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1. Introduction. In 1972, Gary H. Meisters and Wolfgang M. Schmidt [4] proved an elegant and remarkable result concerning Fourier analysis on the circle group $\mathbb T$, which is the set of complex numbers of modulus $1$. We denote the normalised Haar measure on ${\mathbb T}$ by $\mu$. That is, for an integrable, complex-valued function $f$ on ${\mathbb T}$, \[\int_{\mathbb T}f\,d\mu=\frac{1}{2\pi}\int_0^{2\pi}f(e^{it})\,dt.\] The space $L^2({\mathbb T})$ is the family of functions $f$ on ${\mathbb T}$ such that $\int_{\mathbb T}|\,f\,|^2d\mu<\infty.$ Meisters and Schmidt proved that if a function $f$ in $L^2({\mathbb T})$ has the property that $\int_{\mathbb T}\,f\,d\mu=0$, then $f$ is of the form $\sum_{j=1}^3(f_j-\delta_{b_j}\ast f_j)$, for suitable $f_1,f_2,f_3$ in $L^2({\mathbb T})$ and $b_1,b_2,b_3$ in ${\mathbb T}$, where $\delta_b$ denotes the Dirac measure at $b$ and $\ast$ denotes convolution. Note that $\delta_b\ast f$ is the function $z\longmapsto f(b^{-1}z)$ on ${\mathbb T}$ and is called the translation of $f$ by $b$. They deduced that every linear form $T$ on $L^2({\mathbb T})$ such that $T(\delta_a\ast f)=T(f)$ for all $a\in {\mathbb T}$ and all $f$ in $L^2({\mathbb T})$ must be continuous. These results of Meisters and Schmidt were generalized further on ${\mathbb T}$ in [6], and to the non-compact case of $\mathbb R$ in [6, 7]. Meisters and Schmidt actually presented their results in the context of compact abelian groups and spaces of abstract distributions. The present purpose is to give a more accessible proof in the case of the circle group and $L^2({\mathbb T})$ only, with the intention as well of incorporating the more general results in [6]. The motivation for doing this was research work carried out recently with Susumu Okada [9], which makes central use of the main result presented here. The presentation is based on classical measure theory and the Fourier transform in the space $L^2({\mathbb T})$ of the circle group ${\mathbb T}$. The techniques are those of Meisters and Schmidt in all essentials, although they are used to obtain a more general result. As mentioned, ${\mathbb T}$ consists of all numbers $z\in {\mathbb C}$ such that $|z|=1$. We can identify $\mathbb T$ with the interval $[0,2\pi)$ by means of the one-to-one and onto mapping $x\longmapsto e^{ix}$ from $[0,2\pi)$ to $\mathbb T$. Statements about $\mathbb T$ may be expressed as equivalent statements about the interval $[0,2\pi)$. Suitable sources for background knowledge in integration theory and Fourier analysis may be found in [1, 2, 10], for example. The discussion also makes use of the integration of functions in multivariate analysis. The symbols ${\mathbb N}$, ${\mathbb Z}$, ${\mathbb R}$ and ${\mathbb C}$, respectively denote the sets consisting of the positive integers, the integers, the real numbers and the complex numbers. 