Lacunary Fourier Series and Sidon Sets

Expanding $\mathrm{\mid }f{\mathrm{\mid }}^2$ gives [given, algebra] $\mathrm{\mid }f(x){\mathrm{\mid }}^2={\sum }_{j,k\in J}{a}_j\overline{a_k}{e}_{{\lambda }_j-{\lambda }_k}(x)\mathrm{.}$ The frequencies are distinct because the defining sequence is strictly increasing.

For an integer $n$, direct integration gives [step 1.1, algebra] ${\int }_0^1{e}_n(x)dx=1$ when $n=0$ and $0$ otherwise. Thus integration of step 1.1 retains precisely the $j=k$ terms and gives ${\int }_0^1\mathrm{\mid }f(x){\mathrm{\mid }}^{2}dx={\sum }_{j\in J}\mathrm{\mid }{a}_j{\mathrm{\mid }}^2\mathrm{.}$

Suppose the multisets differ, cancel their common entries, and let [given, algebra] ${\lambda }_s$ be the largest remaining frequency. It occurs on only one side. If the other side has a remaining term, its index is at most $s-r$ because all indices lie in one residue class; in particular $s-r\ge 1$, and every such frequency is at most ${\lambda }_{s-r}<{\lambda }_s\mathrm{/}m$. If the other side has no remaining term, the two sums are already unequal.

That other side has at most $m$ remaining terms, so its sum is strictly [step 1.1, algebra] less than $m{\lambda }_s\mathrm{/}m={\lambda }_s$, whereas its opposing side is at least ${\lambda }_s$. This contradicts the equality. Hence nothing remains after cancellation, which is exactly equality of multisets.

For every integer $m\ge 1$, split $J$ into $r=r(m,q)$ residue classes [given, algebra] with $q^r>m$. Expanding the $2m$-th moment of each class, the additive lemma says that only equal index multisets survive integration. Their permutations give at most $(m!{)}^2$ copies, and $\mathrm{\mid }{\sum }_{\nu =1}^r{z}_{\nu }{\mathrm{\mid }}^{2m}\le r^{2m-1}{\sum }_{\nu }\mathrm{\mid }{z}_{\nu }{\mathrm{\mid }}^{2m}$. Consequently $\mathrm{\parallel }f{\mathrm{\parallel }}_{2m}\le B_{m,q}A$ for a constant independent of $J$ and the coefficients.

The cited $L^2$ identity gives $\mathrm{\parallel }f{\mathrm{\parallel }}_2=A$. If $1\le p<2$, Holder [step 1.1, algebra] applied to $\mathrm{\mid }f{\mathrm{\mid }}^2=\mathrm{\mid }f{\mathrm{\mid }}^{\theta p}\mathrm{\mid }f{\mathrm{\mid }}^{2m(1-\theta )}$, with $1\mathrm{/}2=\theta \mathrm{/}p+(1-\theta )\mathrm{/}(2m)$, combines this identity and step 1.1 to give $A\le \mathrm{\parallel }f{\mathrm{\parallel }}_p^{\theta }(B_{m,q}A{)}^{1-\theta }$. If $p\ge 2$, choose $m$ with $p\le 2m$ and apply the same interpolation identity with $1\mathrm{/}p$ between $1\mathrm{/}2$ and $1\mathrm{/}(2m)$ for the upper bound. For the lower bound, normalized Haar measure has mass one, so Holder gives $\mathrm{\parallel }f{\mathrm{\parallel }}_p\ge \mathrm{\parallel }f{\mathrm{\parallel }}_2=A$ whenever $p\ge 2$. Thus the stated two-sided estimate holds for every $p\ge 1$.

Let $0<p<1$. Step 1.1 with $m=2$ and the $L^2$ identity give [step 1.1, step 2.1, algebra] $\int \mathrm{\mid }f{\mathrm{\mid }}^2=A^2$ and $\int \mathrm{\mid }f{\mathrm{\mid }}^4\le B_{2,q}^4{A}^4$. Cauchy--Schwarz applied to $\mathrm{\mid }f{\mathrm{\mid }}^2{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\ge A\mathrm{/}\sqrt{2}\}}$ shows that this set has measure at least $d_q>0$; otherwise its complement contributes at most $A^2\mathrm{/}2$. Hence $\mathrm{\parallel }f{\mathrm{\parallel }}_p^p\ge d_q(A\mathrm{/}\sqrt{2}{)}^p$. Conversely $t^p\le 1+t^2$ for $t\ge 0$, and applying this to $t=\mathrm{\mid }f\mathrm{\mid }\mathrm{/}A$ shows $\mathrm{\parallel }f{\mathrm{\parallel }}_p^p\le 2A^p$ when $A>0$; the case $A=0$ is immediate. This proves both bounds for $0<p<1$.

If $(a_j)\in {\mathrm{\ell }}^2$, its coefficient tails tend to zero. Apply the upper bound in [F1] to each difference $S_m-S_n$ of partial sums. The differences tend to zero in the $L^p$ norm for $p\ge 1$ and in the metric $d_p$ for $0<p<1$, since $d_p(S_m,S_n)=\mathrm{\parallel }{S}_m-S_n{\mathrm{\parallel }}_p^p$.

By [F2], these Cauchy sequences have limits in their respective spaces.

Conversely, convergence makes the partial sums Cauchy in the relevant norm or metric. The lower bound in [F1] applied to every $S_m-S_n$ forces ${\sum }_{n<j\le m}\mathrm{\mid }{a}_j{\mathrm{\mid }}^2\to 0$ uniformly as $m>n\to \mathrm{\infty }$; in the metric case take the $p$-th root first. The nonnegative partial sums ${\sum }_{j\le n}\mathrm{\mid }{a}_j{\mathrm{\mid }}^2$ are then Cauchy and converge, so $(a_j)\in {\mathrm{\ell }}^2$. This proves both implications.

