On the Convergence of Fejér Means in Lp
Definition: A sequence of functions $\{k_m\}_{m=1}^\infty\subset L^1(\mathbb{K}^n)$ for $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{T}$ is called an approximate unit if the following three properties are satisfied:
(i) $\sup_{m\in\mathbb{N}}||k_m||_{L^1}<\infty$
(ii) $\int k_md\mu=1$ for all $m\in\mathbb{N}$
(iii) For any neighborhood $U$ of the origin (that is $(0,0,\dots,0)$ or coset $\mathbb{Z}^n$ respectively) we have $\int_{U^c}|k_m|d\mu\rightarrow 0$ as $m\rightarrow\infty$
Here, $\mu$ denotes the Lebesgue measure on $\mathbb{R}^n$ or the measure defined in pg. 238 of Folland’s text for the $n$-dimensional torus respectively.
The most common and useful types of these functions are ones where $\phi$ is an $L^1$ function with $||\phi||_{L^1}=1$ and we set $k_m:x\mapsto m^n\phi(mx).$ You might have also seen the following common example in probability class:
Note About This Post: In this article, we will prove the following five (actually three) theorems. The proof of the second and fifth theorem is left out (and only a sketch is provided) since it follows easily from the other theorems either directly or through their respective proofs. An astute reader will note that the first three theorems are just generalizations of Theorem 8.14 in Folland’s book. Folland’s “easier” theorem follows from ours. To prove the fourth theorem (the raison d'être of this article) we essentially just show that the Fejér means form an approximate unit on $\mathbb{T}^n.$
In what follows, the generic $\mathbb{K}$ will mean that the results hold equally well for $\mathbb{R}$ and $\mathbb{T}.$ However, in the proofs, we only talk about $\mathbb{R}$ since the proof for $\mathbb{T}$ is the same.
Note About The Note About The Post: I realize how using the words “raison d'être” so casually makes me sound like such an unbearable snob.
Some Properties of Approximate Units
Let $\{k_m\}_{m=1}^\infty\subset L^1(\mathbb{K}^n)$ be any approximate unit.
Theorem 1: For any $p\in[1,\infty),$ the sequence $\{k_m*f\}_{m=1}^\infty$ converges to $f$ in $L^p(\mathbb{K}^n)$ for all $f\in L^p(\mathbb{K}^n).$
Theorem 2: $\{k_m*f\}_{m=1}^\infty$ converges to $f$ uniformly for all $f\in\mathcal{BUC}(\mathbb{K}^n).$ Here, $\mathcal{BUC}$ denotes the class of bounded and uniformly continuous functions.
Theorem 3: $\{k_m*f\}_{m=1}^\infty$ converges to $f$ uniformly in all compact subsets of $U$ for all $f\in L^\infty(\mathbb{K}^n)$ which is continuous on the open set $U.$
Consequences of the Above Theorems
Theorem 4: Let $p\in[1,\infty).$ The Fejér means of any $f\in L^p(\mathbb{T}^n)$ converge to $f$ itself in the $L^p$-norm. That is, $||\mathcal{F}^n_N*f-f||_{L^p}\rightarrow 0$ as $N\to\infty$ for all $f\in L^p(\mathbb{T}^n).$ Here, $\{\mathcal{F}^n_N\}_{N=1}^\infty$ denotes the Fejér kernel.
Theorem 5: Let $p\in[1,\infty).$ The set of test functions $C^\infty_c(\mathbb{K}^n)$ is dense in $L^p(\mathbb{K}^n)$ and, hence, so is the Schwartz class $\mathscr{S}(\mathbb{K}^n).$ Additionaly, the set of trigonometric polynomials on $\mathbb{T}^n,$ denoted $\mathcal{T}(\mathbb{T}^n)$ is dense in $L^p(\mathbb{T}^n)$.
Proofs of the Theorems
Proof of Theorem 1: Let $f\in L^p(\mathbb{R}^n).$ Note that $\{k_m*f\}_{m=1}^\infty\subset L^p(\mathbb{R}^n)$ by Young’s inequality. See Theorem 8.7 (Folland).