2. Notation and statement of the main theorem. Let $M({\mathbb T})$ denote the set of regular Borel measures on $\mathbb T$. Then, $M({\mathbb T})$ has an operation of convolution, which we denote by $\ast$. If $\lambda\in M({\mathbb T})$, $\lambda^n$ denotes the measure $\lambda\ast\lambda\ast\cdots*\lambda$, where $\lambda$ appears $n$ times. If $b\in {\mathbb T}$, $\delta_b\in M({\mathbb T})$ denotes the Dirac measure at $b$. As above, the normalised Haar measure on $\mathbb T$ is denoted by $\mu$, and $\mu\in M({\mathbb T})$. If $\lambda\in M({\mathbb T})$ and $f\in L^2({\mathbb T})$, then $f\ast \lambda\in L^2({\mathbb T})$ and \[ f\ast \lambda (x)=\int_{\mathbb T}f( z^{-1}x)d\lambda(z),\] for almost all $x\in \mathbb T$. If $s\in {\mathbb N}$ and $b\in {\mathbb T}$ put $$\mu_{b,s}=\left(\delta_1-\delta_{b}\right)^s.$$ A function in $L^2({\mathbb T})$ is called a difference of order $s$ if it is of the form \[ \mu_{b,s}\ast g,\] for some $g \in L^2({\mathbb T})$ and some $b\in {\mathbb T}$. The vector space of functions that are finite sums of differences of order $s$ is denoted by ${\mathcal D}_s(L^2({\mathbb T}))$. Note that a difference of order $1$ is a function of the form $g-\delta_b\ast g$ for some $b\in {\mathbb T}$ and $g\in L^2({\mathbb T})$. Also, a difference of order $2$ is a function of the form $g-2^{-1}(\delta_b+\delta_{-b})\ast g$ for some $b\in {\mathbb T}$ and $g\in L^2({\mathbb T})$.
THE MAIN THEOREM. Let $s\in {\mathbb N}$. Then, ${\mathcal D}_s(L^2({\mathbb T}))$ is independent of $s$ and, in fact, \[{\mathcal D}_s(L^2({\mathbb T}))=\Bigg\{f: f\in L^2({\mathbb T})\ \hbox{and}\ \int_{\mathbb T}f\,du=0\Bigg\}.\] Also, if $f\in {\mathcal D}_s(L^2({\mathbb T}))$, $f$ is equal to a sum of $2s+1$ differences of order $s$. The Theorem was proved in the case of $s=1$ by G. Meisters and W. Schmidt [1]. Thus, they proved: a function in $L^2({\mathbb T})$ is a sum of three differences of order $1$ if and only if $\int_{\mathbb T}f\,d\mu=0.$ An immediate consequence of this is: if $T$ is a linear form on $L^2({\mathbb T})$ such that $T(\delta_b\ast f)=T(f)$ for all $f\in L^2({\mathbb T})$ and all $b\in {\mathbb T}$, then $T$ is a multiple of the Haar measure on $\mathbb T$ and $T$ is therefore continuous. However, Meisters showed in [5] that the corresponding statement fails when the non-compact group ${\mathbb R}$ replaces the compact group ${\mathbb T}$. 3. The Fourier transform. The Fourier transform for the circle group is an essential concept in the proof of The Main Theorem. If $\lambda\in M({\mathbb T})$, the Fourier transform of $\lambda$ is the function $\widehat \lambda$ on ${\mathbb Z}$ given by \[{\widehat \lambda}(n)=\int_{\mathbb T}{z}^{-n} \,d\lambda(z),\ \hbox{for}\ n\in {\mathbb Z}.\] Now $L^2({\mathbb T})\subseteq M({\mathbb T})$, and the Fourier transform is a Hilbert space isometry from $L^2({\mathbb T})$ onto $\ell^2({\mathbb Z})$. If $\lambda, \nu\in M({\mathbb T})$, $(\lambda\ast \nu)^{\widehat {\ \ \ }}={\widehat \lambda}\,{\widehat \nu}$. If $b\in {\mathbb T}$, ${\widehat \delta}_b(n)= b^{-n}.