uniform approximation of continuous odd periodic function of period $1$ by $\sum_n a_n f(nx)$ where $f(x)=x-[x]-1/2$

Question

Let $\tilde{f}:[0,1]\to\mathbb{R}$ be defined as $f(x)=x-1/2$ for $x\in(0,1)$ and $f(x)=0$ for $x=0$ or $x=1$ and let $f$ be the $1$-periodic extension of $\tilde{f}$ to $\mathbb{R}$.

I managed to show that every continuous odd periodic function of period $1$ can be arbitrary well uniformly approximated by a finite linear combination of the form $\sum_n a_n f(nx)$ for some $a_n\in\mathbb{R}$ but the proof is rather involved (in that it uses highly non trivial results, the proof itself isn't involved given the highly non trivial result). I'm looking for a short/easy proof. Maybe there is some variant of Weierstrass's Approximation Theorem which handles such functions?

(That we for example can find such a linear combination to arbitrary well approximate any continuous odd periodic function of period $1$ in the $L^2$ norm for example I could show much easier without resorting to anything highly non trivial (of course in this case the $f(x)=0$ for $x=[x]$ part is irrelevant).)

Answer 1

We will give two proofs of the result, one elementary (edit later - incomplete) and one using a non-trivial result of Davenport which immediately implies the result. We use the standard notation $\psi(x)=x-[x]-1/2$

First we have Davenport famous result: $$\sum_{n \ge 1}\frac{\mu(n)}{n}\psi(nx)=-\frac{1}{\pi}\sin 2\pi x$$ with uniform convergence in $x$ (and $\mu$ the Mobius function as usual) which immediately implies that each $\sin 2\pi mx$ is in the uniform closure of $\sum_{k \in I} a_k\psi(kx), I$ finite hence by Feijer's theorem on uniform approximation of continuous periodic functions by their averages of the Fourier partial sums, we immediately get the required result, since odd periodic continous functions on $[0,1]$ have only sine Fourier series (where $\cos 2\pi nx, \sin 2\pi nx$ is the orthogonal basis here).

Note that this is a difficult result and the occasionally cited "simple" proof claimed by Segal is flawed, as was noted by various authors (uniform and absolute convergence of $g(s)=\sum a_n n^{-s}, \Re s =c$ is not enough to imply that we can switch sums and integrals in $\int_{(c)}g(s)f(s)ds$ when $\int_{(c)} f(s)\,ds$ converges only conditionally but not absolutely and $(c)$ is the vertical line passing through $c$; to rigurously prove the switch, one needs to analyze the effect of the remainder $R_N(s)=g(s)-\sum_{n \le N}a_n n^{-s})$ on the integral and in particular on the part from $(c+iT, c+i\infty), T \to \infty$ and precisely that type of analysis is quite difficult here)

Edit later - this is incomplete as the lifting lemma is flawed.

However, an elementary proof can be given by directly showing that $\psi(nx)$ generate the same $L^2$ space as $\sin 2\pi nx$ and then using the standard lifting trick mentioned in the answer by @Jack. Assume there is $f(x)=\sum c_n\sin 2\pi nx, \sum |c_n|^2 < \infty$ orthogonal on the space generated by $\psi(mx)$ in $L^2(0,1)$ and note that since the Fourier series of $\psi$ is boundedly convergent (and to $\psi$ except at integers) we have $$0=-\pi\int_0^1 f(x)\psi(mx)\,dx=-\pi\sum_{k \ge 1} \int_0^1 f(x)\frac{\sin(2\pi mkx)}{mk}\,dx$$ hence $$\sum_{k \ge 1}\frac{c_{km}}{km}=0$$ for all $m=1,2,\ldots$

But now let's consider the double series $$\sum_{k,m}\frac{\mu(m)c_{km}}{km}$$ and note that it is majorized by $$\sum_{k,m}\left|\frac{\mu(m)c_{km}}{km}\right| =\sum_{n}\frac{d(n)c_n}{n}$$ where as usual $d(n)$ is the number of divisors of $n$ and by Cauchy-Schwarz $$\left(\sum_{n}\frac{d(n)c_n}{n}\right)^2 \le \left(\sum |c_n|^2\right)\left(\sum \frac{d^2(n)}{n^2}\right) < \infty$$ since $d(n) \ll n^{\epsilon}$ for any $\epsilon >0$.

This means that the double series is absolutely convergent and we can switch the order as we wish (so in particular summing on $k$ first gives that the double series sums to zero) and if we fix $n=km$ and we sum on $n$, the numerator of each term is given by $c_n\sum_{m|n}\mu(m)=0$ unless $n=1$ hence $c_1=0$.

