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Let $X$ be a $\sigma$-compact metric space and let $\mathcal{A}\equiv\mathrm{span}\{p_\alpha\mid \alpha \in I\}$ ($I$ countable, $(p_\alpha)$ linearly independent) be a subalgebra of $C(X)$ that separates points and contains the constants. Let further $\varphi : X\rightarrow \mathbb{R}$ be a bounded continuous function on $X$.

Then there is an ascending sequence $(K_n)$ of compacts with $X=\bigcup_n K_n$, and Stone-Weierstrass implies the existence of $(f_n)\subset\mathcal{A}$ such that $\|\varphi - f_n\|_{\infty; K_n} \leq 1/n$ for each $n$. Write $f_n = \sum_{\alpha\in I} c_{\alpha, n}p_\alpha$ and $\|(c_{\alpha,n})\|:=\big(\sum_{\alpha\in I} c_{\alpha,n}^2\big)^{1/2}$ (coefficient norm).

Question: Can $(f_n)$ be chosen such that the sequence $(\mathfrak{c}_n)_n:=(\|(c_{\alpha,n})\|)_n$ is bounded?

fsp-b
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1 Answers1

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Let $X=[0,1]$ and take $\mathcal A$ to be the polynomials, where the $p_\alpha$ are just powers of $x$ with coefficient $1$. Let $f:[0,1]\to \Bbb R$ be continuous and suppose $\mathfrak{c}_n\in\ell^2(\Bbb N)$ is a sequence of coefficients so that $\mathfrak c_n \cdot p$ converges uniformly to $f$.

Suppose $\|\mathfrak{c}_n\|_2$ remains bounded by some constant $M$. You then have for $x\in[0,\frac12]$: $$\left|\sum_{\alpha \geq 2}c_{n,\alpha} x^\alpha\right|\leq x^2\sum_{\alpha\geq2}|c_{n,\alpha}|2^{-\alpha+2}\leq x^2 M \frac{\sqrt 3}4$$ The actual numbers don't matter, so absorb the $\frac{\sqrt3}4$ term into $M$. Let $a_0, a_1$ denote the limits of $c_{n,0}, c_{n,1}$. Then if $x\in [0,\frac12]$ $$|f(x)-a_0-a_1x|≤x^2M$$ (Apply the triangle inequality). Which implies that $f$ is differentiable at $x=0$. In particular if $f$ is not differentiable at $1$ then the coefficients of its approximations will not remain bounded in $\ell^2$ (in fact the above shows they will not be bounded in $\ell^\infty$).

s.harp
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