$ Note that the condition on $f$ in The Main Theorem above, namely that $\int_{\mathbb T}f\,d\mu=0$, is equivalent to the condition ${\widehat f}(0)=0$. 4. Preliminary results. The main obstacle to proceeding is that we have no way of deciding whether a function is a finite sum of differences of order $s$. The way out is given by a result of Meisters and Schmidt [4], which changes the problem into a question of whether certain integrals are convergent or divergent. We then obtain estimates on these integrals to obtain our results. Our approach here is the same as in [4], the main difference being that we consider differences of order $s$, not only $s=1$, and a corresponding larger class of integrals. In places like equation (1) below, and elsewhere, formal expressions of the form $0/0$ and $x/0$ for some $x>0$ may occur. We take $0/0$ to be $0$ and $x/0$ to be $\infty$ is such cases. Theorem 1. Let $f\in L^2({\mathbb T})$, let $r\in {\mathbb N}$ and let $\mu_1,\mu_2,\ldots,\mu_r\in M({\mathbb T})$. Then the following statements hold. If \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sum_{n=-\infty}^{\infty}\frac{|{\widehat f}(n)|^2}{\sum_{j=1}^r|{\widehat \mu}_j(n)|^2}<\infty,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\] then $f=\sum_{j=1}^r\mu_j\ast f_j$, for some $f_1,f_2,\ldots,f_r\in L^2({\mathbb T})$. The converse statement is also true. Proof. Let $f\in L^2({\mathbb T})$, let $r\in {\mathbb N}$ and let $\mu_1, \mu_2, \ldots ,\mu_r\in M({\mathbb T})$ be such that (1) holds. Then define \[h(n)=\max \Big\{|{\widehat \mu}_1(n))|,\ |{\widehat \mu}_2(n)|,\ldots ,|{\widehat \mu}_r(n)|\Big\},\] for each $n\in{\mathbb Z}$. Then define subsets $B_0, B_1, \ldots ,B_r$ of ${\mathbb Z}$, as follows: we put \begin{align*} B_0&=\Big\{n:n\in {\mathbb Z}\ \hbox{and}\ h(n) = 0\Big\}, \ \hbox{and then}\\ B_j&=\left\{n:n\in {\mathbb Z} , n\not\in \bigcup_{k=0}^{j-1}B_k\ \hbox{and}\ h(n) =|{\widehat \mu}_j(n)|\right\}, \end{align*} for $j=1,2,\ldots ,r.$ The sets $B_0, B_1, \ldots ,B_r$ are disjoint and their union is ${\mathbb Z}$. Note that if $n\in B_j$ for some $j\in \{1,2,\ldots ,r\}$, then $|{\widehat \mu}_j(n)|=h(n)>0$. Thus, for $j\in \{1,2,\ldots ,r\}$, a complex-valued function $h_j$ may be defined on ${\mathbb Z}$ by \[h_j(n)=\cases{\frac {{\widehat f}(n)}{ {\widehat \mu}_j(n)},&if $n\in B_j$;\\ 0,&if $x\not\in B_j$.\\} \] Then, for $j\in \{1,2,\ldots,r\}$, \begin{align} \sum_{n\in {\mathbb Z}}|h_j(n)|^2 &= \sum_{n\in B_j}\frac{{|\widehat f}(n)|^2}{ |{\widehat \mu_j}(n)|^2},\\ &= r\,\sum_{n\in {\mathbb Z}}\frac{1}{r}\cdot\frac{|{\widehat f}(n)|^2} {[\max\{|{\widehat \mu_1}(n)|^2,\ldots ,|{\widehat \mu}_r(n)|^2 \}] },\\ &\le r\,\sum_{n\in {\mathbb Z}}\frac{|{\widehat f}(n)|^2} {\sum^r_{j=1}|{\widehat \mu}_j(n)|^2},\\ &<\infty. \end{align} Hence $h_j\in {\ell}^2({\mathbb Z})$. By the Riesz-Fischer Theorem, there