Seanie Lee

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Cesaro and Abel Summability

5 minute read

Definition 1.1

Given a sequence $\{c_n\}$, let $s_n:=\sum_{k=0}^nc_k$ be the sequence of partial sums. We define $N$-th Cesaro mean $\sigma_N$ of the sequence $\{s_k\}$ (a.k.a the $N$-th Cesaro sum of the series $\sum_{k=0}^\infty c_k$) by

\[\begin{align*} \sigma_N := \frac{s_0 + s_1 + \cdots + s_{N-1}}{N} \end{align*}\]

Remark

If $\lim_{n\to\infty}s_n = s$, then $\lim_{n\to\infty}\sigma_n=s$.

<Proof>

Let $\epsilon >0$ be given. Since $\lim_{n\to\infty}s_n=s$, there exists $N_1\in\mathbb{N}$ such that if $n>N_1$ then $\lvert s_n - s\rvert <\epsilon/2$. With triangular inequality, for $n> N_1$, we get

\[\begin{align*} \lvert \sigma_n - s\vert &= \left \lvert \sum_{k=0}^{n-1} \frac{s_k}{n} - s \right\rvert \\ &\leq \frac{1}{n}\sum_{k=0}^{n-1}\lvert s_k - s\rvert \\ &= \frac{1}{n}(\lvert s_0-s\rvert + \cdots +\lvert s_{N_1-1}-s\rvert) + \frac{1}{n}(\lvert s_{N_1}-s\rvert + \cdots + \lvert s_{n-1}-s\rvert) \\ &< \frac{N_1}{n}\max_{i\in \{0,1, \ldots, N_1-1\}} \lvert s_i-s\rvert + \frac{n-N_1}{n}\frac{\epsilon}{2} \\ &< \frac{N_1}{n}\max_{i\in \{0,1, \ldots, N_1-1\}} \lvert s_i-s\rvert + \frac{\epsilon}{2} \\ \end{align*}\]

Let $M= \max_{i\in \{0,1, \ldots, N_1-1\}} \lvert s_i-s\rvert\in\mathbb{R}$. By Archimedean property, there is $N_2\in\mathbb{N}$ such that $\frac{MN_1}{N_2} < \frac{\epsilon}{2}$. For $n>\max\{N_1, N_2\}$, we get

\[\begin{align*} \lvert \sigma_n - s\rvert < \frac{\epsilon}{2} + \frac{\epsilon}{2} =\epsilon. \end{align*}\]

$\therefore \sigma_n \to s$ as $n\to\infty$.

\[\tag*{$\square$}\]

Definition 2.1 (Fejer Kernel)

We define the Fejer Kernel, $F_N$, by letting $F_N(x)$ be the $N$-th Cesaro means of $\{D_k(x)\}$; i.e.,

\[\begin{align*} F_n(x) := \frac{1}{N}\sum_{k=0}^{N-1}D_k(x). \end{align*}\]

Remark

Then consider

\[\begin{align*} f*F_N (x) &= f * \left(\frac{1}{N} \sum_{k=0}^{N-1} D_k\right)(x) \\ &= \frac{1}{N}\sum_{k=0}^{N-1}f* D_k (x) \\ &=\frac{1}{N} \sum_{k=0}^{N-1} S_k(f)(x), \end{align*}\]

the average of the first $N$ partial sum of the Fourier series of $f$.

Lemma 2.2

One has,

\[\begin{align*} F_N(x) =\frac{1}{N}\frac{\sin^2(Nx/2)}{\sin^2(x/2)}; \end{align*}\]

further, the Fejer Kernel is a good kernel.

<Proof>

\[\begin{align*} F_N(x) &= \frac{1}{N}\sum_{k=0}^{N-1} D_k(x) \\ &= \frac{1}{N} \sum_{k=0}^{N-1} \frac{\sin (x(k+\frac{1}{2}))}{\sin(x/2)} \\ &=\frac{1}{2N\sin^2(x/2)}\sum_{k=0}^{N-1}2\sin(x(k+1/2))\sin(x/2) \end{align*}\]

By trigometric identity $\sin\alpha \cdot \sin\beta= \frac{1}{2}(\cos (\alpha-\beta) -\cos (\alpha+\beta))$,

\[\begin{align*} \frac{1}{2N\sin^2(x/2)}\sum_{k=0}^{N-1}2\sin(x(k+1/2))\sin(x/2) &=\frac{1}{2N\sin^2(x/2)} \sum_{k=0}^{N-1} (\cos(kx)-\cos ((k+1)x) \\ &= \frac{1- \cos (Nx)}{2N\sin^2(x/2)} \\ &= \frac{1- 1+2\sin^2(Nx/2)}{2N\sin^2(x/2)} \quad ( \because \cos(2\theta) = 1-2\sin^2(\theta)) \\ &=\frac{1}{N}\frac{\sin^2(Nx/2)}{\sin^2(x/2)}. \end{align*}\]