We know that translation is continuous in $L^p(\mathbb{R}^n)$ (see Proposition 8.5 in Folland’s text and note that the proof can be trivially adapted to the torus-case). Let $M=\sup_{m\in\mathbb{N}}||k_m||_{L^1}.$ This is finite by (i) in the definition. Now, for any $\epsilon>0$ there exists a $\delta>0$ such that $||\tau_y f-f||_{L^p}<\frac{\epsilon}{2M}$ if $||y||<\delta.$ Let $U=\mathcal{B}(0,\delta).$
By (iii), there exists an $N\in\mathbb{N}$ such that for all $m\ge N$ we have $\int_{U^c}|k_m(t)|d\mu(t)<\frac{\epsilon}{4||f||_{L^p}}.$ The edge case of $||f||_{L^p}=0$ can be handled by setting $N=1.$ Fix any such $m\ge N.$
Since $\int k_md\mu=1$ by (ii), we have $$k_m*f(x)-f(x)=\int k_m(t)\big[f(x-t)-f(x)\big]d\mu(t)$$$$=\int_{U} k_m(t)\big[f(x-t)-f(x)\big]d\mu(t)+\int_{U^c} k_m(t)\big[f(x-t)-f(x)\big]d\mu(t)$$
We denote the first term as $A(x)$ and the second as $B(x).$ Note that for all $t,$ we have $$||\chi_U(t)k_m(t)(\tau_tf-f)||_{L^p}\le 2|\chi_U(t)k_m(t)||f||_{L^p}<\infty$$
Also, $t\mapsto || \chi_U(t)k_m(t)(\tau_tf-f)||_{L^p}$ is $L^1$ because $$\int ||\chi_U(t)k_m(t)(\tau_tf-f)||_{L^p} d\mu(t)\le \int 2|\chi_U(t)k_m(t)||f||_{L^p} d\mu(t)$$$$=2||f||_{L^p}\int_U|k_m(t)|d\mu(t)\le 2||f||_{L^p}||k_m||_{L^1}<\infty$$
Hence, using Minkowski’s inequality for integrals, we have $$||A||_{L^p}\le \int || \chi_U(t)k_m(t)(\tau_tf-f)||_{L^p} d\mu(t)$$
For $t\in V$, we know that $||\tau_tf-f||_{L^p}<\frac{\epsilon}{2M}.$ Hence, $$||A||_{L^p}\le \frac{\epsilon}{2M}\int_U|k_m(t)|d\mu(t)\le \frac{\epsilon}{2M} ||k_m||_{L^1}\le\frac{\epsilon}{2}$$
We can also apply Minkowski’s inequality to $B$ using a similar reasoning and obtain (since $n\ge N$) $$||B||_{L^p}\le\int || \chi_{U^c}(t)k_m(t)(\tau_tf-f)||_{L^p} d\mu(t)=\int_{U^c}|k_m(t)|\cdot||\tau_tf-f||_{L^p}d\mu(t)$$$$\le 2||f||_{L^p}\int_{U^c}|k_m(t)|d\mu(t)\le\frac{\epsilon}{2}$$
Hence, $||k_m*f-f||_{L^p}\le\epsilon$ for all $m\ge N$ by the triangle inequality and we have our result.
Sketch of Proof of Theorem 2: Let $f\in \mathcal{BUC}(\mathbb{R}^n).$ Note that $\{k_m*f\}_{m=1}^\infty\subset \mathcal{BUC}(\mathbb{R}^n)$ because of Proposition 8.8 (Folland), since the $L^\infty$ norm coincides with the uniform norm for continuous functions. Now, we can adapt the proof of Theorem 1 by simply replacing all $L^p$-norms with the uniform norm. Note that in the “A” case we use uniform continuity and in the “B” case we use boundedness. We don’t even need Minkowski’s inequality here since the logic follows from the elementary fact that $|\int gd\mu|\le \int |g|d\mu$ for all $g\in L^1.$
Proof of Theorem 3: Let $f\in L^\infty(\mathbb{R}^n)$ be continuous on open set $U.$ Note that by Proposition 8.8 (Folland), $\{k_m*f\}_{m=1}^\infty$ is a collection of continuous functions. Let $K$ be any compact subset of $U.$ We can find another compact set $E$ such that $K\subset E^\circ\subset E\subset U.$ To see this, note that we can express $U$ as a countable union of open sets $\{K_j\}_{j=1}^\infty$ such that $K_j\subset K_{j+1}^\circ$ as follows
$$U=\cup_{j=1}^\infty K_j\text{ where }K_j=\left\{x\in\mathbb{R}^n:||x||\le j,\text{dist}(x,U^c)\ge \frac{1}{j}\right\}$$
Hence, $K\subset \cup_{j=1}^\infty K_j^\circ$ and by definition of compact sets and the increasing nature of the collection of compact sets, we have $K\subset K_N^\circ\subset K_N\subset U$ for some $N\in\mathbb{N}.$ Alternatively, one can reason that the Euclidean space is a locally compact Hausdorff space and apply Proposition 4.31 (Folland).