Similarly, for any $q >1$ we can use the subseries $\sum_{k \ge 1}\frac{{c_{qmk}}}{mk}=0, m=1,\ldots$ and get that $c_q=0$ by the Mobius trick, so $f=0$ and $\psi(mx)$ generates the same subspace as $\sin(2\pi nx)$.

Here I would note that Davenport's result in the $L^2$ sense (series converge a.e. and in $L^2$) is actually elementary (though requires some estimate work) and that of course shows the above, but a direct proof without using estimates is quite simple and I think instructive.

Note that using Landau's relations $$\int_0^1\psi(mx)\psi(nx)\,dx=\frac{1}{12}\frac{(m,n)^2}{mn}$$ it is easy to show that $\psi(mx)$ are linearly independent so form a basis of the odd part of $L^2(0,1)$ (in our sense of functions that satisfy $f(1-x)=-f(x)$) but a deep Theorem of Hedelnamn, Lindquist, Seip shows that they do no form a Riesz basis as the operator $\sum a_n \sin(2\pi n x) \to \sum a_n \psi(nx)$ is not bounded (nor its inverse is for that matter) since it corresponds to multiplication by $\zeta(s+1)$ and that is an unbounded function on $\Re s >0$ as is $1/\zeta(s+1)$ which is the multiplier corresponding to the inverse operator $\sum a_n \psi(nx) \to \sum a_n \sin (2\pi nx).$

Answer 2

Given $$\tilde{f}(x)=-\sum_{n\geq 1}\frac{\sin(2\pi n)}{\pi n}=\{x\}-\frac{1}{2}$$ let us define $v_n(x)=f(nx)$ and

$$ w_n(x) = 2 v_{2n}(x) - 4 v_{n}(x).\tag{1} $$

We have that $w_n$ is a square wave with period $\frac{1}{2n}$ oscillating between $1$ and $-1$.

These functions fulfill the almost-orthogonality relation

$$ \langle w_m, w_n \rangle = \int_{0}^{1}w_m(x)w_n(x)\,dx =\left\{\begin{array}{rcl}\frac{\gcd(m,n)^2}{mn}&\text{if}&\frac{m}{\gcd(m,n)}-\frac{n}{\gcd(m,n)}\equiv 0\pmod{2}\\0&\text{if}&\frac{m}{\gcd(m,n)}-\frac{n}{\gcd(m,n)}\equiv 1\pmod{2}\end{array}\right.\tag{2}$$

so the Gram-Schidt process applied to $w_1,w_2,w_3,\ldots$ produces a complete orthonormal base for $L^2(0,1)\cap\{f:f(x)=-f(1-x)\}$.

In order to prove it is enough to compute the squared distance of $g(x)=\sin(2\pi x)$ from the span of $w_1,\ldots,w_N$ through ratios of determinants of Gram matrices, then show that such distance converges to $0$ as $N\to +\infty$. This should not be too hard since $G(w_1,\ldots,w_N)$ is just a small perturbation of the identity matrix. If needed we may replace $g(x)$ with the shifted Legendre polynomial $P_{2M+1}(2x-1)$.

We need an upper bound for $\left|\det G(g,v_1,\ldots,v_N)\right|$ and a lower bound for $\left|\det G(v_1,\ldots,v_N)\right|$. The former can be obtained by considering the product of the $L^2$ norms of the columns, the latter from Gershgorin circle theorem.

Althought this approach requires some pretty heavy computations. We may also consider that

$$\widehat{g}(s)=(\mathcal{L}g)(s)=\frac{2\pi}{4\pi^2+s^2},\qquad \widehat{w}_k(s)=(\mathcal{L}w_k)(s)=\frac{1}{s}\tanh\left(\frac{s}{4k}\right)=\sum_{n\geq 0}\frac{4k}{k^2(2n+1)^2 \pi^2+s^2} \tag{3}$$

and expand them as power series of $\frac{1}{s}$, getting

$$\widehat{g}(s)=\sum_{n\geq 1}\frac{(2\pi)^{2n+1}(-1)^{n+1}}{s^{2n}},\qquad \widehat{w}_k(s)=\frac{1}{4k}-\frac{2}{\pi}\sum_{n\geq 1}\frac{\zeta(2n)(4k)^{2n-1}(-1)^n}{\pi^{2n-1} s^{2n}}.\tag{4}$$