Choose $r$ with $q^r\ge 3$ and split the original indices by residues [given, algebra] modulo $r$. Each resulting frequency class has successive ratio at least $q^r\ge 3$. Each factor of $P_{\nu }$ equals $\mathrm{\mid }1+{\epsilon }_{\lambda }{e}_{\lambda }{\mathrm{\mid }}^2\mathrm{/}2$ and is nonnegative.

On one such class, a nonempty signed sum with coefficients in [step 1.1, algebra] $\{-1,0,1\}$ cannot be zero: its largest frequency exceeds the sum of all smaller possible frequencies, by the ratio-three geometric bound. Therefore the product expansion has constant term only when every factor contributes its $1$. Its integral is consequently $1$.

The same largest-frequency argument says that frequency $-\lambda$ in [step 1.1, step 2.1, algebra] the expansion occurs only by taking $\overline{{\epsilon }_{\lambda }}{e}_{-\lambda }\mathrm{/}2$ from the $\lambda$ factor and $1$ elsewhere. Hence ${\widehat{P}}_{\nu }(-\lambda )=\overline{{\epsilon }_{\lambda }}\mathrm{/}2$, as claimed.

Set $Q=q^r$ and split the original indices of $\mathrm{\Lambda }$ into $r$ residue classes ${\mathrm{\Lambda }}_{\nu }$. Each class has successive ratios at least $Q\ge 3$. Set ${\epsilon }_{\lambda }=a_{\lambda }\mathrm{/}\mathrm{\mid }{a}_{\lambda }\mathrm{\mid }$ when $a_{\lambda }\mathrm{\ne }0$, and set it to $1$ otherwise. By [L1], the product $P_{\nu }={\prod }_{\lambda \in {\mathrm{\Lambda }}_{\nu }}(1+\mathrm{Re}({\epsilon }_{\lambda }{e}_{\lambda }))$ is nonnegative, has integral one, and has the prescribed coefficients on its class. Empty classes use the product $1$.

Every nonzero frequency $m$ in the product expansion is a signed sum of distinct frequencies of ${\mathrm{\Lambda }}_{\nu }$. If $L$ is its largest contributing frequency, the sum of the smaller frequencies is less than $L\mathrm{/}(Q-1)$, so $\mid \mathrm{\mid }m\mathrm{\mid }-L\mid <\frac{L}{Q-1}<(1-\frac{1}{q})L\mathrm{.}$ Every other frequency of the original sequence is at most $L\mathrm{/}q$ or at least $qL$; in the latter case its distance from $L$ is at least $(q-1)L>(1-1\mathrm{/}q)L$. Thus if $\mathrm{\mid }m\mathrm{\mid }$ belongs to the original sequence it must equal $L$, which lies in this class. In particular ${\widehat{P}}_{\nu }(-\lambda )=0$ for $\lambda \in \mathrm{\Lambda }\setminus {\mathrm{\Lambda }}_{\nu }$.

Finite character integration, [L1], and this vanishing give ${\int }_0^{1}F{P}_{\nu }=\frac{1}{2}{\sum }_{\lambda \in {\mathrm{\Lambda }}_{\nu }}\mathrm{\mid }{a}_{\lambda }\mathrm{\mid }\mathrm{.}$ For completeness, if $z=\int F{P}_{\nu }\mathrm{\ne }0$, multiplication by $\bar{z}\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ and taking real parts gives $\mathrm{\mid }z\mathrm{\mid }\le \int \mathrm{\mid }F\mathrm{\mid }{P}_{\nu }$; the same inequality is immediate if $z=0$. Apply [L2] to the real functions $\mathrm{\mid }F\mathrm{\mid }$ and $P_{\nu }$ to obtain $\frac{1}{2}{\sum }_{\lambda \in {\mathrm{\Lambda }}_{\nu }}\mathrm{\mid }{a}_{\lambda }\mathrm{\mid }\le {\int }_0^1\mathrm{\mid }F\mathrm{\mid }{P}_{\nu }\le \mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$

Summing over the $r$ classes yields ${\sum }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }{a}_{\lambda }\mathrm{\mid }\le 2r\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}$. The constant depends only on $q$ and $r$, not on the finite support or coefficients, so this is the Sidon inequality.

The $N$-th Fejer mean is the finite polynomial [given, algebra] ${\sigma }_{N}f={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}(1-\frac{\mathrm{\mid }k\mathrm{\mid }}{N+1})\widehat{f}(k)e_k\mathrm{.}$ Its spectrum lies in $E$. Positivity and mass one of the Fejer kernel give $\mathrm{\parallel }{\sigma }_{N}f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$.

Apply the Sidon inequality to this polynomial: [step 1.1, algebra] ${\sum }_{\begin{array}{c}k\in E\\ \mathrm{\mid }k\mathrm{\mid }\le N\end{array}}(1-\frac{\mathrm{\mid }k\mathrm{\mid }}{N+1})\mathrm{\mid }\widehat{f}(k)\mathrm{\mid }\le C_E\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ For every fixed finite subset of $E$, the displayed weights tend monotonically to $1$.

Letting $N\to \mathrm{\infty }$ first for each finite subset and then taking the [step 2.1, algebra] supremum over finite subsets gives ${\sum }_{k\in E}\mathrm{\mid }\widehat{f}(k)\mathrm{\mid }\le C_E\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Uniform Fejer convergence identifies the same continuous function with these means, so no separate representative is introduced.