Consider any $\epsilon>0.$ Let $M=\sup_{m\in\mathbb{N}}||k_m||_{L^1}$ which is finite by (i).
Since $f$ is continuous on $U,$ it is uniformly continuous on $E$ (by compactness). Hence, there exists a $\delta>0$ such that $|f(x)-f(y)|<\frac{\epsilon}{2M}$ for all $x,y\in E$ with $||x-y||<\delta.$ $(\star)$
For every $x\in K$ there exists a $\delta_x>0$ such that $\mathcal{B}(x,\delta_x)\subset E^\circ$ since $K\subset E^\circ.$ Now, $K\subset \cup_{x\in K}\mathcal{B}(x,\delta_x/2).$ So, by compactness, there exists an $N’\in\mathbb{N}$ such that $K\subset \cup_{j=1}^{N’}\mathcal{B}(x_j,\delta_{x_j}/2).$ Take $\delta’=\frac{1}{2}\min_{j=1}^{N’}\delta_{x_j}.$
Consider any $t\in\mathbb{R}^n$ such that $||t||<\delta’$ and any $x\in K.$ Then, there exists some $x_j\in K$ such that $x\in \mathcal{B}(x_j,\delta_{x_j}/2).$ Hence, $||x-t-x_j||\le ||x-x_j||+||t||<\delta_{x_j}/2+\delta’\le\delta_{x_j}.$ So $x-t\in \mathcal{B}(x_j,\delta_{x_j}/2)\subset E^\circ\subset E.$ $(\star\star)$ Let $V=\mathcal{B}\big(0,\min(\delta,\delta’)\big).$
There exists an $N_0\in\mathbb{N}$ by (iii) such that $\int_{V^c}|k_m|d\mu<\frac{\epsilon}{4||f||_{L^\infty}}$ for all $m\ge N_0.$ The edge case of $||f||_{L^\infty}=0$ is handled by choosing $N_0=1.$ $(\star\star\star)$ Fix any such $m\ge N_0.$
So for any $x\in K,$ by (ii), $$k_m*f(x)-f(x)=\int_{V} k_m(t)\big[f(x-t)-f(x)\big]d\mu(t)+\int_{V^c} k_m(t)\big[f(x-t)-f(x)\big]d\mu(t)$$
Denote the first term as $A(x)$ and the second as $B(x).$ By $(\star)$ and $(\star\star)$, both $x,x-t\in E$ and $||(x-t)-x||<\delta$ so $$|A(x)|\le \int_{V} |k_m(t)|\big|f(x-t)-f(x)\big|d\mu(t)\le \frac{\epsilon}{2M}\int_{V} |k_m(t)|d\mu(t)\le \frac{\epsilon}{2M}||k_m||_{L^1}\le \frac{\epsilon}{2}$$
Moreover, by $(\star\star\star),$ $$|B(x)|\le \int_{V^c} |k_m(t)|\big|f(x-t)-f(x)\big|d\mu(t)\le 2||f||_{L^\infty}\int_{V^c} |k_m(t)|d\mu(t)\le \frac{\epsilon}{2}$$
Since $x\in K$ is arbitrary, we conclude that $\sup_{x\in K}||k_m*f-f||_u\le\epsilon$ for all $m\ge N_0$ by the triangle inequality and we have our result.
Proof of Theorem 4: We will show that the Fejer kernel is an approximate unit and then the result follows from Theorem 1.
We first consider the one-dimensional case $n=1.$
$$\int \mathcal{F}^1_N(x)dx=\frac{1}{N+1}\sum_{n=0}^N\sum_{j=-n}^{j=n}\int_{-1/2}^{1/2}e^{2\pi i jx}dx=\frac{1}{N+1}\sum_{n=0}^N1=1$$
Hence, $(ii)$ is satisfied. We also know that $(\star)$ $$\mathcal{F}^1_N(x)=\frac{1}{N+1}\left[\frac{\sin\big(\pi(N+1\big)x)}{\sin(\pi x)}\right]^2\text{ if }x\notin \mathbb{Z}$$$$\mathcal{F}^1_N(x)=2N+1\text{ if }x\in\mathbb{Z}$$
Hence $\mathcal{F}^1_N\ge 0$ and $(i)$ follows from $(ii).$
To show (iii), we simply need to show that $\int_{\delta\le |x|\le 1/2}|\mathcal{F}^1_N(x)|dx\rightarrow 0$ as $N\rightarrow 0$ for all $\delta>0.$ This follows from the definition of the quotient topology and the measure on $\mathbb{T}.$ From $(\star),$ we know that for all $x\notin\mathbb{Z}$ with $0<\delta\le |x|\le 1/2,$ we have
$$|\mathcal{F}^1_N(x)|\le \frac{1}{N+1}\left[\frac{1}{\sin(\pi x)}\right]^2\le\frac{1}{(N+1)(2\pi x)^2}\le \frac{1}{(N+1)(2\pi \delta)^2}$$
Since $\mathbb{Z}\cap[-1/2,1/2]$ has $0$-measure in $[-1/2,1/2]$, we can use the above bound to get $$\int_{\delta\le |x|\le 1/2}|\mathcal{F}^1_N(x)|dx\le \frac{1}{4\pi^2\delta^2(N+1)}$$
The right hand side approaches $0$ as $N$ approaches infinity so (iii) follows from an application of the Squeeze theorem.
