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Consider the functions $\{e^{2\pi i nx}\}_{n \in \mathbb{Z}}$ defined on the interval $[0,1].$ These are all smooth periodic functions (so functions on $S^1)$ and by the Stone-Weierstrass theorem they are dense in $C^0(S^1)$ when it is given the max norm. Consider the norm on $C^k(S^1)$ given by $$|f|^{C^k}=\sum_{i=0}^k \text{max}_{x \in S^1}|f^{(k)}(x)|.$$

$C^k(S^1)$ is complete in this norm, but is the collection $\{e^{2\pi i nx}\}_{n \in \mathbb{Z}}$ a dense subset?

Bruno B
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mck
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  • Do you mean the linear subspace spanned by ${e^{2\pi i x}}_{n\in\mathbb{Z}}$ is dense? – 23rd May 04 '13 at 16:24

2 Answers2

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As Landscape noted, the question should be about the linear span of $\{e^{2\pi i n x}\}$. The answer is yes, it is dense. Given $f\in C^k(S^1)$, approximate its derivative $f^{(k)}$ to within $\epsilon$ by some trigonometric polynomial $p_k(x)=\sum_n c_n e^{2\pi i n x}$. Then consider the antiderivative $p_{k-1}(x) =b+ \sum_{n\ne 0} \frac{c_n}{2\pi i n} e^{2\pi i n x}$ where $b$ is chosen so that $p_{k-1}(0)=f^{(k-1)}(0)$. By the fundamental theorem of calculus, the difference $|p_{k-1}-f^{(k-1)}|$ does not exceed $\int_0^1 \epsilon \,dx=\epsilon$. Therefore, $p_{k-1}$ is within $\epsilon$ of $f^{(k-1)}$. Repeat the process until you reach $p_0$, an $\epsilon$-approximation for $f$. Since $(p_0)^{(j)}=p_j$ for $j=1,\dots,k$, it follows that $\|p_0-f\|_{ C^k}\le (k+1)\epsilon$.

75064
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    When you take the antiderivative of $p_{k-1}$, wouldn’t you get some linear term $bx$? I fail to see how $p_0$ is a trigonometric polynomial. At first glance it seems to be a polynomial + trigonometric polynomial. What am I overlooking? – Doofenshmert Apr 20 '24 at 14:19
  • @Doofenshmert Could you please look at my answer to see if your concern was correctly addressed, or if I missed something. – Bruno B May 05 '24 at 23:56
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I would like to address a legitimate concern from Doofenshmert, who commented on 75064's answer (however please excuse my choice of $k$ for the Fourier coefficients instead of the $k$ of the question for the regularity, I only realised mid-writing. I'm using $p$ for that instead).
As noticed by Doofenshmert, he issue with simply taking antiderivatives of a "random" approximation of $f$ by a trigonometric polynomial is the apparition of a non-trigonometric polynomial term, which should have appeared at the first antiderivative but will inevitably arrive on the second one if $b \neq 0$ in 75064's construction.
75064's account hasn't been logged in since 10 years ago though, looking at their profile page, hence my decision to write an answer instead of just commenting.

I think a good way to correct this is to refine the selction of said approximation.
Let $c_k(f)$ be the $k$-th Fourier coefficient of $f$. Then we know (by successive integrations by parts) that $c_k(f') = ik c_k(f)$, so that: $$\left(c_k(f) e^{ikx}\right)'(x_0) = c_k(f) ik e^{ikx_0} = c_k(f') e^{ikx_0}$$ Moreover, by Fejér's theorem, we know that, if $f$ is continuous, $(\sigma_N(f))_{N \geq 0}$ converges uniformly to $f$ as $N \to \infty$, where $\sigma_N(f)$ is defined by, for $N \in \mathbb{N}$: $$\sigma_N(f) : x \mapsto \frac{1}{N+1}\sum_{n = 0}^N \sum_{k = -n}^{n} c_k(f) e^{ikx}$$ We can rewrite $\sigma_N(f)$ by swapping the indices, although it's not really required for this answer: $$\sigma_N(f)(x) = \frac{1}{N+1}\sum_{k = -N}^N \sum_{n = |k|}^{N} c_k(f) e^{ikx} = \sum_{k = -N}^N \frac{N - |k| + 1}{N + 1} c_k(f) e^{ikx}$$ Thus, thanks to our previous observation, we can see that: $$\sigma_N(f') = (\sigma_N(f))'$$ which means that not only does $\sigma_N(f)$ converge uniformly to $f$, but all its derivatives also do to the corresponding derivatives of $f$ (by simple induction), which means that $(\sigma_N(f))_N$ converges to $f$ in $\mathcal{C}^p(\mathbb{S}^1)$ if $f \in \mathcal{C}^p(\mathbb{S}^1)$.

Bruno B
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