is $f_j\in L^2({\mathbb R})$ such that ${\widehat {f_j}}=h_j$, for all $j=1,2,\ldots ,r$. The definition of $h_j$ gives \[{\widehat f}(n)={\widehat \mu}_j(n) h_j(n)={\widehat \mu}_j(n) {\widehat {f_j}}(n),\] for all $n\in B_j$. Now, note that $B_0, B_1, \ldots ,B_r$ are disjoint and have union equal to $\mathbb Z$. Also, ${\widehat {f_j}}=0$ on $B_k$ if $j\ne k$, and ${\widehat f}=0$ on $B_0$. It follows that, on $\mathbb Z$, \[{\widehat f}=\sum^r_{j=1} {\widehat \mu}_j\, {\widehat {f_j}}=\sum^r_{j=1} (\mu_j\ast f_j)^{\widehat {\ \ \ }}=\left(\sum^r_{j=1} \mu_j\ast f_j\right)^{\widehat {\ \ \ }}.\] As functions that have equal Fourier transforms must be equal, we have the required identity \begin{equation}{\ }\hskip 7cm f=\sum^r_{j=1} \mu_j\ast f_j.\hskip 10cm\square\end{equation} As mentioned above, the converse statement in Theorem 1 is true, but it is not needed here and is not proved (for further details see [4, pages 411-412] and [6, pages 77-78]). Lemma 1. If $x\in [0,\pi]$, \[\frac{x}{\pi}\le \sin\frac{x}{2}.\] Proof. Observe that $x\longmapsto \sin x/x$ is decreasing on $(0,\pi/2]$. So, we see that for all $0\le x\le \pi$, \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sin \frac{x}{2} \ge \frac{x/2}{\pi/2}= \frac{x}{\pi}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \square\] Lemma 2. Let $s,m\in {\mathbb N}$ with $m>2s$. Then we have: \[\int_{[0,1)^{m}}\frac{dx_1dx_2\ldots dx_m}{\sum_{j=1}^mx_j^{2s}}<\infty\ \ \hbox{and}\ \ \int_{(0,2\pi)^{m}}\frac{dx_1dx_2\ldots dx_{m}} { \sum_{j=1}^{m} \sin^{2s} (x_j/2)} <\infty. \] Proof. Let $m>2s$, let $S^{m-1}$ denote the surface of the unit sphere in ${\mathbb R}^m$, and let $\sigma_{m-1}$ denote the usual surface measure on $S^{m-1}$. Observe that if $(x_1,x_2,\ldots,x_m)\in [0,1)^m$, then \[ \left(\sum_{j=1}^mx_j^2\right)^{1/2}\le {\sqrt m}.\] Also, there is $c>0$ such that \[\left(\sum_{j=1}^mx_j^{2}\right)^s\le c\cdot \sum_{j=1}^mx_j^{2s},\ \hbox{for all}\ (x_1,x_2,\ldots,x_m)\in {\mathbb R}^m.\] Thus, we have \begin{align*} \int_{[0,1)^{m}}\frac{dx_1dx_2\ldots dx_m}{\sum_{j=1}^mx_j^{2s}}&\le c\cdot\int_{[0,1)^m}\frac{dx_1dx_2\ldots dx_m}{\left(\sum_{j=1}^mx_j^{2}\right)^s}\\ &\le c\cdot \int_0^{\sqrt m}\int_{S^{m-1}}\frac{r^{m-1}}{r^{2s}}d\sigma_{m-1}\,dr\\ &=c\cdot \sigma_{m-1}(S_{m-1})\cdot \int_0^{\sqrt m}r^{m-2s-1}\, dr \\ &<\infty,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \end{align*} because $m>2s$. So we that the first-mentioned integral is finite. When we come to the second integral, observe that for any function $f$ on $[0,2\pi)^m$, \begin{equation*}\int_{[0,2\pi)^m}f(x_1,x_2,\ldots,x_m)dx_1dx_2\ldots dx_m=\sum_{J_1,J_2,\ldots, J_m}\int_{J_1\times\cdots\times J_m}f(x_1,x_2,\ldots,x_m)dx_1dx_2\ldots,dx_m,\ \ \ \ \ \ \ (3)\end{equation*} where the sum is taken over all $2^m$ combinations of intervals $J_1,J_2,\ldots,J_m$ where for each $j$, $J_j=[0,\pi)$ or $J_j=[\pi,2\pi)$. Now if $p+q=m$ we have \begin{align*} \int_{[0,\pi)^p\times[\pi,2\pi)^q}\frac{dx_1dx_2\ldots dx_m} {\sum_{j=1}^m\sin^{2s} (x_j/2)} &=\int_{[0,\pi)^p\times [0,\pi)^q} \frac{dx_1\ldots dx_pdu_1\ldots du_q} {\sum_{j=1}^p\sin^{2s} (x_j/2)+\sum_{j=1}^q\sin^{2s} ((2\pi-u_j)/2)}\\ &=\int_{[0,\pi)^{p+q}}\frac{dx_1\ldots dx_m} {\sum_{j=1}^m\sin^{2s} (x_j/2)}\\ &\le\pi^{2s}\int_{[0,\pi)^m}\frac{dx_1\ldots dx_m} {\sum_{j=1}^m x_j^{2s}}, \ \hbox{using Lemma 1},\\ &=\pi^{m+2s}\int_{[0,1)^m}\frac{dx_1\ldots dx_m} {\sum_{j=1}^m x_j^{2s}}, \ \hbox{on putting}\ x_j/\pi\ \hbox{in placed of}\ x_j,\\ &<\infty,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4) \end{align*} upon using (2) and as $m>2s$. We now see from (3) and (4) that for $m>2s$, \[\ \ \ \ \ \ \ \int_{[0,2\pi)^m}\frac{dx_1dx_2\ldots dx_m} {\sum_{j=1}^m\sin^{2s} (x_j/2)} \le 2^m\pi^{m+2s}\int_{[0,1)^m}\frac{dx_1\ldots dx_m} {\sum_{j=1}^m x_j^{2s}}<\infty. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \square\]
5. Proof of The Main Theorem. We have to show that a function $f$ in $L^2({\mathbb T})$ is a finite sum of differences of order $s$ if and only if ${\widehat f}(0)=0$. When this holds we have to show we need at most $2s+1$ differences of order $s$. First: the easy part. Let $f\in {\mathcal D}_s(L^2({\mathbb T}))$; we prove ${\widehat f}(0)=0$. For $n\in {\mathbb Z}$, $s\in {\mathbb N}$ and $b\in {\mathbb T}$, \begin{equation} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\widehat {\mu}}_{b,s}(n)=\left[\left(\delta_1-\delta_{b}\right)^s\right]^{\widehat {\ \ \ }}(n)=\Bigl((\delta_1-\delta_{b})^{\widehat {\ \ \,} }(n)\Bigr)^s =(1-b^{-n})^s. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) \end{equation} However, if $f$ is a sum of $q$ differences of order $s$, there are $b_1,b_2,\ldots,b_q\in {\mathbb T}$ and $f_1,f_2,\ldots,f_q\in L^2({\mathbb T})$ such that $f=\sum_{j=1}^q \mu_{b_j,s}\ast f_j$. Then, by (5), \begin{equation} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\widehat f}(n)=\sum_{j=1}^q{\widehat \mu}_{b_j,s}(n)\, {\widehat f}_j(n)\ =\ \sum_{j=1}^q\,(1-b_j^{-n})^s\, {\widehat {f}_j}(n). \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6) \end{equation} Now, putting $n=0$ means that $1-b_j^{-n}=1-1=0$, so we deduce from (6) that ${\widehat f}(0)=0$. Thus, if $f$ is a finite sum of differences of order $s$, ${\widehat f}(0)=0$. Alternatively we can say that $f\in {\mathcal D}_s(L^2({\mathbb R}))$ implies that ${\widehat f}(0)=0$, as required. Second: we prove the hard part. We show that if $f\in L^2({\mathbb T})$ and ${\widehat f}(0)=0$, then $f$ is a sum of $2s+1$ differences of order $s$. Let $n\in {\mathbb Z}$ and $n\ne 0$. Observe that if $m\in {\mathbb N}$, and if $\mu_m$ is the normalised Haar measure on ${\mathbb T}^m$ then for all $n\in {\mathbb Z}$ with $n\ne 0$, and for $f\in L^1({\mathbb T}^m)$, \[\int_{{\mathbb T}^m} f(z_1^n,z_2^n,\ldots,z_m^n)\,d\mu_m(z_1,z_2,\ldots,z_m)=\int_{{\mathbb T}^m} f(z_1,z_2,\ldots,z_m)\,d\mu_m(z_1,z_2,\ldots,z_m).