For the general case, the Fejér kernel is defined as $$\mathcal{F}^n_N(x_1,x_2,\dots,x_n)=\prod_{j=1}^n\mathcal{F}^1_N(x_j)$$
Since $\mathcal{F}^1_N\ge 0$ we have $\mathcal{F}^n_N\ge 0$ as well. Now (ii) follows from the $1$-dimensional case using Tonelli’s theorem and hence (i) follows too. Let $Q^n=[-1/2,1/2]^n.$ To show (iii) we need to show $\int_{x\in Q^n,||x||\ge \delta}|\mathcal{F}^n_N(x)|dx\rightarrow 0$ as $N\rightarrow 0$ for all $\delta>0.$ If $||x||\ge \delta$ then there exists some $j\le n$ such that $|x_j|\ge\delta/\sqrt{n}.$ Hence, $$\int_{x\in Q^n,||x||\ge \delta}|\mathcal{F}^n_N(x)|dx\le\int_{\cup_{j=1}^n\{x\in Q^n,|x_j|\ge\delta/\sqrt{n}\}}|\mathcal{F}^n_N(x)|dx\le \sum_{j=1}^n \int_{x\in Q^n,|x_j|\ge\delta/\sqrt{n}}|\mathcal{F}^n_N(x)|dx$$
$$=\sum_{j=1}^n \left[\int_{\delta/\sqrt{n}\le |x_j|\le 1/2}|\mathcal{F}^1_N(x_j)|dx_j\prod_{k\neq j}\int_{|x_k|\le 1/2}|\mathcal{F}^1_N(x_k)|dx_k\right]\text{ (Tonelli’s theorem) }$$
Now we use the inequality obtained in the single dimensional case to get $$\int_{x\in Q^n,||x||\ge \delta}|\mathcal{F}^n_N(x)|dx\le \sum_{j=1}^n \left[\frac{1}{4\pi^2(\delta/\sqrt{n})^2(N+1)}\prod_{k\neq j}1\right]$$$$=\frac{n}{4\pi^2(\delta/\sqrt{n})^2(N+1)}=\frac{1}{4\pi^2\delta^2(N+1)}$$
Hence, (iii) follows similarly for the general case too.
Sketch of Proof of Theorem 5:
(a) General $L^p$ functions can be approximated by simple $L^p$ functions.
(b) For any measurable set $E$ with finite measure, there exists an open set $U\supset E$ and a compact set $K\subset E$ such that $\mu(U\setminus K)$ is arbitrarily small. Using Urysohn’s lemma, there exists a compactly supported continuous function that is $1$ on $K$ and $0$ outside $U.$ Thus, simple $L^p$ functions (and, hence, general $L^p$ functions) can be approximated by $\mathcal{C}_c$ functions.
(c) Let $g\in \mathcal{C}_c$ approximate $f\in L^p$ arbitrarily well. Choose any $\phi\in \mathcal{C}^\infty_c$ such that $||\phi||_{L^1}=1.$ Then, $\{k_m\}_{m=1}^\infty$ defined as $k_m:x\mapsto m^n\phi(mx)$ forms an approximate unit and by Theorem 1, $k_m*g$ approximates $g$ (and, hence, $f$) in $L^p$ arbitrarily well for large enough $m$. One can verify that $k_m*g\in\mathcal{C}^\infty_c.$
(d) For $\mathbb{T}^n,$ use Theorem 4 to show the density of trigonometric polynomials in $L^p.$
An Additional Exercise For Readers
For the torus-case, one can derive Theorem 1 from Theorem 2 in two lines!
