On The Lp Boundedness of Reisz Projections and Its Implications on The Fourier Series

Introduction

Before we state the two theorems that are the main objective of this article, we will define Reisz projections and the “conjugate” of smooth functions on the torus T. It is not immediately clear why the following functions are well-defined. We will talk about it in the “Preliminary Proofs” section.

Definition: For any $f\in\mathcal{C}^\infty(\mathbb{T})$, we define its Reisz Projections $P_+$ and $P_−$ as

$$P_+(f)(x)=\sum_{m=1}^\infty \hat{f}(m)e^{2\pi imx}$$

$$P_−(f)(x)=\sum_{m=−\infty}^{−1}\hat{f}(m)e^{2\pi imx}$$

We define the conjugate of f as $$\tilde{f}(x)=−i\sum_{m=−\infty}^\infty\text{sgn}(m)\hat{f}(m)e^{2\pi imx}=−iP_+(f)(x)+iP_−(f)(x)$$

Theorem I: For any $p\in (1,\infty)$, there exists an $A_p\ge 0$ such that for all trigonometric polynomials $f$ on $\mathbb{T}$,

$$||\tilde{f}||_{L_p}\le A_p||f||_{L_p}$$

Theorem II: For any $p\in (1,\infty)$ and $f\in L^p(\mathbb{T})$, its Fourier series $\{D_N∗f\}_{N=1}^\infty$ converges to $f$ in $L^p.$

Preliminary Proofs

Lemma 1: Any $h\in C^\infty(\mathbb{T}^n)$, we have $\sum_{m\in\mathbb{Z}^n}|\hat{h}(m)|<\infty$.

Proof: Consider any multi-index $\alpha$. Of course, $\partial^\alpha h$ exists and is continuous. Since $\mathbb{T}^n$ is compact, $\partial^\alpha h$ is bounded and since $\mathbb{T}^n$ has finite measure, $\partial^\alpha h\in L^1$. Further, for any $\epsilon>0$, the set $\{x\in\mathbb{T}^n:|\partial^\alpha h(x)|\ge\epsilon\}$ is closed (being the pre-image of a closed set under the continuous function $\partial^\alpha h$) and hence compact (since $\mathbb{T}^n$ is compact). So $\partial^\alpha h\in \mathcal{C}_0$.

Let $V=\{m\in\mathbb{Z}^n:m_j\neq 0\text{ for all }j\in\{1,2,\dots,n\}\}.$ Using Theorem 8.22(e) (Folland), for any $m\in V,$

$$|\hat{h}(m)|=\left|\frac{\widehat{(\partial^\alpha h)}(m)}{(2\pi im)^\alpha}\right|\le\frac{||\partial^\alpha h||_{L^1}}{(2\pi)^{|\alpha|}\big|m^\alpha\big|}$$

Let $k(m)\in\text{argmax}_{j}\{|m_j|\}.$ Then, $|m^\alpha|=\left|\prod_{j=1}^nm_j^{\alpha_j}\right|\ge |m_{k(m)}|^{\alpha_{k(m)}}$ since $|m_j|\ge 1$ for all $j.$ But $|m|\le \sqrt{n}|m_{k(m)}|.$ So, $$2^{|\alpha|}|m^\alpha|\ge 2^{|\alpha|} \left(\frac{|m|}{\sqrt{n}}\right)^{\alpha_{k(m)}}\ge \left(\frac{|2m|}{\sqrt{n}}\right)^{\alpha_{k(m)}}\ge \left(\frac{1+|m|}{\sqrt{n}}\right)^{\alpha_{k(m)}}$$

Combining these results, we have $$|\hat{h}(m)|\le \frac{(\sqrt{n})^{\alpha_{k(m)}}||\partial^\alpha h||_{L^1}}{\pi^{|\alpha|}(1+|m|)^{\alpha_{k(m)}}}=\frac{C_{(h,\alpha,{k(m)})}}{(1+|m|)^{\alpha_{k(m)}}}\text{ for all }m\in V\text{ where }C_{(h,\alpha,{k(m)})}=\frac{(\sqrt{n})^{\alpha_{k(m)}}||\partial^\alpha h||_{L^1}}{\pi^{|\alpha|}}<\infty$$

