The Marcel Riesz conjugate-function theorem on the circle

Let $g$ be a real mean-zero trigonometric polynomial. By [F3] and the coefficient rule of [F1], $\mathrm{\parallel }Cg{\mathrm{\parallel }}_2^2={\sum }_k\mathrm{\mid }\widehat{Cg}(k){\mathrm{\mid }}^2={\sum }_{k\mathrm{\ne }0}\mathrm{\mid }\widehat{g}(k){\mathrm{\mid }}^2={\sum }_k\mathrm{\mid }\widehat{g}(k){\mathrm{\mid }}^2=\mathrm{\parallel }g{\mathrm{\parallel }}_2^2$, where $\widehat{g}(0)=0$ is used in the middle equality; again by [F1], $Cg$ is real-valued with $\widehat{Cg}(0)=0$, so $Cg$ is real with zero mean. Moreover [F3] with the coefficient rule gives ${\int }_{T}gCg={\sum }_k\widehat{g}(k)\overline{\widehat{Cg}(k)}=i{\sum }_k\mathrm{sgn}(k)\mathrm{\mid }\widehat{g}(k){\mathrm{\mid }}^2$, which is $i$ times the real number ${\sum }_k\mathrm{sgn}(k)\mathrm{\mid }\widehat{g}(k){\mathrm{\mid }}^2$; since $gCg$ is real-valued, its integral is real, so ${\int }_{T}gCg=0$ and $gCg$ is a real mean-zero trigonometric polynomial.

For a trigonometric polynomial $h$ put $P_{+}h:={\sum }_{k\ge 0}\widehat{h}(k)e_k$, and for $a\in \mathbb{Z}$ put $M_{a}h:=e_{a}h$. Then $P_{+}$ is complex-linear, $(M_{a}h)\widehat{}(k)=\widehat{h}(k-a)$ for every $k$, and for every trigonometric polynomial $h$ and every $N\ge 1$ one has $S_{N}h=M_{-N}{P}_{+}{M}_{N}h-M_{N+1}{P}_{+}{M}_{-(N+1)}h$. Indeed the definitions and $e_a{e}_b=e_{a+b}$ of [F1] give $(M_{-N}{P}_{+}{M}_{N}h)\widehat{}(k)={\mathbf{1}}_{\{k\ge -N\}}\widehat{h}(k)$ and $(M_{N+1}{P}_{+}{M}_{-(N+1)}h)\widehat{}(k)={\mathbf{1}}_{\{k\ge N+1\}}\widehat{h}(k)$ for every $k$, and subtracting gives ${\mathbf{1}}_{\{\mathrm{\mid }k\mathrm{\mid }\le N\}}\widehat{h}(k)=\widehat{S_{N}h}(k)$ for every $k$, which identifies the two trigonometric polynomials.

For $f\in L^1(T)$ and $g\in L^{\mathrm{\infty }}(T)$, ${\int }_T(S_{N}f)gdm={\int }_{T}f(S_{N}g)dm$. Indeed [F7] gives $S_{N}f(x)={\int }_0^{1}f(x-t)D_N(t)dt$ and $S_{N}g(s)={\int }_0^{1}g(s-u)D_N(u)du$ with $D_N$ real and even, so the double integral of $\mathrm{\mid }f(x-t)D_N(t)g(x)\mathrm{\mid }$ is at most $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{D}_N{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$ and [F14] applies; the substitution $s=x-t$ and evenness of $D_N$ turn $\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int f(x-t)D_N(t)g(x)dtdx$ into $\int f(s)(\int D_N(s-x)g(x)dx)ds=\int f(s)S_{N}g(s)ds$.

For $f\in L^1(T)$, each ${\sigma }_{j}f$ is a trigonometric polynomial with $\mathrm{\parallel }{\sigma }_{j}f{\mathrm{\parallel }}_1\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$ and $\mathrm{\parallel }{\sigma }_{j}f-f{\mathrm{\parallel }}_1\to 0$; for $f\in L^{\mathrm{\infty }}(T)$, each ${\sigma }_{j}f$ is a trigonometric polynomial with $\mathrm{\parallel }{\sigma }_{j}f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $\mathrm{\parallel }{\sigma }_{j}f-f{\mathrm{\parallel }}_1\to 0$. This is [F5] with $p=1$ together with [F6]: from $F_j\ge 0$ and ${\int }_0^1{F}_j=1$ one gets $\mathrm{\mid }{\sigma }_{j}f\mathrm{\mid }=\mathrm{\mid }f\ast F_j\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }\ast F_j$, hence $\mathrm{\parallel }{\sigma }_{j}f{\mathrm{\parallel }}_1\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$ and, for $f\in L^{\mathrm{\infty }}$, $\mathrm{\mid }{\sigma }_{j}f(x)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $x$.

