Let $\lambda:[0, 1]\to \mathbb{R}$, and $b_{1j}, b_{2j}:[0, 1] \to \mathbb{R}$, $j = 1, \ldots, m$ be smooth functions. Consider the following two sets

$$\begin{align*} S_1 &= \left\{ \begin{bmatrix} P(\lambda)b_{1j} \\ P(\lambda)b_{2j} \end{bmatrix} \bigg\vert \:j = 1, \ldots, m, P(\lambda) \text{ is a polynomial of } \lambda\right\}\\ S_2 &= \left\{ \begin{bmatrix} P(\lambda)b_{1j} + P'(\lambda)b_{2j} \\ P(\lambda)b_{2j} \end{bmatrix} \bigg\vert \:j = 1, \ldots, m, P(\lambda) \text{ is a polynomial of } \lambda\right\}. \end{align*}$$

I'm stuck proving or disproving the following statement:

If $S_2$ is dense in $C([0, 1], \mathbb{R}^2)$ with respect to sup-norm, then $S_1$ is uniformly dense in $C([0, 1], \mathbb{R}^2)$ as well.

The major issue was that there is no guarantee that we can use $\{P_n\} \subset C^1([0, 1])$ to uniformly approximate $f\in C^1([0, 1])$ with $f'\in C([0, 1])$ be uniformly approximated by $\{P'_n\}\subset C([0, 1])$. I tend to believe the answer to the above statement is negative. But I cannot figure out a counter-example to it. Or is there any way to provide a positive answer to the statement?

  • $\begingroup$ I don't understand. what is meant by "$P(\lambda)$ is a polynomial"? Is $P$ (without $\lambda$) a polynomial? I'm irritated since $\lambda$ is a function, not a parameter. $\endgroup$ Nov 8, 2020 at 22:34
  • $\begingroup$ @DieterKadelka Sorry I should have been clearer. I’ve edited the OP. I meant that $P$, as a function, can be written as a polynomial of $\lambda$. For example, if we have $\lambda(x)=\cos (x)$, then $P(x)= \cos^2(x)=\lambda^2$ is a polynomial of $\lambda$. $\endgroup$ Nov 8, 2020 at 22:52

1 Answer 1


$\newcommand\la\lambda\newcommand\R{\mathbb R}$It is true that, if $S_2$ is dense in $C([0,1],\R^2)$ with respect to the sup-norm $\|\cdot\|$, then $S_1$ is uniformly dense in $C([0,1],\R^2)$ as well.

However, the reason for this may disappoint you: it is that $S_2$ is actually never dense in $C([0,1],\R^2)$ with respect to $\|\cdot\|$ -- for any continuous functions $\la$ and $b_{1j},b_{2j}$ with $j\in[m]:=\{1,\dots,m\}$.

Indeed, there are two cases to consider:

Case 1: $\la(t)=c$ for some real $c$ and all $t$. Then $S_2$ is finite dimensional and therefore cannot be dense in $C([0,1],\R^2)$.

*Case 2: there are $t_1$ and $t_2$ such that $0\le t_1<t_2\le1$ and $\la(t_1)=:u\ne v:=\la(t_2)$. Let $x_2(t):=1$ for $t\in[0,1]$. Take any real $$w>\max_{j\in[m]}\Big(2\|b_{1j}\|+\frac4{|v-u|}\,\|b_{2j}\|\Big) \tag{0}$$ and let $x_1(t):=w$ for $t\in[0,1]$.

Suppose the contrary to our claim: that $S_2$ is dense in $C([0,1],\R^2)$. Then for each natural $n$ there exist a polynomial $P_n$ and some $j_n\in[m]$ such that $$P_n(\la)b_{1j_n}+P'_n(\la)b_{2j_n}\to w,$$ $$P_n(\la)b_{2j_n}\to1,$$ where the convergence is uniform. Passing to a subsequence, without loss of generality (wlog) we may assume that $j_n$ does not depend on $n$, and then wlog $j_n=1$ for all $n$. For brevity, let then $b_1:=b_{11}$ and $b_2:=b_{21}$. So, $$P_n(\la)b_1+P'_n(\la)b_2\to w,\tag{1}$$ $$P_n(\la)b_2\to1.\tag{2}$$ Since the convergence is uniform, it follows from (2) that the function $b_2$ does not take the zero value on $[0,1]$. Therefore and because $b_2$ is continuous, wlog $|b_2|\ge1$, whence, again by (2), $$\|P_n(\la)\|\le2 \tag{4}$$ eventually (for all large enough $n$). So, by the mean value theorem and the intermediate value theorem, eventually $$4\ge|P_n(\la(t_2))-P_n(\la(t_1))| \\ =|P'_n(\la(s_n))|\,|\la(t_2)-\la(t_1)| \\ =|P'_n(\la(s_n))|\,|v-u| \tag{5}$$ for some $s_n\in(t_1,t_2)$, whence, in view of (4), (5), and (0), $$P_n(\la(s_n))b_1(s_n)+P'_n(\la(s_n))b_2(s_n) \le2\|b_1\|+\frac4{|v-u|}\,\|b_2\|< w,$$ which contradicts the uniform convergence in (1). $\Box$

  • 1
    $\begingroup$ @potionowner : the denseness of $S_2$ implies the denseness of $S_1$ because, whenever $S_2$ is dense (which never happens), $S_1$ is dense. Generally, implication $A\implies B$ holds (by definition) iff $A$ is false or $B$ is true. In particular, $A\implies B$ holds whenever $A$ is false (no matter whether $B$ is true or not). In our case, $A$ is the statement "$S_2$ is dense", which is always false. So, the statement "$S_2$ is dense" implies any statement whatsoever. $\endgroup$ Nov 10, 2020 at 3:56
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    $\begingroup$ @potionowner : Your original question was fully answered. If you actually meant to ask something else, you should now assume the responsibility for your original post (as for any other posts of yours) and post any other/further questions in separate posts. You should not change your question so as to invalidate a valid answer and nullify other people's time and effort (to help you, actually). $\endgroup$ Nov 10, 2020 at 4:05
  • $\begingroup$ Sure. I'll change the OP back and delete the comment. Thanks. $\endgroup$ Nov 11, 2020 at 2:50

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