Now, we simply choose $\alpha=(n+1,n+1,\dots,n+1).$ Then, $C_{(h,\alpha,{k(m)})}$ is independent of $k(m).$ We denote it $C’.$ Letting $C=\max\big\{C’,\max_{m\in V^c}|\hat{h}(m)|\big\},$ we conclude that $$|\hat{h}(m)|\le\frac{C}{(1+|m|)^{n+1}}\text{ for all }m\in\mathbb{Z}^n$$

Now, $\sum_{m\in \mathbb{Z}^n}\frac{1}{(1+|m|)^{n+1}}$ converges (compare with related integral and use Corollary 2.52 (Folland)). Hence, $\sum_{m\in\mathbb{Z}^n}|\hat{h}(m)|$ converges by the comparison test.

Lemma 2: The Reisz projections and the conjugate of any $h\in\mathcal{C}^\infty(\mathbb{T}^n)$ are well-defined. In fact, the respective series converge uniformly and in $L^p$ for all $p\in [1,\infty]$.

Proof: Uniform convergence (and, hence, convergence in $L^\infty$) follows from Lemma 1 and the Weierstrass M-test. Note that this means that the Reisz projections and the conjugate function are continuous. Convergence in $L^p$ for any $p\in [1,\infty)$ follows from Proposition 6.12 (Folland).

Lemma 3: For any $h\in \mathcal{C}^\infty(\mathbb{T}^1)$, we have $\hat{\tilde{h}}(k)=−i\text{sgn}(k)\hat{h}(k)$ for all $k\in\mathbb{Z}$.

Proof: By Lemma 2, $\tilde{h}_N:x\mapsto \sum_{m=−N}^N(−i\text{sgn}(m))\hat{h}(m)e^{2\pi imx}$ converges to $\tilde{h}$ in $L^1$ as $N\rightarrow\infty$. Hence,

$$|\hat{\tilde{h}}_N(k)−\hat{\tilde{h}}(k)|=\left|\int_{-1/2}^{1/2} \tilde{h}_N(x)e^{−2\pi ikx}dx−\int_{-1/2}^{1/2} \tilde{h}(x)e^{−2\pi ikx}dx\right|\le \int_{-1/2}^{1/2}|\tilde{h}_N(x)−\tilde{h}(x)||e^{−2\pi ikx}|dx\le||\tilde{h}_N−\tilde{h}||_{L^1}$$

Since $\lim_{N\to\infty}||\tilde{h}_N−\tilde{h}||_{L^1}=0$, we have by the Squeeze theorem,

$$\hat{\tilde{h}}(k) = \lim_{N\to\infty}\hat{\tilde{h}}_N(k)=\lim_{N\to\infty}\sum_{m=−N}^N(−i\text{sgn}(m))\hat{h}(m)\int_{-1/2}^{1/2} e^{2\pi i(m−k)x}dx$$$$=\lim_{N\to\infty}\sum_{m=−N}^N(−i\text{sgn}(m))\hat{h}(m)\mathbb{I}_{\{m=k\}}=−i\text{sgn}(k)\hat{h}(k)$$

Proof of Theorem I

Let $f$ be a trigonometric polynomial of degree $d\in\mathbb{N}$ on $\mathbb{T}$. We first assume that

(a) $\hat{f}(0)=0$.

(b) $f$ is real-valued.

Using assumption (a), $\overline{\hat{f}(m)}=\overline{\int_{-1/2}^{1/2} f(x)e^{−2\pi imx}dx}=\int_{-1/2}^{1/2} f(x)\overline{e^{-2\pi imx}}dx=\int_{-1/2}^{1/2} f(x)e^{2\pi imx}dx=\hat{f}(−m)$. Hence, $$\tilde{f}(x)=i\sum_{m=1}^\infty \hat{f}(−m)e^{−2\pi imx}−i\sum_{m=1}^\infty \hat{f}(m)e^{2\pi imx}$$

$$=\sum_{m=1}^\infty\overline{(−i\hat{f}(m)e^{2\pi imx})}+\sum_{m=1}^\infty(−i\hat{f}(m)e^{2\pi imx})=2\sum_{m=1}^\infty\text{Re}(−i\hat{f}(m)e^{2\pi imx})\in\mathbb{R}$$