For $g\in L^1(T)$ and every real $x$ one has $\mathrm{\mid }{S}_{N}g(x)\mathrm{\mid }\le \mathrm{\parallel }{D}_N{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }g{\mathrm{\parallel }}_1$ by [F7], hence $\mathrm{\parallel }{S}_{N}g{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{D}_N{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }g{\mathrm{\parallel }}_1$: on the unit ball of $L^1$, every $S_N$ is bounded by $\mathrm{\parallel }{D}_N{\mathrm{\parallel }}_{\mathrm{\infty }}$.

Put $A_m:=\sup \{\mathrm{\parallel }Cg{\mathrm{\parallel }}_{2^m}\mathrm{/}\mathrm{\parallel }g{\mathrm{\parallel }}_{2^m}:g$ a real mean-zero trigonometric polynomial, $g\mathrm{\ne }0\}$. Then $A_1=1$ by 1.1, and $A_{m+1}\le A_m+\sqrt{A_m^2+1}\le 2A_m+1$ for every $m\ge 1$, so every $A_m$ is finite. Indeed fix $m$, put $p:=2^m$ and let $g\mathrm{\ne }0$ be real with zero mean; the square identity [F2], the identification of $gCg$ as a real mean-zero trigonometric polynomial in 1.1, and the definition of $A_m$ give $\mathrm{\parallel }Cg{\mathrm{\parallel }}_{2p}^2=\mathrm{\parallel }(Cg{)}^2{\mathrm{\parallel }}_p\le \mathrm{\parallel }g{\mathrm{\parallel }}_{2p}^2+2\mathrm{\parallel }C(gCg){\mathrm{\parallel }}_p\le \mathrm{\parallel }g{\mathrm{\parallel }}_{2p}^2+2A_m\mathrm{\parallel }gCg{\mathrm{\parallel }}_p$, while [F4] applied with exponents $2,2$ to the functions $\mathrm{\mid }g{\mathrm{\mid }}^p$ and $\mathrm{\mid }Cg{\mathrm{\mid }}^p$ gives $\mathrm{\parallel }gCg{\mathrm{\parallel }}_p\le \mathrm{\parallel }g{\mathrm{\parallel }}_{2p}\mathrm{\parallel }Cg{\mathrm{\parallel }}_{2p}$. Dividing by $\mathrm{\parallel }g{\mathrm{\parallel }}_{2p}^2>0$ and writing $u:=\mathrm{\parallel }Cg{\mathrm{\parallel }}_{2p}\mathrm{/}\mathrm{\parallel }g{\mathrm{\parallel }}_{2p}\ge 0$ yields $u^2\le 1+2A_{m}u$, hence $u\le A_m+\sqrt{A_m^2+1}$; taking the supremum over $g$ gives the recursion.

Suppose a bounded linear $V:L^{\mathrm{\infty }}(T)\to L^{\mathrm{\infty }}(T)$ satisfies $Vp=Cp$ for every trigonometric polynomial $p$; put $M:=\mathrm{\parallel }V\mathrm{\parallel }$ and $Af:=\frac{1}{2}(({\int }_{T}fdm)\mathbf{1}+f+iVf)$ for $f\in L^{\mathrm{\infty }}$. Then $\mathrm{\parallel }A{\mathrm{\parallel }}_{L^{\mathrm{\infty }}\to L^{\mathrm{\infty }}}\le 1+\frac{1}{2}M$ and $Ap=P_{+}p$ for every trigonometric polynomial $p$, by 1.2: the multiplier of $\frac{1}{2}(f+iVf)$ is $1$ on positive frequencies, $0$ on negative frequencies and $1\mathrm{/}2$ at zero, and the half-mean term supplies the remaining $1\mathrm{/}2$ at zero. For $h\in L^{\mathrm{\infty }}$ with $\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ let $h_j:={\sigma }_{j}h$: by 1.4 these are trigonometric polynomials with $\mathrm{\parallel }{h}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ and $\mathrm{\parallel }{h}_j-h{\mathrm{\parallel }}_1\to 0$, and 1.2 gives $S_N{h}_j=M_{-N}A{M}_N{h}_j-M_{N+1}A{M}_{-(N+1)}{h}_j$ because $M_{\pm a}{h}_j$ is again a trigonometric polynomial on which $A$ acts as $P_{+}$. Hence $\mathrm{\parallel }{S}_N{h}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2\mathrm{\parallel }A{\mathrm{\parallel }}_{L^{\mathrm{\infty }}\to L^{\mathrm{\infty }}}\mathrm{\parallel }{h}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2+M$, while 1.5 with $\mathrm{\parallel }{h}_j-h{\mathrm{\parallel }}_1\to 0$ gives $\mathrm{\parallel }{S}_N{h}_j-S_{N}h{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$, so $\mathrm{\parallel }{S}_{N}h{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2+M$. Therefore $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_{L^{\mathrm{\infty }}\to L^{\mathrm{\infty }}}\le 2+M$ for every $N$, and in particular the continuous functions give $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_{C\to C}\le 2+M$.

