Absolute Convergence and the Wiener Algebra — Examples

Character orthogonality extracts exactly the three displayed coefficients and makes all others zero.

Their absolute values sum to $2+3+1=6$, so the defining coefficient series is summable.

Since ${\sum }_{k\mathrm{\ne }0}\mathrm{\mid }k{\mathrm{\mid }}^{-2}<\mathrm{\infty }$, the coefficient sequence is in ${\mathrm{\ell }}^1$, so the displayed function belongs to $A(\mathbb{T})$.

If $f$ were $C^2$, two integrations by parts would give $\widehat{f^{\mathrm{\prime }\mathrm{\prime }}}(k)=(2\pi ik{)}^2\widehat{f}(k)=-4{\pi }^2$ for every $k\mathrm{\ne }0$.

This contradicts [L1] for the continuous, hence integrable, function $f^{\mathrm{\prime }\mathrm{\prime }}$; therefore $f\mathrm{\notin }{C}^2(\mathbb{T})$.

Summation by parts and [F1] give, uniformly in $x$ and $L\ge M\ge 2$, $\mid {\sum }_{k=M}^L{u}_k(x)\mathrm{/}k\mid \le C_1{M}^{-1\mathrm{/}2}$. Hence the series defining $g$ converges uniformly and $g$ is continuous.

In fact $g$ is Hölder-one-half. Let $0<H\le 1\mathrm{/}2$ be the circular size of a displacement $h$, choose its representative with $\mathrm{\mid }h\mathrm{\mid }=H$, put $N=\lfloor H^{-1}\rfloor$, and set $b_k=(e^{2\pi ikh}-1)\mathrm{/}k$. Then $\mathrm{\mid }{b}_N\mathrm{\mid }\le 2\pi H$ and $\mathrm{\mid }{b}_k-b_{k+1}\mathrm{\mid }\le 4\pi H\mathrm{/}k$ for $2\le k<N$. Summation by parts and [F1] now bound ${\sum }_{k=2}^N{u}_k(x)b_k$ by $C_{2}H\sqrt{N}\le C_2\sqrt{H}$.

Uniform convergence permits termwise integration, so $\widehat{g}(k)=e^{ik\log k}\mathrm{/}k$ for $k\ge 2$ and $\widehat{g}(k)$ vanishes otherwise. Therefore ${\sum }_k\mathrm{\mid }\widehat{g}(k)\mathrm{\mid }={\sum }_{k=2}^{\mathrm{\infty }}1\mathrm{/}k=\mathrm{\infty }$, and $g\mathrm{\notin }A(\mathbb{T})$.

The terms with $k>N$ equal the difference of the two series tails at $x+h$ and $x$. Step 1.1 bounds their sum by $2C_1{N}^{-1\mathrm{/}2}\le 2\sqrt{2}{C}_1\sqrt{H}$.

For $H=0$ the difference vanishes; for $H>0$, steps 1.2 and 2.2 give $\mathrm{\mid }g(x+h)-g(x)\mathrm{\mid }\le C_3{H}^{1\mathrm{/}2}$. Thus this continuous $g$ has non-absolutely-summable Fourier coefficients and refutes the statement.

Summation by parts and [F1] give, uniformly in $x$ and $L\ge M\ge 2$, $\mid {\sum }_{k=M}^L{u}_k(x)\mathrm{/}k\mid \le C_1{M}^{-1\mathrm{/}2}$. Hence the series defining $g$ converges uniformly and $g$ is continuous.

Let $0<H\le 1\mathrm{/}2$ be the circular size of a displacement $h$, choose its representative with $\mathrm{\mid }h\mathrm{\mid }=H$, put $N=\lfloor H^{-1}\rfloor$, and set $b_k=(e^{2\pi ikh}-1)\mathrm{/}k$. Then $\mathrm{\mid }{b}_N\mathrm{\mid }\le 2\pi H$ and $\mathrm{\mid }{b}_k-b_{k+1}\mathrm{\mid }\le 4\pi H\mathrm{/}k$ for $2\le k<N$. Summation by parts and [F1] now bound ${\sum }_{k=2}^N{u}_k(x)b_k$ by $C_{2}H\sqrt{N}\le C_2\sqrt{H}$.

Uniform convergence permits termwise integration, so $\widehat{g}(k)=e^{ik\log k}\mathrm{/}k$ for $k\ge 2$ and $\widehat{g}(k)$ vanishes otherwise. Therefore ${\sum }_k\mathrm{\mid }\widehat{g}(k)\mathrm{\mid }={\sum }_{k=2}^{\mathrm{\infty }}1\mathrm{/}k=\mathrm{\infty }$, and $g\mathrm{\notin }A(\mathbb{T})$.

The terms with $k>N$ equal the difference of the two series tails at $x+h$ and $x$. Step 1.1 bounds their sum by $2C_1{N}^{-1\mathrm{/}2}\le 2\sqrt{2}{C}_1\sqrt{H}$.

For $H=0$ the difference vanishes; for $H>0$, steps 1.2 and 2.2 give $\mathrm{\mid }g(x+h)-g(x)\mathrm{\mid }\le C_3{H}^{1\mathrm{/}2}$, so $g\in C^{1\mathrm{/}2}(\mathbb{T})$ but $g\mathrm{\notin }A(\mathbb{T})$.

The two nonzero Fourier coefficients of $f$ are $1$ and $-1$, so $f\in A(\mathbb{T})$.

At the identity $0\in \mathbb{T}$, $f(0)=1-e^0=0$.

A pointwise reciprocal would be unbounded near this zero and cannot be continuous; [L1] says that every member of $A(\mathbb{T})$ is represented by a continuous function. Thus no reciprocal belongs to $A(\mathbb{T})$.