Thus $\tilde{f}$ and $f$ are both real-valued. $(\star)$

Since $f$ is a trigonometric polynomial (or more generally use Lemma 1 and the Weierstrass M-test), we have by assumption (a), $$f(x)=\sum_{m=−\infty}^\infty\hat{f}(m)e^{2\pi imx}=\sum_{m=1}^\infty\hat{f}(−m)e^{−2\pi imx}+0+\sum_{m=1}^\infty\hat{f}(m)e^{2\pi imx}$$

We also know, by definition, that

$$i\tilde{f}(x)=−\sum_{m=1}^\infty\hat{f}(−m)e^{−2\pi imx}+\sum_{m=1}^\infty\hat{f}(m)e^{2\pi imx}$$

Adding these and letting $\text{deg}(f)=d$, we get $$f(x)+i\tilde{f}(x)=2\sum_{m=1}^d\hat{f}(m)e^{2\pi imx}$$

Hence, for any $k\in\mathbb N$, each term in the trigonometric polynomial $(f+i\tilde{f})^{2k}$ has a positive frequency. That is, $(f+i\tilde{f})^{2k}=\sum_{j=1}^{N_k}\lambda_j e^{2\pi ijx}$ for some $\lambda_j\in\mathbb R$ and some $N_k\in\mathbb N$. Hence, $\int_{-1/2}^{1/2}(f+i\tilde{f})^{2k}(x)dx=0.$ Using binomial expansion, $$\int_{-1/2}^{1/2}\left[\sum_{j=0}^{2k}i^{2k-j}{{2k}\choose {j}}\tilde{f}(x)^{2k-j}f(x)^{j}\right]dx=0$$

Using $(\star)$, $\tilde{f}(x)^{2k-j}f(x)^{j}$ is real for all $x$ and all $0\le j\le 2k$. Thus, $$0=\text{Re}\left(\int_{-1/2}^{1/2}\left[\sum_{j=0}^{2k}i^{2k-j}{{2k}\choose {j}}\tilde{f}(x)^{2k-j}f(x)^{j}\right]dx\right)=\int_{-1/2}^{1/2}\sum_{j=0}^{k}(-1)^{k-j}{{2k}\choose {2j}}\tilde{f}(x)^{2k-2j}f(x)^{2j}dx$$

This, combined with the fact that $|\tilde{f}(x)|^{2k}=\tilde{f}(x)^{2k}$ from $(\star)$, gives $$||\tilde{f}||_{L^{2k}}^{2k}=\left|\int_{-1/2}^{1/2}(-1)^{k-0}{{2k}\choose {2\cdot 0}}\tilde{f}(x)^{2k-2\cdot 0}f(x)^{2\cdot 0}dx\right|$$$$=\left|−\int_{-1/2}^{1/2}\sum_{j=1}^{k}(-1)^{k-j}{{2k}\choose {2j}}\tilde{f}(x)^{2k-2j}f(x)^{2j}dx\right|\le \sum_{j=1}^{k}{{2k}\choose {2j}}\int_{-1/2}^{1/2}|\tilde{f}(x)^{2k-2j}f(x)^{2j}|dx$$

Let $p=2k/(2k−2j)$ and $q=2k/2j$ adopting the convention that $2k/0=\infty$. Using Hölder’s inequality, $\int_{-1/2}^{1/2}|\tilde{f}(x)^{2k-2j}f(x)^{2j}|dx\le \left(\int_{-1/2}^{1/2}|\tilde{f}(x)^{p(2k-2j)}|dx\right)^{1/p}\left(\int_{-1/2}^{1/2}|f(x)^{q(2j)}|dx\right)^{1/q}=||\tilde{f}||_{L^{2k}}^{2k-2j}||f||_{L^{2k}}^{2j}.$ Hence,

$$||\tilde{f}||_{L^{2k}}^{2k}\le \sum_{j=1}^{k}{{2k}\choose {2j}}||\tilde{f}||_{L^{2k}}^{2k-2j}||f||_{L^{2k}}^{2j}\text{ }(\dagger)$$