By [F8], $\mathrm{\parallel }{S}_N:C(T)\to C(T)\mathrm{\parallel }={\int }_0^1\mathrm{\mid }{D}_N\mathrm{\mid }\ge \frac{1}{3\pi }\log (N+1)$ for every $N\ge 1$. Write $L_N:={\int }_0^1\mathrm{\mid }{D}_N\mathrm{\mid }$. Given $\delta >0$, choose a continuous one-periodic $g$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ and $\mathrm{\parallel }{S}_{N}g{\mathrm{\parallel }}_{\mathrm{\infty }}\ge L_N-\delta \mathrm{/}2$. Choose $x_0$ with $\mathrm{\mid }{S}_{N}g(x_0)\mathrm{\mid }=\mathrm{\parallel }{S}_{N}g{\mathrm{\parallel }}_{\mathrm{\infty }}$, let $c$ have modulus one with $c{S}_{N}g(x_0)=\mathrm{\mid }{S}_{N}g(x_0)\mathrm{\mid }$ (take $c=1$ if this value is zero), and set ${\psi }_j:=c{F}_j(\cdot -x_0)$. Then $\mathrm{\parallel }{\psi }_j{\mathrm{\parallel }}_1=1$ by [F6], and ${\int }_T{\psi }_j(S_{N}g)dm=c{\sigma }_j(S_{N}g)(x_0)\to \mathrm{\parallel }{S}_{N}g{\mathrm{\parallel }}_{\mathrm{\infty }}$ because $S_{N}g$ is continuous and its Cesaro means converge uniformly by [F5]. By pairing symmetry from 1.3 and $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, for all sufficiently large $j$ we have $\mathrm{\parallel }{S}_N{\psi }_j{\mathrm{\parallel }}_1\ge \mathrm{\mid }{\int }_T(S_N{\psi }_j)gdm\mathrm{\mid }=\mathrm{\mid }{\int }_T{\psi }_j(S_{N}g)dm\mathrm{\mid }\ge \mathrm{\parallel }{S}_{N}g{\mathrm{\parallel }}_{\mathrm{\infty }}-\delta \mathrm{/}2\ge L_N-\delta$. Since $\mathrm{\parallel }{\psi }_j{\mathrm{\parallel }}_1=1$, this gives $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_{L^1\to L^1}\ge L_N-\delta$ for every $\delta >0$, hence $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_{L^1\to L^1}\ge L_N\ge \frac{1}{3\pi }\log (N+1)$.

Suppose a bounded linear $U:L^1(T)\to L^1(T)$ satisfies $Up=Cp$ for every trigonometric polynomial $p$; put $M:=\mathrm{\parallel }U\mathrm{\parallel }$ and define $A$ on $L^1$ by the same formula $Af:=\frac{1}{2}(({\int }_{T}fdm)\mathbf{1}+f+iUf)$. Then $\mathrm{\parallel }A{\mathrm{\parallel }}_{L^1\to L^1}\le 1+\frac{1}{2}M$ and $Ap=P_{+}p$ for every trigonometric polynomial $p$, by the frequency check in 2.2. For $f\in L^1$ with $\mathrm{\parallel }f{\mathrm{\parallel }}_1\le 1$ and its Cesàro means $f_j={\sigma }_{j}f$, which are trigonometric polynomials with $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_1\le 1$ by 1.4, identity 1.2 gives $S_N{f}_j=M_{-N}A{M}_N{f}_j-M_{N+1}A{M}_{-(N+1)}{f}_j$ and hence $\mathrm{\parallel }{S}_N{f}_j{\mathrm{\parallel }}_1\le 2\mathrm{\parallel }A{\mathrm{\parallel }}_{L^1\to L^1}\le 2+M$; by 1.5, $\mathrm{\parallel }{S}_N{f}_j-S_{N}f{\mathrm{\parallel }}_1\le \mathrm{\parallel }{D}_N{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_1\to 0$, so $\mathrm{\parallel }{S}_{N}f{\mathrm{\parallel }}_1\le 2+M$. Therefore $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_{L^1\to L^1}\le 2+M$ for every $N$.