\] The reason is that both sides define an invariant integral on ${\mathbb T}^m$ and so they must be equal by the uniqueness of the Haar integral. So, for all $n\in {\mathbb Z}$ with $n\ne 0$, and for $f\in L^1({\mathbb T}^m)$, \begin{equation}\int_{{\mathbb T}^m} f(z_1^n,z_2^n,\ldots,z_m^n)\,d\mu_m(z_1,z_2,\ldots,z_m)=\frac{1}{{(2\pi)}^m}\int_{(0,2\pi)^m}f(e^{ix_1},e^{ix_2},\ldots,e^{ix_m})dx_1dx_2\ldots dx_m.\ \ \ \ \ \ \ \ (7)\end{equation} We will use this result with \[f(z_1,z_2,\ldots,z_m)=\frac{1}{\sum_{j=1}^m\left|1-z^{-1}_j\right|^{2s}}.\] NOW, let $n\in {\mathbb Z}$ with $n\ne 0$ be given. We have \begin{align} \int_{{\mathbb T}^m}\frac{d\mu_m(y_1,y_2,\ldots,y_m)}{\sum_{j=1}^m|{\widehat \mu}_{y_j,s}(n)|^2 } &=\int_{{\mathbb T}^m}\frac{d\mu_m(y_1,y_2,\ldots,y_m)}{\sum_{j=1}^m \left|1-y_j^{-n}\right|^{2s} }\nonumber\\ &=\int_{{\mathbb T}^m}f(z_1^{n},z_2^{n},\ldots,z_m^{n})\,d\mu_m(z_1,z_2\ldots, z_m)\nonumber\\ &=\frac{1}{{(2\pi)}^m}\int_{(0,2\pi)^m}f(e^{ix_1},e^{ix_2},\ldots,e^{ix_m})\,dx_1dx_2\ldots dx_m, \ \hbox{using}\ (7),\nonumber\\ &=\frac{1}{{(2\pi)}^m}\int_{(0,2\pi)^m}\frac{dx_1dx_2\ldots dx_m}{\sum_{j=1}^m\left|1-e^{-ix_j}\right|^{2s}}\nonumber\\ &=\frac{1}{4}\cdot\frac{1}{{(2\pi)}^m}\int_{(0,2\pi)^m}\frac{dx_1dx_2\ldots dx_m}{\sum_{j=1}^m \sin^{2s}(x_j/2)}\nonumber\\ &<\infty, \end{align} using Lemma 2 and the assumption that $m>2s$. Taking $m=2s+1$ we get that for all $n\in {\mathbb Z}$ with $n\ne 0$, \begin{equation} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \int_{{\mathbb T}^{2s+1}} \frac{d\mu_{2s+1}(y_1,y_2,\ldots,y_{2s+1})} {\sum_{j=1}^{2s+1}\left|{\widehat \mu}_{y_j,s}(n)\right|^2}=M<\infty.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8) \end{equation} Note that $M$ does not depend upon $n$. Now, \begin{align} &\int_{{\mathbb T}^{2s+1}} \left(\sum_{n=-\infty,\, n\ne 0}^{\infty}\,\frac{|{\widehat f}(n)|^2} {\sum_{j=1}^{2s+1}|{\widehat \mu}_{y_j,s}(n)|^2}\right) d\mu_{2s+1}(y_1,y_2,\ldots,y_{2s+1})\nonumber\\ &=\sum_{n=-\infty,\, n\ne 0}^{\infty} \,|{\widehat f}(n)|^2\,\left(\int_{{\mathbb T}^{2s+1}} \frac{d\mu_{2s+1}(y_1,y_2,\ldots,y_{2s+1})} {\sum_{j=1}^{2s+1}|{\widehat \mu}_{y_j,s}(n)|^2}\right)\nonumber\\ &\le M\, \sum_{n=-\infty,\, n\ne 0}^{\infty}\, |{\widehat f}(n)|^2,\ \hbox{by}\ (8), \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,\nonumber\\ &<\infty. \end{align} We deduce from this that \[\sum_{n=-\infty,\, n\ne 0}^{\infty}\,\frac{|{\widehat f}(n)|^2} {\sum_{j=1}^{2s+1}|{\widehat \mu}_{y_j,s}(n)|^2}<\infty,\] for almost all $(y_1,y_2,\ldots,y_{2s+1})$ in ${\mathbb T}^{2s+1}$. So, by Theorem 1, for almost all $(y_1,y_2,\ldots,y_{2s+1})$ in ${\mathbb T}^{2s+1}$, there are $g_1,g_2,\ldots,g_{2s+1}\in L^2({\mathbb T})$ such that \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f=\sum_{j=1}^{2s+1}(g_j-\mu_{y_j,s}\ast g_j).\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)\] That is, we have shown that $f$ is a sum of $2s+1$ differences of order $s$.