Consider the algebraic polynomial $P:x\mapsto x^{2k}−\sum_{j=1}^k{{2k}\choose {2j}}x^{2k−2j}$ on $\mathbb{R}$. Note that $\sum_{j=1}^k{{2k}\choose {2j}}x^{−2j}=1−\frac{P(x)}{x^{2k}}$

Since the left hand side tends to $0$ as x grows to infinity, we conclude that any for large enough (positive) $x$,

$$0\le\sum_{j=1}^k{{2k}\choose {2j}}x^{−2j}\le\frac{1}{2}\text{ and hence }\frac{P(x)}{x^{2k}}=1−\left(1−\frac{P(x)}{x^{2k}}\right)\ge\frac{1}{2}$$

Thus, one can show $\lim_{x\to\infty}P(x)=\infty$. If $||f||_{L^{2k}}\neq 0,$ let $t=\frac{||\tilde{f}||_{L^{2k}}}{||f||_{L^{2k}}}\ge 0.$ Since $P(t)\le 0$ from $(\dagger),$ the intermediate value theorem implies that there exists an $r\ge t\ge 0$ such that $P(r)=0$. Hence, $t\le \max\{r\in\mathbb{R}:P(r)=0\}=R_{k}<\infty.$ Thus, $||\tilde{f}||_{L^{2k}}\le R_k||f||_{L^{2k}}$ for all $f$ satisfying (a) and (b). Note that if $||f||_{L^{2k}}=0,$ the result still holds and follows trivially from $(\dagger).$

Relaxation 1: Now, let $f$ satisfy only (a). Say, $f(x)=\sum_{m=-d}^d\lambda_me^{2\pi imx}.$ Let $$g(x)=\sum_{m=-d}^d\frac{\lambda_m+\overline{\lambda_{-m}}}{2}e^{2\pi imx}\text{ and }h(x)=\sum_{m=-d}^d\frac{\lambda_m-\overline{\lambda_{-m}}}{2i}e^{2\pi imx}$$

Now, $f=g+ih.$ One can check that $g$ and $h$ are trigonometric polynomials on $\mathbb{T}$ satisfying both (a) and (b). Hence, by our previous result, and using the the fact that $g=\text{Re}(f)$ and $h=\text{Im}(f),$ $$||\tilde{f}||_{L^{2k}}=||\tilde{g}+i\tilde{h}||_{L^{2k}}\le R_k(||g||_{L^{2k}}+||h||_{L^{2k}})\le 2R_k||f ||_{L^{2k}}$$

Relaxation 2: Now, let $f$ satisfy neither (a) nor (b). Let $g(x)=f(x)-\hat{f}(0)$ and $h=\hat{f}(0).$ One can check that $g$ and $h$ are trigonometric polynomials on $\mathbb{T}$ satisfying (a). Firstly, $\hat{h}(m)=0$ for all $m\neq 0.$ Hence, $\tilde{h}=0.$ Secondly, $||h||_{L^{2k}}\le|\hat{f}(0)|\left(\int_{-1/2}^{1/2}dx\right)^{1/p}\le ||f||_{L^{2k}}$ so $||g||_{L^{2k}}\le ||f||_{L^{2k}}+||h||_{L^{2k}}\le 2||f||_{L^{2k}}.$ Using these facts and the previous result, $$||\tilde{f}||_{L^{2k}}=||\tilde{g}+\tilde{h}||_{L^{2k}}=||\tilde{g}||_{L^{2k}}\le 2R_k||g||_{L^{2k}}\le 4R_k||f||_{L^{2k}}$$

Setting $A_{2k}=4R_k$ for all $k\in\mathbb{N},$ we have proved the theorem for positive even integers. Now, we have to extend the result to a general $p\in (1,\infty).$

