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Let $m \in N$ be fixed and let $C_{2\pi}^m$ be a class of functions $f : R \rightarrow R$ of class $C^m$ and periodic with a period $2\pi$ with the following metric $$d(f,g)=\sum_{k=0}^m \sup_{\{x \in R \}}|f^{(k)}(x)-g^{(k)}(x)|$$ for $f,g \in C_{2\pi}^{m}$. Is the set of trigonometric polynomials dense in this metric space?

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I try for $m=1$ in the following way: 1. To take arbitrary $f \in C_{2\pi}^m$; 2. To take (by Weierstrass appr. thr) a sequence $(P_n)$ of trig polynomials s.t. $P_n(x)\rightrightarrows f'(x)$; 3. To put $Q_n(x)=f(0)+\int_0^{x} P_n(t)dt$ (for $n \in N, x\in R$). Then $Q_n(0)$ is convergent, consequently $Q_n(x) \rightrightarrows f(x)$ ; But, is $Q_n$ a trig polynomial? –  L.T May 17 '12 at 19:05
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As it stands, not necessarily: $P_n$ could have a constant term, and then $Q_n$ would have a linear term. But with a little more work, you can show that if the constant term of $P_n$ is $c_n$, then $P_n - c_n$ also converges uniformly to $f'$. (Hint: note that $\int_0^{2 \pi} (P_n(x) - f'(x)) \,dx \to 0$. What's the value of this integral?) Now if you instead take $Q_n(x) = f(0) + \int_0^x (P_n(t)-c_n)\,dx$, $Q_n$ will be a trig polynomial. –  Nate Eldredge May 17 '12 at 19:52

2 Answers 2

up vote 4 down vote accepted

Yes they are, because you can restrict to an interval $[0, 2 \pi]$ of function $f(0) = f(2 \pi)$, and use that they are dense in $C[0,2 \pi]$. Integration $k$-times and observing carefully what gets added will give you an affirmative answer. It is a continuous, surjective operator $C \mapsto C^k$.

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I have an another idea using Fourier series. Let $f \in C_{2\pi}^m$ and let $(S_n)_{n=0}^\infty$ be the sequence of partial sum of the Fourier series of $f$. Then it is known that $(S_n^{(k)})_{n=0}^\infty$, for each $k\in \{0,\ldots m\}$, (that is sequuences of $k$-th derivatives) is a sequence of the partial sum for $f^{(k)}$. By continuity and $2\pi$-periodicity of each $f^{(k)}$, for $k=0,\ldots,m$, and the Feier Theorem we have that for each $k\in \{0,\ldots ,m \}$:

$$\left(\frac{S_0(x)+S_1(x)+\ldots S_n(x)}{n+1}\right)^{(k)}=\frac{S_0^{(k)}(x)+S_1^{(k)}(x)+\ldots S_n^{(k)}(x)}{n+1} \rightrightarrows f^{(k)}(x) $$ as $n\rightarrow \infty$, for $x \in R$. Obviously $\left(\frac{S_0+S_1+\ldots S_n}{n+1}\right)_{n=0}^\infty$ is a sequence of trigonometric polynomials.

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