$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \square$ Note that the statement in (9), in that for almost all $(y_1,y_2,\ldots,y_{2s+1})\in {\mathbb T}^{2s+1}$ there are suitable functions $g_1,g_2,\ldots,g_{2s+1}$, has proved more than the original statement in The Main Theorem. Summarizing, we have the following. Let $s\in {\mathbb N}$ and $f\in L^2({\mathbb T})$. Then, ${\widehat f}(0)=0$ if and only if $f$ is a sum of $2s+1$ differences of order $s$. In this case, for almost all $(b_1,b_2,\ldots,b_{2s+1})\in {\mathbb T}^{2s+1}$, there are $f_1,f_2,\ldots,f_{2s+1}\in L^2({\mathbb T})$ such that $f=\sum_{j=1}^{2s+1}(\delta_1-\delta_{b_j})^s\ast f_j$. 6. Comments and conclusion. The main result mentioned here has analogues on other compact abelian groups apart from $\mathbb T$ (see [3, 4, 6]). There are also analogous results for non-compact groups (see [6, 7]). There is also the the question as to the sharpness of the result in The Main Theorem: if ${\widehat f}(0)=0$ do we need $2s+1$ differences in general in order to express $f$ as sum of differences of order $s$, or will fewer suffice? For the space ${\mathcal D}_1({\mathbb T})$, Meisters and Schmidt showed that $3$ differences are needed, in general. For the corresponding spaces on the real line, it is shown in [8] that $2s+1$ differences are needed, in general. Acknowledgements. I wish to thank Susumu Okada for his suggestion to write an exposition of this result, and for his assistance in its preparation. Also, I wish to thank my former multivariate analysis student, Michael Van Der Kolff, for kindly pointing out some errors in a draft of this proof. References [1] R. Edwards, Fourier Series, volumes I and II,, Holt, Rinehart and Winston, [2] E. Hewitt and K. Ross, Abstract Harmonic Analysis volume 1, Springer Verlag, Berlin, 1963. [3] B. Johnson, A proof of the translation invariant form conjecture for $L^2(G)$, Bull. Sci. Math. de France, 107 (1983), 301-310. [4] G. Meisters and W. Schmidt, Translation-invariant linear forms on $L^2(G)$ for compact abelian groups $G$, Journ. Func. Anal., 11 (1972), 407-424. [5] G. Meisters, Some discontinuous translation-invariant linear forms, Journal of Functional Analysis, 12 (1973), 199-210. [6] R. Nillsen, Banach spaces of functions and distributions characterized by singular integrals involving the Fourier transform, Journal of Functional Analysis, 110 (1992), 73-95. [7] R. Nillsen, Difference Spaces and Invariant Linear Forms, Lecture Notes in Mathematics 1586, Springer Verlag, Berlin, 1994. [8] R. Nillsen and S. Okada, Sharp results concerning the expression of functions as sums of finite differences, Proc. Lond. Math. Soc. 88 (2004), 479-504. [9] R. Nillsen and S. Okada, Aspects of the Fourier transform and finite differences on the circle group, preprint. [10] W. Rudin, Fourier Analysis on Groups, second edition, Interscience McGraw-Hill, New York, 1991.
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