Case 1: $p\in (2,\infty)\cap\mathbb{N}^c$

There exists some $k\in\mathbb{N}$ such that $p\in[2k,2(k+1)].$ We want to apply Reisz-Thorin Interpolation, but we don’t have a common linear operator defined in all of and both of $L^{2k}$ and $L^{2(k+1)}.$ To solve this issue, note that the class of trigonometric polynomials on $\mathbb T$ is dense in both $L^{2k}(\mathbb T)$ and $L^{2(k+1)}(\mathbb T)$ (proved in the previous post titled “On the Convergence of Fejér Means in Lp”). So the linear operator $f\mapsto \tilde{f}$ on trigonometric polynomials can be extended to bounded operators $A$ and $B$ respectively. The bounds are $A_{2k}$ and $A_{2(k+1)}$ respectively. For any $f\in L^{2(k+1)}$ let $\{p_m\}_{m=1}^\infty$ be a sequence of trigonometric polynomials that converge to $f$ in $L^{2(k+1)}.$ $(\dagger_1)$ Using Proposition 6.12 (Folland), it also converges to $f$ in $L^{2k}.$ $(\dagger_2)$ Now, $(\dagger_1)$ implies that $\{\tilde{p}_m\}_{m=1}^\infty$ converges to $Bf$ in $L^{2(k+1)}$ and hence (using the same proposition) also converges to $Bf$ in $L^{2k}.$ However, $(\dagger_2)$ implies that $\{\tilde{p}_m\}_{m=1}^\infty$ converges to $Af$ in $L^{2k}.$ So $Bf=Af$ a.e. So we can apply interpolation on $A.$

Since $||Af||_{L^{2k}}\le A_{2k}||f||_{L^{2k}}$ for all $f\in L^{2k}$ and $||Af||_{L^{2(k+1)}}=||Bf||_{L^{2(k+1)}}\le A_{2(k+1)}||f||_{L^{2(k+1)}}$ for all $f\in L^{2(k+1)}$ we have $||Af||_{L^p}\le A_{p}||f||_{L^p}$ for all $f\in L^{p}$ by for some $A_p<\infty$ by interpolation. Hence, $||\tilde{f}||_{L^p}\le A_{p}||f||_{L^p}$ for all trigonometric polynomials on $\mathbb{T}.$ Note that, $L^{2k}(\mathbb T)+L^{2(k+1)}(\mathbb T)=L^{2k}(\mathbb T)$ by the same proposition. This was implicitly used in our invocation of the interpolation theorem.

Case 2: $p\in (1,2)$

Let $q=p/(p-1).$ Note that $q\in (2,\infty).$ Fix any trigonometric polynomial $f.$

For any trigonometric polynomial $g,$ we have $f,g,\tilde{f},\tilde{g}\in L^2(\mathbb T)$ since the functions are continuous (see proof of Lemma 2) and $\mathbb{T}$ is compact with finite measure. We know that $\{x\mapsto e^{2\pi imx}\}_{m=-\infty}^\infty$ is an orthonormal basis of Hilbert space $L^2(\mathbb T).$ Hence, using Theorem 5.27 (c) (Folland) and Proposition 5.21 (Folland), we have

$$\langle \tilde{f},g\rangle_{L^2}=\lim_{N\to\infty}\left\langle\sum_{j=-N}^N \hat{\tilde{f}}(j)e^{2\pi ijx},\sum_{k=-N}^N \hat{g}(k)e^{2\pi ikx}\right\rangle=\sum_{m=-\infty}^\infty \hat{\tilde{f}}(m)\overline{\hat{g}(m)}=\sum_{m=-\infty}^\infty −i\text{sgn}(m)\hat{f}(m)\overline{\hat{g}(m)}$$

The last equality follows from Lemma 3. Similarly,

$$\langle f,\tilde{g}\rangle_{L^2}=\sum_{m=-\infty}^\infty \hat{f}(m)\overline{\hat{\tilde{g}}(m)}=\sum_{m=-\infty}^\infty \hat{f}(m)(i\text{sgn}(m)\overline{\hat{g}(m)})$$

