part in your comment is that $\int_a^x h(z)dz$ does not necessarily lie in $S$. $\endgroup$
    – mw19930312
    Commented Oct 7, 2020 at 19:37
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    $\begingroup$ Still, what I said reduces the problem to the case where $f$ is itself a polynomial. $\endgroup$
    – Algernon
    Commented Oct 7, 2020 at 19:57

3 Answers 3

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The span of $\{x^{2k}\colon k=0,1,2,\dots\}$ is dense in $C([0,1])$. But all their derivatives vanish at $0$.

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    $\begingroup$ To see the density: math.stackexchange.com/q/2733385/34287 $\endgroup$ Commented Oct 7, 2020 at 21:34
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    $\begingroup$ @StevenGubkin, or just because it's an algebra that separates points! $\endgroup$
    – LSpice
    Commented Oct 7, 2020 at 22:14
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    $\begingroup$ @LSpice Sure, sure just recording an easy reference for anyone who doesn't know this stuff. $\endgroup$ Commented Oct 7, 2020 at 22:19
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    $\begingroup$ So you may create a separate question, if you wish many people ro see it. $\endgroup$ Commented Oct 8, 2020 at 5:19
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Any prior assumption that makes it true. A simple case where your statement is true is the case of an interval $K\subset\mathbb{R}_+ $, and $(p_k)_{k\ge0}$ are monomials, $p_k(x)=x^{n_k}$, of degrees $0=n_0<n_1<\dots$. For in this case, by the Müntz-Szász Theorem, the $(p_k)_{k\ge0}$ span a dense subspace in $C^0(I)$, if and only if their derivatives $(p'_k)_{k\ge0}$ do (this of course because $\sum_{k=1}^\infty\frac{1}{n_k}=+\infty$ if and only if $\sum_{k=2}^\infty\frac{1}{n_k-1}=+\infty$). If by the assumption $S$ is uniformly dense in $C^1(I)$, it is also uniformly dense in $C^0(I)$, therefore the span of the $(p'_k)_{k\ge0}$ is also uniformly dense in $C^0(I)$. Hence, for any $f\in C^1(K)$ there is a sequence $g_n\in S$ such that $g_n'$ converges uniformly to $f'$; since $S$ contains the constants, we can assume $g_n$ converges to $g$ at least on a pont of $K$, but then also $g_n$ converges uniformly to $f $ by the theorem of limit under the sign of derivative.

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    $\begingroup$ Note I assumed $0\notin K$, as it is needed to apply the Muntz-Szász Theorem. If $0\in K$, the trivial exceptions can be ruled out e.g. assuming $ n_0=0$ and $n_1=1 $ . $\endgroup$ Commented Oct 7, 2020 at 21:57
  • $\begingroup$ Thanks for the reply! I know of M\"{u}ntz-Szasz theorem. May I ask if there are any extensions on the cases where $p_k$'s are not monomials? The cases that I'm mostly interested in are (1) $p_k = a^k b$ for some fixed polynomials $a$ and $b$ (2) $p_k$'s are vector-valued polynomials. But these cases seem to be more complicated and less-addressed in the literature of polynomial approximation. $\endgroup$
    – mw19930312
    Commented Oct 7, 2020 at 22:16
  • $\begingroup$ I don't know of the extensions to polynomials, but I guess there are some. An obvious extension is the case of polynomials $p_k$ that maybe are not monomials but whose span coincides with a span of monomials $x^{n_k}$ satisfying MS hypothesis. Or also, polynomials $p_k=L(x^{n_k})$ for $L$ an invertible operator on $C^0(K)$ $\endgroup$ Commented Oct 7, 2020 at 22:36
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[Update: As Pietro Majer pointed out in the comments, the following does not answer the OP's question, but a variant in which $S=\{p_1,p_2,\ldots\}$ instead of $S=\operatorname{span}(\{p_1,p_2,\ldots\})$.]

Not necessarily.

For simplicity, let $K=[0,1]$. Let $T_0=\{u_1,u_2,\ldots\}$ be a countable dense set in $C(K)$ whose elements are differentiable, say the set of polynomials with rational coefficients. For each $n$, let \begin{align*} v_n(x) &:= u_n + \frac{1}{n}\sin(\beta(n)x) \;. \end{align*} for some function $\beta:\mathbb{N}\to\mathbb{R}$ that grows rapidly to $\infty$. Note that

  • $T_1:=\{v_1,v_2,\ldots\}$ is still a countable dense set in $C(K)$;
  • Each $v_n$ is differentiable;
  • Assuming $\beta$ grows fast enough, $\|v'_n\|\to\infty$ as $n\to\infty$.

Finally, for each $n$, let $p_n$ be a polynomial such that $\|p_n-v_n\|<2^{-n}$ and $\|p'_n-v'_n\|<2^{-n}$. Let $S:=\{p_1,p_2,\ldots\}$.

Now, clearly $S$ is dense in $C(K)$, but $S':=\{p'_1,p'_2,\ldots\}$ cannot be dense in $C(K)$ because $\|p'_n\|\to\infty$ as $n\to\infty$.

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    $\begingroup$ Note that $S$ is the span of the $p_k$, so there are many other converging sequences $g_n$ in $S$ (for instance $S$ could be the whole ring $\mathbb{R}[x]$, in which case the OP statement is true), so it seems more complicated (I also think it's not true, for similar reasons) $\endgroup$ Commented Oct 7, 2020 at 20:59
  • $\begingroup$ You are right, I missed the fact that $S$ is supposed to be the span of $p_1,p_2,\ldots$. $\endgroup$
    – Algernon
    Commented Oct 7, 2020 at 21:12

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