# Uniformly approximating a function and its derivative using polynomials

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.

• 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.) – LSpice Oct 7 at 18:58
• @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... – mw19930312 Oct 7 at 19:07
• 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 $\delta:=\varepsilon |b-a|$. – Algernon Oct 7 at 19:31
• @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 $S$. – mw19930312 Oct 7 at 19:37
• Still, what I said reduces the problem to the case where $f$ is itself a polynomial. – Algernon Oct 7 at 19:57

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

• To see the density: math.stackexchange.com/q/2733385/34287 – Steven Gubkin Oct 7 at 21:34
• @StevenGubkin, or just because it's an algebra that separates points! – LSpice Oct 7 at 22:14
• @LSpice Sure, sure just recording an easy reference for anyone who doesn't know this stuff. – Steven Gubkin Oct 7 at 22:19
• So you may create a separate question, if you wish many people ro see it. – Ilya Bogdanov Oct 8 at 5:19

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 $$0=n_0. For in this case, by the Müntz-Szász Theorem, the $$(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.

• 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 $n_1=1$ . – Pietro Majer Oct 7 at 21:57
• 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. – mw19930312 Oct 7 at 22:16
• 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 $C^0(K)$ – Pietro Majer Oct 7 at 22:36

[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 $$n\to\infty$$.

• 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) – Pietro Majer Oct 7 at 20:59
• You are right, I missed the fact that $S$ is supposed to be the span of $p_1,p_2,\ldots$. – Algernon Oct 7 at 21:12