(1)

\[\begin{align*} \frac{1}{2\pi}\int_{-\pi}^{\pi} F_N(x) dx &= \frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{1}{N}\sum_{k=0}^{N-1} D_k(x)dx \\ &= \sum_{k=0}^{N-1}\frac{1}{2N\pi}\int_{-\pi}^{\pi} D_k(x) dx \\ &= \sum_{k=0}^{N-1}\frac{1}{2N\pi}\int_{-\pi}^{\pi} \sum_{m=-k}^k e^{imx}dx \\ &=\frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{2\pi} \sum_{m=-k}^k\int_{-\pi}^\pi \mathbf{1}\{m=0 \}dx \\ &= 1 \end{align*}\]

(2) Since $F_N(x)>0$ for all $x\in[-\pi, \pi]$ and

\[\begin{align*} \frac{1}{2\pi}\int_{-\pi}^{\pi} F_N(x)dx =1, \end{align*}\]

\(\begin{align*} \frac{1}{2\pi}\int_{-\pi}^{\pi}\lvert F_N(x)\rvert dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}F_N(x)dx = 1 \end{align*}\) for all $x\in[-\pi, \pi]$.

(3) Let $\delta>0$ be given. Since $\sin^2 x \leq 1$ and $\sin^2(x/2) \geq \sin^2(\delta/2)$ for $x\in [\delta, \pi]$,

\[\begin{align*} \frac{1}{2\pi}\int_{\delta \leq \lvert x \rvert \leq \pi}\frac{1}{N} \frac{\sin^2(Nx/2)}{\sin^2 (x/2)}dx &\leq \frac{1}{2\pi}\int_{\delta \leq \lvert x \rvert \leq \pi} \frac{1}{N\sin^2(\delta/2)}dx \\ &=\frac{\pi-\delta}{2\pi} \frac{1}{N\sin^2(\delta/2)} \\ &\to 0 \text{ as } N\to\infty. \end{align*}\]

Theorem 2.3

Let $f$ be integrable function on the circle. If $f$ is continuous at $\theta_0$, then the Fourier series of $f$ is Cesaro summable to $f$ at $\theta_0$. In other words, $\lim_{N\to\infty}f*F_N(\theta_0) = f(\theta_0)$. Note that

\[\begin{align} \begin{split} \sigma_N(f)(x) &:=\frac{1}{N}\sum_{k=0}^{N-1} \sum_{n=-k}^k\hat{f}(n)e^{inx} \\ &= \frac{1}{N}\sum_{k=0}^{N-1}S_k(f)(x) \\ &=\frac{1}{N}\sum_{k=0}^{N-1}f*D_k(x) \\ &= f*\left(\frac{1}{N}\sum_{k=0}^{N-1}D_k \right)(x) \\ &=f*F_N(x). \end{split} \label{eq:1} \end{align}\]

Further, if $f$ is continuous on the entire circle, then the convergence of the Cesaro sums is uniform.

<Proof>

Since the Fejer kernel $F_N$ is a good kernel, $\lim_{N\to\infty}f*F_N(\theta_0)=f(\theta_0)$ by the previous theorem. \(\tag*{$\square$}\)

Corollary 2.4

Let $f$ be integrable function on the circle. If $\hat{f}(n)=0$ for all $n\in\mathbb{Z}$, then $f=0$ at all points of continuity of $f$.

<Proof>

By Equation $\ref{eq:1}, f * F_N(x)=0$ for all $N\in\mathbb{N}_0$. Thus,

\[\begin{align*} f(x)=\lim_{N\to\infty}f*F_N(x) = 0. \end{align*}\] \[\tag*{$\square$}\]

Corollary 2.5

Any continuous function on the circle can be uniformly approximated by trigometric polynomials. Note that it is the periodic analogue of Stone-Weierstrass approximation Theorem

<Proof>

It is obvious since

\[f*F_N(x) = \frac{1}{N}\sum_{k=0}^{N-1} \sum_{n=-k}^k \hat{f}(n) e^{inx}\]

for all $x\in [-\pi, \pi]$ and $f*F_N\rightrightarrows f$.

\[\tag*{$\square$}\]

Definition 3.1 (Abel Means and Abel summable)

Let $\sum_{k=0}^\infty c_k$ be a series of complex numbers.

i. We define the Abel Means $A(r)$ of the series $\sum c_k$ by

\[A(r) := \sum_{k=0}^\infty c_k r^k\]

ii. If for every $0\leq r <1$, the abel means $A(r)$ converges, and

\[\lim_{r\to 1} A(r) =s\]

then we say the series $\sum_{k=0}^\infty c_k$ is Abel summable to $s$.

