May 06, 2026
Let \(\mathbf{P}\) denote the Riesz projection on the unit circle \(\mathbb{T}\) and suppose that \(1<p<\infty\). We present a simple proof of the bound \(\|\mathbf{P}f\|_p \leq \max(p,q) \|f\|_p\), where \(f\) is in \(L^p(\mathbb{T})\) and \(p^{-1}+q^{-1}=1\). Our proof is a variation of a classical argument due to M. Riesz demonstrating that the Hilbert transform is bounded on \(L^p(\mathbb{T})\).
This note contains a simple proof of an important result in Fourier analysis. A well-known consequence of Fejér’s theorem is that the set of trigonometric polynomials is dense in \(L^p(\mathbb{T})\) for \(1 \leq p<\infty\), where \(\mathbb{T}\) is the unit circle endowed with the normalized Lebesgue arc length measure. This implies that the Riesz projection \(\mathbf{P}\) can be densely defined on \(L^p(\mathbb{T})\) by \[\mathbf{P}\left(\sum_{n \in \mathbb{Z}} \widehat{f}(n)\, e^{in\theta}\right) = \sum_{n=0}^\infty \widehat{f}(n) \,e^{in\theta}.\] The Hilbert transform \(\mathbf{H}\) is the quintessential example of a singular integral operator, and it can be related to the Riesz projection via the formula \[\label{eq:conjfunc} \mathbf{H}f = -i\left(2\mathbf{P}f - f - \widehat{f}(0)\right).\tag{1}\] A celebrated result due to M. Riesz Riesz1928? is that \(\mathbf{H}\) defines a bounded linear operator on \(L^p(\mathbb{T})\) for \(1<p<\infty\). We refer to Littlewood Littlewood1986?*pp. 194–195 and Gårding Garding1970?*pp. III–IV for historical context and to Grafakos Grafakos2014?*pp. 247–248 for a modern presentation of the original proof.
The crux of the present note lies in the observation that this proof can be made significantly shorter and more elegant by shifting the focus from \(\mathbf{H}\) to \(\mathbf{P}\), while following the exact same blueprint: establish the result first for even integers \(p\), then extend to the general case by interpolation and duality.
Lemma 1. If \(k=1,2,3,\ldots\) and if \(f\) is a trigonometric polynomial, then \[\|\mathbf{P}f\|_{2k} \leq k \|f\|_{2k}.\]
Proof. Let \(\mathbf{P}_\perp f = f-\mathbf{P}f\) and assume that \(\|\mathbf{P}f\|_{2k} \geq \|\mathbf{P}_\perp f\|_{2k}\). Since \(f\) is a trigonometric polynomial and \(k\) is a positive integer, it is plain that \[\label{eq:mstrick} (\mathbf{P}f)^k \perp (-\mathbf{P}_\perp f)^k\tag{2}\] in \(L^2(\mathbb{T})\). Since \(\|\mathbf{P}f\|_{2k}^k = \|(\mathbf{P}f)^k\|_2\), this can be parlayed into the estimate \[\label{eq:step1} \left\|\mathbf{P}f\right\|_{2k}^k \leq \left\|(\mathbf{P}f)^k - (-\mathbf{P}_\perp f)^k\right\|_2 = \left\| f \sum_{j=0}^{k-1} (-1)^j (\mathbf{P} f)^{k-1-j}(\mathbf{P}_\perp f)^j \right\|_2.\tag{3}\] Using Hölder’s inequality and the triangle inequality, we infer from 3 that \[\label{eq:step2} \left\|\mathbf{P}f\right\|_{2k}^k \leq \left\|f\right\|_{2k} \sum_{j=0}^{k-1} \left\|(\mathbf{P}f)^{k-1-j} (\mathbf{P}_\perp f)^j \right\|_{\frac{2k}{k-1}}.\tag{4}\] More applications of Hölder’s inequality demonstrate that the terms in this sum are each bounded by \(\|\mathbf{P}f\|_{2k}^{k-1-j} \|\mathbf{P}_\perp f\|_{2k}^j \leq \left\|\mathbf{P}f\right\|_{2k}^{k-1}\). Hence \[\left\|\mathbf{P}f\right\|_{2k}^k \leq k \left\|f\right\|_{2k} \left\|\mathbf{P}f\right\|_{2k}^{k-1},\] which implies the stated bound. If \(\|\mathbf{P}f\|_{2k} \leq \|\mathbf{P}_\perp f\|_{2k}\), then the same argument yields that \(\|\mathbf{P}f\|_{2k} \leq \|\mathbf{P}_\perp f\|_{2k} \leq k \|f\|_{2k}\). ◻
The case \(k=2\) of the main trick 2 in the above proof is from Marzo and Seip MS2011?*Theorem 1. The algebraic identity used in 3 plays a similar role in the proof of a result due to Forelli Forelli1963?*Lemma 4, while our use of Hölder’s inequality to handle the terms of the sum in 4 follows M. Riesz Riesz1928?.
In what follows \(q\) will be the conjugate exponent of \(p\), so that \(p^{-1}+q^{-1}=1\).
Theorem 1. If \(1<p<\infty\), then \(\mathbf{P}\) extends to a bounded linear operator on \(L^p(\mathbb{T})\) satisfying \[\left\|\mathbf{P}f\right\|_p \leq \max(p,q) \left\|f\right\|_p.\]
Proof. If \(k=1,2,3,\ldots\), then the lemma shows that \(\mathbf{P}\) extends by density and continuity to a bounded linear operator on \(L^{2k}(\mathbb{T})\) with norm at most \(k\). If \(2k \leq p \leq 2(k+1)\), then we use the Riesz-Thorin interpolation theorem to infer from this that \(\mathbf{P}\) extends to a bounded linear operator on \(L^p(\mathbb{T})\) with norm at most \(k^{1-\theta} (k+1)^\theta\) for some \(0 \leq \theta \leq 1\) that depends only on \(p\). Since \[k \leq k+1 \leq 2k \leq p,\] we obtain the stated estimate for the case \(2 \leq p<\infty\). Since \(\langle \mathbf{P}f,g\rangle = \langle f, \mathbf{P}g \rangle\) for every pair of trigonometric polynomials \(f\) and \(g\), the case \(1<p<2\) can be established via duality. ◻
We have made no effort to optimize the constant \(\max(p,q)\) in the above bound. It is possible to extract \(C\max(p,q)\) for \(C=1/(e\log{2})=0.5307\ldots\) from the same argument by a small calculus computation. The best constant is \(1/\sin(\pi/p)\) by a result due to Hollenbeck and Verbitsky HV2000?.
The following result is immediate from 1 and the theorem.
Corollary 1 (M. Riesz Riesz1928?). If \(1<p<\infty\), then \(\mathbf{H}\) extends to a bounded linear operator on \(L^p(\mathbb{T})\) satisfying \[\|\mathbf{H}f\|_p \leq 2\left(\max(p,q)+1\right) \|f\|_p.\]
The best constant in this bound is due to Pichorides Pichorides1972?.
Research supported by Grant 354537 of the Research Council of Norway.↩︎