Hence, $\langle \tilde{f},g\rangle_{L^2}=-\langle f,\tilde{g}\rangle_{L^2}.$ So, using the result from Case 2 and Hölder’s inequality, we have $$\left|\int_{-1/2}^{1/2} \tilde{f}(x)\overline{g(x)}dx\right|\le \left|\int_{-1/2}^{1/2} f(x)\overline{\tilde{g}(x)}dx\right|\le (A_q||g||_{L^q})||f||_{L^p}$$

Now consider an arbitrary $g\in L^q.$ Again, by density of trigonometric polynomials in $L^p$ (shown in previous post), for any $\epsilon>0,$ there exists some such polynomial $P$ such that $||g-P||_{L^q}<\min\left\{\frac{\epsilon}{2||\tilde{f}||_{L^p}},\frac{\epsilon}{2A_q||f||_{L^p}}\right\}.$ In taking the minimum, we assume the convention $\epsilon/0=\infty.$ Note that if both $||f||_{L^p}=||\tilde{f}||_{L^p}=0$ then we mean that any $g$ works. In fact, in that case most of the terms in the following arguments become $0$ and the result holds trivially. Now,

$$\left|\int_{-1/2}^{1/2} \tilde{f}(x)\overline{g(x)}dx\right|\le \left|\int_{-1/2}^{1/2} \tilde{f}(x)\left[\overline{g(x)-P(x)}\right]dx\right|+\left|\int_{-1/2}^{1/2} \tilde{f}(x)\overline{P(x)}dx\right|\le ||\tilde{f}||_{L^p}||g-P||_{L^q}+(A_q||P||_{L^q})||f||_{L^p}$$

$$\le ||\tilde{f}||_{L^p}||g-P||_{L^q}+(A_q||P-g||_{L^q})||f||_{L^p}+(A_q||g||_{L^q})||f||_{L^p}\le \frac{\epsilon}{2}+\frac{\epsilon}{2}+(A_q||g||_{L^q})||f||_{L^p}=\epsilon+(A_q||g||_{L^q})||f||_{L^p}$$

Since $\epsilon$ is arbitrary, we have $\left|\int_{-1/2}^{1/2} \tilde{f}(x)\overline{g(x)}dx\right|\le (A_q||g||_{L^q})||f||_{L^p}$ for all $g\in L^q.$ Hence, using Proposition 6.13 (Folland),

$$||\tilde{f}||_{L^p}=\sup\left\{\left|\int_{-1/2}^{1/2} \tilde{f}(x)g(x)dx\right|:g\in L^q,||g||_{L^q}=1\right\}\le A_q||f||_{L^p}$$

Hence, setting $A_p=A_q$ works.

Thus, we have shown the theorem for all $p\in (1,\infty)$.

Proof of Theorem II

Let $p\in (1,\infty).$ Define $S_N:\mathbb{L}^p(\mathbb T)\rightarrow\mathbb{L}^p(\mathbb T)$ for all $N\in\mathbb{N}$ as $S_Ng=\mathcal{D}_N*g.$ We know that $||S_Ng||_{L^p}\le||\mathcal{D}_N||_{L^1}||g||_{L^p}$ for all $g\in L^p$ $(\dagger)$ by Proposition 8.7 (Folland) i.e Young’s Inequality. However, we wish to find a bound for $||S_N||_{L^p\to L^p}$ independent of $N.$

Consider any trigonometric polynomial $f$ on $\mathbb{T}.$ Then,

$$S_Nf(x)=\left(\sum_{m=1}^\infty\hat{f}(m)e^{2\pi imx}-\sum_{m=N+1}^\infty\hat{f}(m)e^{2\pi imx}\right)+\left(\sum_{m=-\infty}^{-1}\hat{f}(m)e^{2\pi imx}-\sum_{m=-\infty}^{-N-1}\hat{f}(m)e^{2\pi imx}\right)+\hat{h}(0)$$

Note that every “infinite” sum is actually finite because $\hat{f}(m)=0$ for all $|m|>\text{deg}(f).$ Now,

$$\sum_{m=N+1}^\infty\hat{f}(m)e^{2\pi imx}=e^{2\pi iNx}\sum_{m=1}^\infty\hat{f}(m+N)e^{2\pi imx}$$

