Recently I've been investigating results from approximation theory, especially the uniform approximation by polynomials. I find most of the interesting results are for one-dimensional, uni-variate case, i.e., using $\text{span}\:\{x^{\lambda_i} \:|\: \lambda_i \in S\}$ to approximate a function in $C(K, \mathbb{R})$, where $K \subset \mathbb{R}$ is an interval. Several extensions have been made for the one-dimensional, multivariate case, i.e., using $\text{span}\:\{\prod_{i=1}^n x_i^{\lambda_i} \:|\: \lambda_i \in S_i, i = 1, \ldots, n\}$ to approximate a function in $C(K, \mathbb{R})$, where $K \subset \mathbb{R}^n$.

I'm wondering whether there exists results for multi-dimensional, uni-variate case. The simplest question would be whether $$\text{span}\:\{ \begin{bmatrix}1\\ 0\end{bmatrix}, \begin{bmatrix}0\\ 1\end{bmatrix}, \begin{bmatrix}x^2\\ x\end{bmatrix}, \begin{bmatrix}-x\\ x^2\end{bmatrix}, \begin{bmatrix}x^4\\ x^3\end{bmatrix} , \begin{bmatrix}-x^3\\ x^4\end{bmatrix} , \ldots\}$$ is dense in $C([-1, 1], \mathbb{R}^2)$. Can anyone give me an idea or a reference?

  • $\begingroup$ What is $b$ in your question? Should it be $x?$ $\endgroup$ – zhw. Nov 22 '19 at 18:47
  • $\begingroup$ @zhw. It's just a dummy variable. Anyway it's better for me to change it into $x$. $\endgroup$ – mw19930312 Nov 22 '19 at 20:18
  • $\begingroup$ You need "span" in your first two examples. $\endgroup$ – zhw. Nov 22 '19 at 21:40
  • $\begingroup$ @zhw.Thanks! Already edited the OP. $\endgroup$ – mw19930312 Nov 25 '19 at 14:47

The Hahn-Banach theorem + Riesz representation theorem says that the span of $\begin{bmatrix}p_n\\ q_n\end{bmatrix}$ fails to be dense in $C([-1,1],\mathbb R^2)$ iff there are finite signed measures $\mu_1,\mu_2$ on $[-1,1],$ not both zero, such that $\int p_nd \mu_1=\int q_nd\mu_2=0$ for all $n.$ In this case there happens to be suitable measures supported on $\{-1,0,1\}.$

All the elements $\begin{bmatrix}p\\ q\end{bmatrix}$ of your basis satisfy $p(1)+q(-1)-p(0)-q(0)=0.$ So anything in their span is at sup-norm distance at least $\tfrac14$ from $\begin{bmatrix}x\\ 0\end{bmatrix}$ for example.

Here is more detail on why a continuous linear functional on $C([0,1],\mathbb R^2)$ can be written in the form $(p,q)\mapsto \int p\; d\mu_1 + \int q\; d\mu_2.$

Consider an arbitrary $h\in C([0,1],\mathbb R^2)^*.$ This takes as input continuous functions $f\in C([-1,1],\mathbb R^2),$ or equivalently, pairs $(p,q)\in C([-1,1]).$ The norm on $\mathbb R^2$ doesn't make a difference, but we can make $C([0,1],\mathbb R^2)$ a Banach space by setting $\|f\|=\max(\|p\|,\|q\|).$ Define $h_1,h_2\in C([0,1])^*$ by $h_1(p)=h((p,0))$ and $h_2(q)=h((0,q))$ - with the Banach space norm above, these are continuous because $\|h_1\|,\|h_2\|\leq \|h\|.$ By what Wikipedia calls the Riesz-Markov theorem, each continuous linear functional $h_i\in C([0,1])^*$ is represented by a finite signed measure measure $\mu_i$:

$$\forall p\in C([0,1]),\qquad h_i(g)=\int p\; d\mu_i.$$

So $h((p,q))=\int p\; d\mu_1 + \int q\; d\mu_2.$

  • $\begingroup$ Hi Dap, thanks for the answer here! Can you elaborate a bit more on how the iff condition comes from the combination of Hahn-Banach theorem and Riesz representation theorem? $\endgroup$ – mw19930312 Dec 2 '19 at 14:58
  • $\begingroup$ I should elaborate my confusion. I'm not so clear about the Riesz representation part. By Hahn-Banach theorem, the span is not dense iff there exists a linear functional $h\in (C([-1, 1], \mathbb{R}^2))*$ such that $h$ vanishes on the span. However $C([-1, 1], \mathbb{R}^2)$ is not a Hilbert space. I don't see any way to represent $h$ using element in $C([-1, 1], \mathbb{R}^2)$. $\endgroup$ – mw19930312 Dec 4 '19 at 21:02
  • $\begingroup$ @mw19930312: I've added some explanation of how to represent $h$ by measures $\endgroup$ – Dap Dec 5 '19 at 20:22
  • $\begingroup$ Thanks for the explanation. I thought you meant the Riesz representation theorem in Hilbert space. Just one more question, when are the signed measures $h_1, h_2$ absolutely continuous to Lebesgue measure? Otherwise it's very difficult for me to generalize the above analysis to other cases. $\endgroup$ – mw19930312 Dec 6 '19 at 15:31
  • $\begingroup$ @mw19930312: I don't know a condition that would let you reduce to the case of absolutely continuous measures. $\endgroup$ – Dap Dec 6 '19 at 19:39

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