No bounded $V:L^{\mathrm{\infty }}(T)\to L^{\mathrm{\infty }}(T)$ satisfies $Vp=Cp$ on trigonometric polynomials: such a $V$ would give $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_{C\to C}\le 2+\mathrm{\parallel }V\mathrm{\parallel }$ for every $N$ by 2.2, while $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_{C\to C}\ge \frac{1}{3\pi }\log (N+1)$ grows without bound by 2.3, a contradiction for $N$ large.

For every $m\ge 1$ there is a finite constant $K_m$ with $\mathrm{\parallel }Cf{\mathrm{\parallel }}_{2^m}\le K_m\mathrm{\parallel }f{\mathrm{\parallel }}_{2^m}$ for every complex trigonometric polynomial $f$: writing $f=\mathrm{Re}f+i\mathrm{Im}f$ and applying 2.1 to the real mean-zero parts gives $\mathrm{\parallel }C(\mathrm{Re}f){\mathrm{\parallel }}_{2^m}\le A_m\mathrm{\parallel }\mathrm{Re}f-\overline{\mathrm{Re}f}{\mathrm{\parallel }}_{2^m}\le 2A_m\mathrm{\parallel }\mathrm{Re}f{\mathrm{\parallel }}_{2^m}\le 2A_m\mathrm{\parallel }f{\mathrm{\parallel }}_{2^m}$, and likewise for the imaginary part, so $K_m:=4A_m$ works; here $\mathrm{\parallel }\mathrm{Re}f-\overline{\mathrm{Re}f}\mathrm{\parallel }\le 2\mathrm{\parallel }\mathrm{Re}f\mathrm{\parallel }$ uses $\mathrm{\mid }\overline{\mathrm{Re}f}\mathrm{\mid }\le \mathrm{\parallel }\mathrm{Re}f{\mathrm{\parallel }}_1\le \mathrm{\parallel }\mathrm{Re}f{\mathrm{\parallel }}_{2^m}$ and [F4]. Since the trigonometric polynomials are dense in $L^{2^m}(T;\mathbb{C})$ by [F5] and that space is complete by [F9], $C$ therefore has a unique extension to a bounded complex-linear operator $C^{(m)}$ on $L^{2^m}$ with $\mathrm{\parallel }{C}^{(m)}\mathrm{\parallel }\le K_m$; uniqueness holds because two continuous extensions of one map agree on the dense polynomial core.

No bounded $U:L^1(T)\to L^1(T)$ satisfies $Up=Cp$ on trigonometric polynomials: such a $U$ would give $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_{L^1\to L^1}\le 2+\mathrm{\parallel }U\mathrm{\parallel }$ for every $N$ by 3.1, while $\mathrm{\parallel }{S}_N{\mathrm{\parallel }}_{L^1\to L^1}\ge \frac{1}{3\pi }\log (N+1)$ grows without bound by 2.3, a contradiction for $N$ large.

For all $f,g\in L^2(T;\mathbb{C})$ one has ${\int }_T(C^{(1)}f)gdm=-{\int }_{T}f(C^{(1)}g)dm$ for the bilinear pairing. Indeed, for trigonometric polynomials $f,g$ the coefficient identity ${\int }_T(Cf)g={\sum }_k\widehat{Cf}(k)\widehat{g}(-k)$ and the rule $\widehat{Cf}(k)=-i\mathrm{sgn}(k)\widehat{f}(k)$ of [F1] give ${\int }_T(Cf)g=-i{\sum }_k\mathrm{sgn}(k)\widehat{f}(k)\widehat{g}(-k)=-{\int }_{T}f(Cg)$, since $\widehat{Cg}(-k)=-i\mathrm{sgn}(-k)\widehat{g}(-k)=i\mathrm{sgn}(k)\widehat{g}(-k)$; both pairings are bounded bilinear functionals on $L^2\times L^2$ (bounded by $\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2$ times the norm of $C^{(1)}$), and they agree on the dense polynomial core, so they agree everywhere.