One can rewrite the fourth “infinite” sum similarly. Hence, we have

$$S_Nf(x)=\left(\sum_{m=1}^\infty\hat{f}(m)e^{2\pi imx}-e^{2\pi iNx}\sum_{m=1}^\infty\hat{f}(m+N)e^{2\pi imx}\right)+\left(\sum_{m=-\infty}^{-1}\hat{f}(m)e^{2\pi imx}-e^{-2\pi iNx}\sum_{m=-\infty}^{-1}\hat{f}(m-N)e^{2\pi imx}\right)+\hat{f}(0)$$

$$=P_+\big(f\big)(x)-e^{2\pi iNx}P_+\big(t\mapsto e^{-2\pi iNt}f(t)\big)(x)+P_-\big(f\big)(x)-e^{-2\pi iNx}P_-\big(t\mapsto e^{2\pi iNt}f(t)\big)(x)+\hat{f}(0)\text{ }(\star)$$

Note that $f=\sum_{m=-\infty}^{m=\infty}\hat{f}(m)e^{2\pi imx}$ because $f$ is a trigonometric polynomial. Hence,

$$P_+(f)=\frac{1}{2}(f+i\tilde{f})-\frac{1}{2}\hat{f}(0)$$

Hence, using Theorem 1, one can show that $||P_+(f)||_{L^p}\le (1+A_p)||f||_{L^p}.$ Similarly, $||P_-(f)||_{L^p}\le (1+A_p)||f||_{L^p}.$ We used the fact that $||\hat{f}(0)||_{L^p}\le ||f||_{L^p}\left(\int_{-1/2}^{1/2}dx\right)^{1/p}=||f||_{L^p}.$ So, using this same fact, we have from $(\star),$

$$||S_Nf||_{L^p}\le 4(1+A_p)||f||_{L^p}+||f||_{L^p}=(5+4A_p)||f||_{L^p}$$

This holds for all trigonometric polynomials $f.$

Now, consider any $g\in L^p.$ By density of the class of trigonometric polynomials in $L^p$ (shown in previous post), there exists such a polynomial $f$ such that $||f-g||_{L^p}<\min\left\{\frac{\epsilon}{2||\mathcal{D}_N||_{L^1}},\frac{\epsilon}{2(5+4A_p)}\right\}.$ Then, using the previous result and the naïve bound in $(\dagger),$ we have

$$||S_Ng||_{L^p}\le ||S_N(g-f)||_{L^p}+||S_Nf||_{L^p}\le ||\mathcal{D}_N||_{L^1}||g-f||_{L^p}+(5+4A_p)||f||_{L^p}$$

$$\le ||\mathcal{D}_N||_{L^1}||g-f||_{L^p}+(5+4A_p)||f-g||_{L^p}+(5+4A_p)||g||_{L^p}\le \frac{\epsilon}{2}+\frac{\epsilon}{2}+(5+4A_p)||g||_{L^p}=\epsilon+(5+4A_p)||g||_{L^p}$$

Since this holds for all $\epsilon>0,$ we have $||S_Ng||_{L^p}\le (5+4A_p)||g||_{L^p}$ for all $g\in L^p.$ $(\ddagger)$

Now we are ready to show convergence. Take any $g\in L^p.$ For any $\epsilon>0$ find a trigonometric polynomial $f$ such that $||f-g||_{L^p}<\frac{\epsilon}{2(5+4A_p)}.$ Take any $N\ge \text{deg}(f).$ Then, one can easily check that $S_Nf=f$ itself. Hence, by $(\ddagger),$

$$||S_Ng-g||_{L^p}\le ||S_Ng-S_Nf||_{L^p}+||S_Nf-g||_{L^p}=||S_Ng-S_Nf||_{L^p}+||f-g||_{L^p}$$

$$\le (5+4A_p)||f-g||_{L^p}+||f-g||_{L^p}<\frac{\epsilon}{2}+\frac{\epsilon}{2(5+4A_p)}<\epsilon$$

Thus, $S_Ng\rightarrow g$ in $L^p$ as $N\rightarrow \infty$ for all $g\in L^p(\mathbb{T})$ for $p\in(1,\infty).$ This proves the theorem.