Proving that $\widehat{g}$ is arbitrarily well approximated by a linear combination of $\widehat{w}_1,\ldots,\widehat{w}_N$ still requires to bound the norm of the inverse of a linear operator, but in the context the direct operator is more clearly structured with respect to $(2)$. If we replace $\zeta(2n)$ with $1$ in the RHS of $(4)$ we get

$$\widehat{W_k}(x)=\frac{1}{4k}-\frac{2}{\pi}\sum_{n\geq 1}\frac{(4k)^{2n-1}(-1)^n}{\pi^{2n-1} s^{2n}}=\frac{1}{4k}+\frac{8k}{\pi^2 s^2+16k^2}$$ $$ W_k(x)=\left(\mathcal{L}^{-1}\widehat{W_k}\right)(x) = \frac{\delta(x)}{4k}+\frac{2}{\pi}\sin\left(\frac{4k x}{\pi}\right)\tag{5}$$

and the subspace generated by $kW_k-(k+1)W_{k+1}$ is clearly dense in $L^2_{\text{odd}}(0,1)$.

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The last step is to lift the convergence in $L^2$ to a convergence in $L^{\infty}$, which can be done by a standard trick. For any $f\in C^{0}_{\text{odd}}[0,1]$ there exists a $f_{\varepsilon}\in C^{2}_{\text{odd}}(0,1)$ such that $\|f-f_{\varepsilon}\|\leq\varepsilon$: it is enough to consider the convolution between $f$ and a suitable kernel (for instance, Jackson's kernel). By the Weierstrass approximation theorem $f'_{\varepsilon}$ can be uniformly approximated by a sequence of polynomials, so there is some even $p_N(x)$ with degree $N$ such that $\|p_N(x)-f'_\varepsilon\|\leq\varepsilon$. If $p_N(x)=a_N x^N+\ldots a_1 x +a_0$ we consider approximations $x^{N+1}_{\text{app}},x^N_{\text{app}},x_{\text{app}},1_{\text{app}}$ in the $L^2$ norm of (the odd parts of) $x^{N+1},x^N,\ldots,x,1$, then take

$$ f_{\text{app}}(x) = \frac{a_{N}}{N+1} x^{N+1}_{\text{app}}+\frac{a_{N-1}}{N} x^{N}_{\text{app}} +\ldots + a_0 x^{1}_{\text{app}}.\tag{6}$$

In simpler terms, if we are able to uniformly approximate $f'$ in the $L^2$ norm we are also able to approximate $f$ in the uniform norm, since

$$ \left\|f-f_{\text{app}}\right\|_{\infty}=\sup_{x\in(0,1)}\left|\int_{1/2}^{x}(f'-f'_{\text{app}})\,dx\right| \stackrel{\text{CS}}{\leq}\|f'-f'_{\text{app}}\|_2. $$

In order to avoid even-ness/odd-ness issues we may also approximate $f''$ in the $L^2$ norm, then "integrate it twice" (more precisely, suitably change the coefficients of $(6)$) to get a uniform approximation of $f$.

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Now I am wondering if $(3)$ and Moebius inversion formula allow us to find explicit coefficients $c_k$ such that $\widehat{g}\stackrel{L^2}{=}\sum_{k\geq 1}c_k \widehat{w}_k$. The residues of $\widehat{g}$ at $\pm 2\pi i$ equal $\mp\frac{i}{2}$. For any odd positive integer $d$, the residues of $\widehat{w_k}$ at $\pm kd\cdot 2\pi i$ equal $\mp \frac{2i}{d\pi}$. By considering only the poles in the upper-half plane, the identity $\widehat{g}=\sum_{k\geq 1}c_k \widehat{w}_k$ gives $c_1=\frac{\pi}{4}$ and for any integer $M>1$

$$0=\text{Res}\left(\widehat{g}(s),s=2M\pi i\right)=-\frac{2i}{\pi}\sum_{\substack{d\text{ odd}\\d\mid M}}\frac{c_{M/d}}{d}$$ hence by Moebius inversion we get the $L^2$-version of Davenport's identity

$$\sum_{n\geq 1}\frac{\mu(n)}{n}\psi(nx)\stackrel{L^2}{=}-\frac{1}{\pi}\sin(2\pi x)$$

which finishes the proof. And yes, the convergence is pretty slow, since the partial sums of $\mu$, even assuming $RH$, can grow as large as $\sqrt{N}\log(N)$. Neverthess the LHS is pointwise convergent by summation by parts (to some function which is bounded, but not necessarily continuous) and equality holds in $L^2$, so we are able to lift the approximation as described in the previous paragraph (actual polynomials or trigonometric polynomials do not make much of a difference).

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A highly non-trivial corollary is that $\det G(v_1,\ldots v_N)$ has a closed form in terms of the tail of the series giving a $L$-function evaluated at $s=2$, greatly improving the lower bound given by Gershgorin theorem.