Rudin PMA theorem 8.14.

Question

Code borrowed from here (This question has been asked there before.).

A few Definitions that needed:

Dirichlet kernel: $$\tag{77}D_N(x) = \sum_{n=-N}^Ne^{inx} = \frac{\sin\left( (N+\frac12)x \right)}{\sin(x/2)} $$

$$\tag{78}s_N(f; x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x - t) D_N(t)\, dt $$

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Theorem 8.14:

If, for some $x$, there are constants $\delta > 0$ and $M < \infty$ such that: $$\tag{79}\left| f(x + t) - f(x) \right| \leq M|t| $$

for all $t \in (-\delta, \delta)$, then $$\tag{80}\lim_{N\rightarrow\infty} s_N(f; x) = f(x) $$

And the proof goes as follows:

Define $$\tag{81}g(t) = \frac{f(x-t) - f(x)}{\sin(t/2)} $$

for $0 < |t| \leq \pi$, and put $g(0) = 0$. By the definition $(77)$, $$\frac{1}{2\pi} \int_{-\pi}^{\pi}D_N(x)\, dx = 1.$$

Hence $(78)$ shows that \begin{align}s_N(f; x) - f(x) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} g(t) \sin\left((N +\frac12)t\right)\\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left[ g(t)\cos\frac{t}2 \right]\sin(Nt) \space dt \\ &\qquad+ \frac{1}{2\pi} \int_{-\pi}^{\pi} \left[ g(t)\sin\frac{t}2 \right]\cos(Nt) \, dt. \end{align}

by $(79)$, $(81) $, $g(t)\cos(t/2)$ and $g(t)\sin(t/2)$ are bounded. The last two integrals thus tend to $0$ as $N \rightarrow \infty$ by $(74)$.

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equation $(74 )$ is

$$\tag{74}\lim_{n \to \infty} c_n = 0,$$ where $$c_n =\int_{-\pi}^{\pi} f(x)e^{-inx}dx$$

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I don't know why "the last two integrals thus tend to $0$ as $N \rightarrow \infty$ by $(74)$"?

I don't like the answer on the original question because 1) it didn't explain why $\int_{-\pi}^\pi{[g(t) \;\cos(t/2)]\;\sin (Nt) dt}\le M\delta\int_{-\pi}^\pi{\sin (Nt) dt}=0$ as N does to infinity 2) equation $(74)$ tells us $\lim_{n \to \infty} c_n = 0$ and the answer didn't make a use of that, instead it mistakenly used equation $(72)$ as if it is $(74)$. Nether the OP nor the answerer are active so I decided to ask a new question.

I know that Riemann Lebesgue lemma could be used here but I don't want a solution with this lemma because I want to see what is the prove that make use of $74$ and the boundness of $g(t)\cos(t/2)$ and $g(t)\sin(t/2)$ without Riemann Lebesgue lemma (That lemma is not in this book before this chapter so rudin proof should work without it)

Answer 1

It is $(74)$ that is used. You have, since $e^{-i nt}=\cos(nt)+i \sin(nt)$ and with $h(t)=g(t)\,\cos t/2$, $$ c_n=\frac1{2\pi}\int_{-\pi}^{\pi}h(t)e^{-i nt}\,dt =\frac1{2\pi}\int_{-\pi}^{\pi}h(t)\cos nt\,dt -i \frac1{2\pi}\int_{-\pi}^{\pi}h(t)\sin nt\,dt $$ From $(74)$ we know that $c_n\to0$, and so its real and imaginary parts go to zero. In particular, $$ \lim_{N\to \infty}\int_{-\pi}^{\pi} \left[ g(t)\cos\frac{t}2 \right]\sin(Nt) \,dt=0, $$ and the same for the other integral.

The answer in the question you linked is wrong, because the estimates ignore that the sine and the cosine are not positive.

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Edit: Details on Riemann integrability of $g(t)\sin\frac t2$ and $g(t)\cos\frac t2$.

For $t\in[-\pi,\pi]$, the function $\sin \frac t2$ has a single zero at $t=0$. So, on any interval $[-\pi,-\delta]\cup[\delta,\pi]$, the function $1/\sin\frac t2$ is continuous. So the situation is, we have a function $h:[-\pi,\pi]\to\mathbb R$, bounded, and such that $h$ is Riemann integrable on $[-\pi,-\delta]\cup[\delta,\pi]$ for any $\delta>0$. We want to show that $h$ is Riemann integrable.

Say $|h|\leq M$. Fix $\def\e{\varepsilon}\e>0$. Let $\delta=\frac\e{8M}$. As $h$ is integrable on $[-\pi,-\delta]\cup[\delta,\pi]$, there exists a partition $P=\{x_1,\ldots,x_m\}$ of $[-\pi,-\delta]\cup[\delta,\pi]$ such that $$ U(f,P)-L(f,P)<\frac\e2. $$ Consider $P$ as a partition $P'$ of $[-\pi,\pi]$. This means that on each of $U(f,P)$ and $L(f,P)$ we add a term corresponding to the interval $[-\delta,\delta]$. The term we add is $$ M_0\,\Delta x_0=\sup\{f(x):\ x\in[-\delta,\delta]\}\,(\delta-(-\delta))\leq 2\delta M. $$ Then we have $$ U(f,P')=U(f,P)+M_0\Delta x_0\leq U(f,P)+2\delta M. $$ Similarly, using that $h(x)\geq-M$ for all $x$, $$ L(f,P')=L(f,P)+m_0\,\Delta x_0\geq L(f,P)-2\delta M. $$ Thus $$ U(f,P')-L(f,P')\leq U(f,P)-L(f,P)+4\delta M\leq\frac\e2+4\delta M \leq\frac\e2+\frac\e2=\e. $$ As $\e$ was arbitrary, this shows that $h$ is integrable on $[-\pi,\pi]$.

Answer 2

On page 187, it is stated that the theorems for the rest of the chapter are for Riemann-integrable functions. So in order to apply Theorem 8.12, which involves equation (74) in your question, you need to know that the functions $\varphi_{1}:[-\pi , \pi ]\to\mathbb{C}$ defined by $\varphi_{1}(t) := g(t)\cos (t/2)$ and $\varphi_{2}:[-\pi , \pi ]\to\mathbb{C}$ defined by $\varphi_{2}(t) := g(t)\sin (t/2)$ are Riemann-integrable on $[-\pi, \pi ]$. This is where you need the assumption that $\varphi_{1}$ and $\varphi_{2}$ are bounded.

However, it should be emphasised that the fact that $\varphi_{1}$ and $\varphi_{2}$ are bounded cannot be used alone to conclude that those functions are Riemann-integrable on $[-\pi , \pi ]$, but that this must be used in conjunction with a partition argument and the fact that the functions are Riemann-integrable on any closed subinterval of $[-\pi , \pi ]$ which does not contain $0$. This is one of those arguments which can be straightforward to see in your head why it is true, but at the same time be quite complicated to write down in detail. This question has some more information about such an argument.

Once you have shown that $\varphi_{1}$ and $\varphi_{2}$ are Riemann-integrable, you can use that scalar multiples of the sine and cosine functions, which $\varphi_{1}$ and $\varphi_{2}$ respectively are being multiplied by in the integrals, form an orthogonal system of functions on $[-\pi , \pi]$. Then you can apply Theorem 8.12 in order to conclude that the Fourier coefficients in equation (74) go to zero. As the Fourier coefficients are non-zero multiples of the integrals you are looking at, it follows that both of those integrals go to zero as $N$ increases.