Lacunary Fourier Series and Sidon Sets — Examples

For every $j\ge 1$, ${\lambda }_{j+1}=2^{j+1}=2{\lambda }_j$.

Thus ${\lambda }_{j+1}\ge 2{\lambda }_j$ with one fixed ratio $2>1$, exactly [step 1.1] as required by Hadamard-lacunary sequences and lacunary trigonometric series.

Its successive gaps are [given, algebra] ${\lambda }_{j+1}-{\lambda }_j=(j+1{)}^2-j^2=2j+1⟶\mathrm{\infty }\mathrm{.}$

But [step 1.1, algebra] $\frac{{\lambda }_{j+1}}{{\lambda }_j}={(1+\frac{1}{j})}^2⟶1.$ Hence no fixed $q>1$ can satisfy the ratio condition eventually, much less at every index. This violates the definition in Hadamard-lacunary sequences and lacunary trigonometric series.

Each factor equals $\mathrm{\mid }1+{\epsilon }_n{e}_n{\mathrm{\mid }}^2\mathrm{/}2$, so $P\ge 0$. In an [given, algebra] expansion, each chosen frequency is a signed sum of $1,3,9$ with coefficients in $\{-1,0,1\}$.

Such a signed sum is zero only when all three coefficients are zero: [step 1.1, algebra] the largest nonzero term has magnitude larger than the sum of all smaller ones. Thus the constant coefficient is $1$, so ${\int }_0^{1}P=1$.

The same observation shows that frequency $-n$ has only the first-order [step 2.1, algebra] contribution $\overline{{\epsilon }_n}{e}_{-n}\mathrm{/}2$. Therefore $\widehat{P}(-n)=\overline{{\epsilon }_n}\mathrm{/}2$, agreeing with Riesz-product witnesses for a Hadamard-lacunary set.

Inductively, $P_m$ and $Q_m$ have $2^m$ coefficients, all $\pm 1$, on [given, algebra] the frequencies $0,\dots ,2^m-1$. The two supports in each recursion are disjoint, so this is immediate from the displayed construction.

On $\mathrm{\mid }z\mathrm{\mid }=1$, the parallelogram identity gives [step 1.1, algebra] $\mathrm{\mid }{P}_{m+1}{\mathrm{\mid }}^2+\mathrm{\mid }{Q}_{m+1}{\mathrm{\mid }}^2=2\mathrm{\mid }{P}_m{\mathrm{\mid }}^2+2\mathrm{\mid }{Q}_m{\mathrm{\mid }}^2\mathrm{.}$ Starting from $\mathrm{\mid }{P}_0{\mathrm{\mid }}^2+\mathrm{\mid }{Q}_0{\mathrm{\mid }}^2=2$, induction yields $\mathrm{\mid }{P}_m{\mathrm{\mid }}^2+\mathrm{\mid }{Q}_m{\mathrm{\mid }}^2=2^{m+1}$, hence $\mathrm{\parallel }{P}_m{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2^{(m+1)\mathrm{/}2}$.

The coefficient ${\mathrm{\ell }}^1$ mass of $P_m$ is $2^m$. If the defining [step 2.1, algebra] inequality of Sidon sets in the integer dual held for $\mathbb{Z}$ with one constant $C$, it would give $2^m\le C{2}^{(m+1)\mathrm{/}2}$ for every $m$, which is impossible. Hence $\mathbb{Z}$ is not Sidon.