Example

1) Consider $\sum_{k=0}^\infty(-1)^k$. It diverges, but is Abel summable to $s=\frac{1}{2}$.

\[\begin{align*} A(r) &= \sum_{k=0}^\infty (-1)^k r^k \\ &= \frac{1}{1+r}. \end{align*}\]

2) Consider $\sum_{k=0}^\infty(-1)^k (k+1)$. It diverges, but is Abel summable to $s=\frac{1}{4}$.

\[\begin{align*} A(r) = \sum_{k=0}^\infty (-1)^k (k+1)r^k = \frac{1}{(1+r)^2}. \end{align*}\]

Definition 4.1

Suppose we know the Fourier series of a function $f$:

\[\begin{align*} f(\theta)\sim \sum_{n=-\infty}^\infty a_n e^{in\theta}. \end{align*}\]

We define the Abel means of $A_r(f)(\theta)$ of the Fourier Series of the function $f$ by

\[\begin{align*} A_r(f)(\theta) = \sum_{n=-\infty}^\infty r^{\lvert n \rvert}a_n e^{in\theta} \end{align*}\]

Remarks

  1. If we let $c_0=a_0$ and $c_n = a_n e^{in\theta} + a_{-n}e^{-in\theta}$, then the Abel means of the Fourier series above equals the Abel means of the series $\sum_{k=0}^\infty c_k$.

  2. For integrable $f, \lvert a_n \rvert$ is uniformly bounded in $n\in\mathbb{N}$. So $A_r(f)(\theta)$ converges absolutely. By Weierstrass M-test, it also uniformly converges for each fixed $0\leq r <1$.

\[\lvert a_n \rvert= \left\lvert\frac{1}{2\pi}\int_{-\pi}^\pi f(\theta) e^{-in\theta}d\theta \right\rvert \leq \frac{1}{2\pi}\int_{-\pi}^\pi \lvert f(\theta)\rvert d\theta \leq B\]

Lemma 4.2 (Abel means as a convolution)

\[\begin{align*} A_r(f)(\theta) = (f*P_r)(\theta) \end{align*}\]

where $P_r(\theta) = \sum_{n=-\infty}^\infty r^{\lvert n \rvert} e^{in\theta}$ is the Poisson Kernel.

<Proof>

\[\begin{align*} A_r(f)(\theta) &= \sum_{n=-\infty}^\infty r^{\lvert n \rvert} a_ne^{in\theta} \\ &= \sum_{n=-\infty}^\infty r^{\lvert n \rvert}\left( \frac{1}{2\pi}\int_{-\pi}^\pi f(\phi) e^{-in\phi}d\phi\right)e^{in\theta} \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi f(\phi)\sum_{n=-\infty}^\infty r^{\lvert n\rvert} e^{in(\theta-\phi)}d\phi \\ &=\frac{1}{2\pi}\int_{-\pi}^\pi f(\phi) P_r(\theta-\phi)d\phi \\ &= (f*P_r)(\theta). \end{align*}\] \[\tag*{$\square$}\]

Lemma 4.3

The Poisson Kernel is an approximation of the identity as ($r\uparrow 1$).

<Proof>

We know that $P_r(\theta)=\sum_{n=-\infty}^\infty r^{\lvert n \rvert}e^{in\theta}$ converges absolutely and uniformly and already showed that

\[\begin{align*} P_r(\theta) = \frac{1-r^2}{1-2r\cos\theta + r^2} = \frac{1-\lvert \omega \rvert}{\lvert 1 -\omega\rvert^2} > 0 \end{align*}\]

for $0<r<1$, where $\omega = re^{i\theta}$.

1).

\[\begin{align*} \frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta)d\theta &= \frac{1}{2\pi}\int_{-\pi}^\pi\sum_{n=-\infty}^\infty r^{\lvert n\rvert} e^{in\theta}d\theta \\ &=\sum_{n=-\infty}^\infty \frac{1}{2\pi}\int_{-\pi}^\pi r^{\lvert n\rvert} e^{in\theta}d\theta \quad (\because \text{ uniform convergence}) \\ &= \sum_{n=-\infty}^\infty \frac{1}{2\pi} 2\pi\mathbf{1}\{n=0\}r^{\lvert n\rvert} \\ &=1 \end{align*}\]

2). Since $P_r(\theta)>0$ for all $\theta \in [-\pi, \pi]$,

\[\begin{align*} \frac{1}{2\pi}\int_{-\pi}^\pi\lvert P_r(\theta) \rvert d\theta= \frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta)d\theta=1. \end{align*}\]

3). Note that $1-2r\cos\theta + r^2\geq 1-2r +r^2 =(1-r)^2 \geq 1/4 =: c_\delta$. For $\frac{1}{2} \leq r <1$ and $\delta \leq \lvert \theta \rvert \leq \pi$,

which implies that $P_r(\theta) \leq (1-r^2)/ c_\delta$.

Thus,

\[\lim_{r\to 1^{-}} \frac{1}{2\pi}\int_{\delta \leq \lvert \theta \rvert \leq \pi} P_r(\theta)d\theta =0.\]

Corollary 4.4

Let $f$ be integrable function on the circle. Then the Abel means of the Fourier series of $f$ pointwise to $f$ at every point of continuity. If, further, $f$ is continuous on the circle, then the convergence is uniform.

\[A_r(f) = f * P_r \rightrightarrows f.\]

Reference

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