Let $p\in [2,\mathrm{\infty })$. If $p=2^k$, set $C_p:=C^{(k)}$. Otherwise choose $m\ge 1$ with $2^m<p<2^{m+1}$ and let $T$ be $C^{(1)}$ restricted to the complex finite simple functions on $T$. For $r=2^m$ and $r=2^{m+1}$, approximate a simple function $h$ by trigonometric polynomials in $L^r$ using [F5]; since $r\ge 2$ and $m(T)=1$, [F13] gives convergence in $L^2$ as well. The extensions therefore satisfy $C^{(1)}h=C^{(m)}h$ and $C^{(1)}h=C^{(m+1)}h$ as $L^2$ classes, so $T$ has the endpoint bounds $K_m,K_{m+1}$. Applying [F10] gives an extension $C_p$ with $\mathrm{\parallel }{C}_p\mathrm{\parallel }\le K_m^{1-\theta }{K}_{m+1}^{\theta }$, where $\frac{1}{p}=\frac{1-\theta }{2^m}+\frac{\theta }{2^{m+1}}$. It agrees with $C$ on polynomials: uniformly approximate a polynomial $P$ by finite simple functions $h_j$; then $C_p{h}_j=C^{(m+1)}{h}_j\to C^{(m+1)}P=CP$ in $L^{2^{m+1}}$, hence in $L^p$ by [F13], while boundedness gives $C_p{h}_j\to C_{p}P$ in $L^p$.

Let $1<p<2$ and $q:=\frac{p}{p-1}>2$, so that 4.3 gives a bounded operator $C_q$ on $L^q$ with norm $K_q$. For $f\in L^p$ the formula ${\mathrm{\Lambda }}_f(\psi ):=-{\int }_{T}f(C_q\psi )dm$ defines a complex-linear functional on $L^q$, bounded by $\mathrm{\parallel }{\mathrm{\Lambda }}_f\mathrm{\parallel }\le K_q\mathrm{\parallel }f{\mathrm{\parallel }}_p$ because [F4] bounds $\mathrm{\mid }{\mathrm{\Lambda }}_f(\psi )\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }{C}_q\psi {\mathrm{\parallel }}_q\le K_q\mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }\psi {\mathrm{\parallel }}_q$; by [F11] there is a unique $h\in L^p$ with ${\mathrm{\Lambda }}_f(\psi )={\int }_T\psi hdm$ for all $\psi \in L^q$ and $\mathrm{\parallel }h{\mathrm{\parallel }}_p=\mathrm{\parallel }{\mathrm{\Lambda }}_f\mathrm{\parallel }\le K_q\mathrm{\parallel }f{\mathrm{\parallel }}_p$. Put $C_{p}f:=h$: then $C_p$ is complex-linear and bounded with $\mathrm{\parallel }{C}_p\mathrm{\parallel }\le K_q$. It extends the polynomial core, because for a trigonometric polynomial $f$ and any $\psi \in L^q$ the defining identity, the consistency of 4.3 and the skew-adjointness of 4.2 give ${\int }_T(C_{p}f)\psi dm=-{\int }_{T}f(C_q\psi )dm=-{\int }_{T}f(C^{(1)}\psi )dm={\int }_T(Cf)\psi dm$, so $C_{p}f=Cf$ by the norm recovery [F12] applied to the difference in $L^p$.

Collecting: for every $1<p<\mathrm{\infty }$ the operator $C_p$ of 4.3 (for $p\ge 2$) and of 5.1 (for $1<p<2$) is a bounded complex-linear extension of $C$ to $L^p(T;\mathbb{C})$, and it is the only such extension because trigonometric polynomials are dense in $L^p$ for $p<\mathrm{\infty }$ and continuous extensions of one map agree on a dense set. If $1<p\le q<\mathrm{\infty }$ and $f\in L^p\cap L^q$, choose trigonometric polynomials ${\varphi }_k\to f$ in $L^q$ using [F5]; [F13] gives convergence in $L^p$, and the bounded extensions agree on polynomials, so their images converge to the same $L^p$ limit. Thus $C_{p}f=C_{q}f$. Also $C_{p}1=0$ because $C$ kills constants by [F1]. By 3.2 there is no bounded compatible $L^{\mathrm{\infty }}$ extension and by 4.1 no bounded compatible $L^1$ extension. This proves all three assertions.