Lemma 11

For every $N\geqslant 1$, let $\mu_N$ be the measure on $\mathbf{T}^n$ with density the continuous mapping

The sequence of measures $(\mu_N)_{N\geqslant 1}$ converges to $\varepsilon_0$ in the space $\mathscr{M}^1(\mathbf{T}^n)$ endowed with the topology of compact convergence in $\mathscr{C}(\mathbf{T}^n)$.

We are reduced to the case $n= 1$ by observing that $\mu_N$ is the product of measures of the same type for $n= 1$. It is then enough to verify that the sequence $(\mu_N)$ satisfies the hypotheses of lemma 4 of INT, VIII, §2, No.$^o7$ with $a= 0$.

For this, let us first observe that $F_N$ is the Fourier cotransform of the mapping $\varphi_N:h\mapsto (1- |h|/N)$ on $\mathbf{Z}$. This may be written $\varphi_N=$ $N^{-1}\psi_N*\widetilde{\psi}_N$, where $\psi_N$ is the characteristic function of the set defined by $-N/2<|h|\leqslant N/2$. Consequently, $F_N= N^{-1}|\mathscr{F}(\psi_N)|^2\geqslant 0$. Thus, $\mu_N$ is a positive measure; we have $\mu_N(\mathbf{T}) = 1$, which proves (i) and (iii) in loc. cit.

Let us prove condition (ii) of loc. cit. Let U be an open neighbourhood of 0 in $\mathbf{T}$. It is enough to prove that $\mu_N(U)\rightarrow 1$ when $N\rightarrow +\infty$. Let K be a symmetric compact neighbourhood of 0 such that $K^2\subset U$ and let $\psi$ be the characteristic function of K. Put $\varphi =\psi *\psi$. It is an element of $A(\mathbf{T})$ with support contained in U. The real number $m=\varphi (0)$ is the measure of the set K and hence $m >0$. Moreover, we have $0\leqslant \varphi \leqslant m$ since $\varphi (x)$ is the measure of the set $K\cap xK$. We have

$h\in \mathbf{Z}$

by the transposition properties of the Fourier transform (prop. 13 of II, p. 221). Since $\varphi \in A(\mathbf{T})$, its Fourier transform belongs to $L^1(\mathbf{Z})$ and $\varphi$ satisfies the fourier inversion formula (prop. 11 of II, p. 217). Since $\varphi_N(h)\rightarrow 1$ for every $h\in \mathbf{Z}$ and $|\varphi_N(h)|\leqslant 1$, the Lebesgue theorem (INT, IV, §3, n$^o7$, th. 6) and the fourier inversion formula imply that

lim inf$_{N\rightarrow+\infty}\mu_N(U)\geqslant \frac{1}{m}$ lim$_{N\rightarrow+\infty}\sum_{h\in\mathbf{Z}}\mathscr{F}(\varphi )(h)\varphi_N(h) =$

1 $\sum\mathscr{F}(\varphi )(h) =1\varphi (0) = 1$.

Let us prove the proposition. We have $f*F_N=f_N$ for $N\geqslant 1$. The regular representation $\boldsymbol{\gamma }$ of $\mathbf{T}^n$ in $\mathscr{C}(\mathbf{T}^n)$ (INT, VIII, §2, No.$^o3$) is continuous and satisfies $f*F_N=\boldsymbol{\gamma }(\mu_N)f$ (INT, VIII, §4, No.$^o5$, prop. 5 (iv)). The mapping $\mu\mapsto \boldsymbol{\gamma }(\mu)f$ is continuous from $\mathscr{M}^1(\mathbf{T}^n)$ into $\mathscr{C}(\mathbf{T}^n)$ (INT, VI, §1, No.$^o6$, prop. 14). By the lemma, we therefore have

lim$_{N\rightarrow+\infty}f_N=$ lim$_{N\rightarrow+\infty}f*F_N=$ lim$_{N\rightarrow+\infty}\boldsymbol{\gamma }(\mu_N)f=\boldsymbol{\gamma }(\varepsilon_0)(f) =f$

in $\mathscr{C}(\mathbf{T}^n)$.