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In the Manfred Stoll, Introduction to Real Analysis he states 9.5.3 Theorem as follows :

9.5.3 THEOREM ( Dirichlet ) Let $f$ be a real-valued periodic function of $\mathbb{R}$ of period $2\pi$ with $f\in \mathcal{R}[-\pi ,\pi]$. Suppose $x_0 \in \mathbb{R}$ is such that (a) $f(x_0^{+})$ and $f(x_0^{-})$ both exist, and (b) there exists a constant $M$ and a $\delta >0$ such that $$ |f(x_0 +t)-f(x_0^{+})| \le Mt , |f(x_0-t)-f(x_0^{-})| \le Mt$$ for all $t, 0<t \le \delta$, then $$ \lim_{n\to \infty} S_n(x_0)= \frac{1}{2}[f(x_0^{+})+f(x_0^{-})],$$ where $S_n(x_0)$ is the $n$th partial sum of Fourier series.

If needed, I will upload full proof. In the proof, second paragraph, he set $h(t):= f(x_0+t)-f(x_0^{+})$. And he notes as " since $f\in \mathcal{R}[-\pi,\pi]$ and is periodic of period $2\pi$, $h$ is Riemann integrable on $[0,\pi]$." It seems easy, but I don't understand his argument. Can anyone explain more in detail? Can anyone help?

Plantation
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    $f(x_0^{+})$ is just a constant. Can you show that $f(x_0+t)$ is Riemann integrable? – geetha290krm Mar 17 '24 at 06:10
  • @geetha290krm : O.K. I noted that. I am trying to show that.. – Plantation Mar 17 '24 at 06:12
  • @geetha290krm : My first strategy is, first noting that $f(x_0 +t)= f|{[x_0, x_0 +\pi]} \circ g$, where $g(t):= x_0 +t$ on $[0, \pi]$ and by second using answer of https://math.stackexchange.com/questions/3029512/integrability-of-composite-functions?noredirect=1&lq=1. To use the answer, maybe we need that $f|{[x_0, x_0 +\pi]}$ is integrable on $[x_0, x_0 +\pi]$. And I think that here is point that we may use the $2 \pi$-periodicity of $f$.. Working argument? Uhm..Can we really show that? Or is there any other ( simpler ) route, by cathing some point ? – Plantation Mar 17 '24 at 06:20
  • @geetha290krm : Note that $[x_0, x_0 + \pi]$ is the image of $g$ in my comment. Perhaps, does there exists $n \in \mathbb{Z}$ such that $[x_0, x_0 +\pi] \subseteq [-\pi + 2n \pi, \pi+2n \pi ]$ ? If so, since $f \in \mathcal{R}[-\pi,\pi]$ and has period $2 \pi$, $f$ is integrable on $[-\pi + 2n \pi, \pi+2n \pi ]$ (?) and we may apply the answer in the linked question in my comment.. – Plantation Mar 17 '24 at 06:38
  • @geetha290krm : To show the inclustion, it suffices to show the existence of $n\in \mathbb{Z}$ such that $(2n-1) \pi \le x_0 \le 2n \pi$. For any $x_0 \in \mathbb{R}$, such existence of $n \in \mathbb{Z}$ is true? – Plantation Mar 17 '24 at 06:48

1 Answers1

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From $h(t):= f(x_0+t)-f(x_0^{+})$, since $f(x_0^{+})$ is constant, it suffices to show that $f(x_0 +t)$ is Riemann integrable on $[0, \pi]$. We will use answer of Integrability of composite functions :

Lemma. If $f$ is a Riemann integrable function defined on $[a,b],\ g$ is a differentiable function with non-zero continuous derivative on $[c,d]$ and the range of $g$ is contained in $[a,b]$, then $f\circ g$ is Riemann integrable on $[c,d]$.

First, for each $n\in \mathbb{Z}$, we define $$U_n:=[(2n-1)\pi , 2n \pi] , U'_n:=[2n \pi, (2n+1)\pi] .$$

Note that $\mathbb{R}= \cup_{n\in \mathbb{Z}} U_n \cup U'_n$.

Second, note that since $f \in \mathcal{R}[=\pi ,\pi]$ and is periodic of period $2 \pi$, for each $n\in \mathbb{Z}$, $$f\in \mathcal{R}[-\pi + 2n\pi , \pi +2n\pi]= \mathcal{R}[(2n-1)\pi , (2n+1) \pi]. \tag{1}$$

And, from this (by cutting domain) we obtain $f\in \mathcal{R}[2n\pi, (2n+1)\pi]$. Furthermore, by replacing $n$ to $n+1$, we obtain $f\in \mathcal{R}[(2n+1)\pi, (2n+3)\pi]$, from which (again by cutting domain) we obtain $f\in \mathcal{R}[(2n+1)\pi, 2(n+1)\pi]$. By 6.2.3. theorem of the Manfred's book, we conclude that $$f\in \mathcal{R}[2n \pi , 2(n+1)\pi]. \tag{2} $$

Third, now fix $x_0 \in \mathbb{R}$ as in the statement of the theorem in our question. Then by the first argument there exists $n\in \mathbb{Z}$ such that $x_0 \in U_n$ or $x_0\in U'_n$.

  • Case 1) $x_0 \in U_n$ : In this case, $x_0 \in [(2n-1)\pi,2n\pi]$ so that $[x_0 , x_0+\pi] \subseteq [(2n-1)\pi , (2n+1)\pi]$, on which $f$ is integrable by the $(1)$. By noting that $f(x_0+t) = f|_{[(2n-1)\pi , (2n+1)\pi]} \circ g$ where $g(t):=x_0 +t$ and $[x_0 , x_0+\pi]$ is the image of $g$ on $[0,\pi]$, by the above lemma, $f(x_0+t) \in \mathcal{R}[0,\pi]$.

  • Case 2) $x_0 \in U_n'$ : In this case, $x_0 \in [2n \pi, (2n+1)\pi]$ so that
    $[x_0 , x_0+\pi] \subseteq [2n\pi , 2(n+1)\pi]$, on which $f$ is integrable by $(2)$. Then by similar argument for the case 1), we get $f(x_0+t) \in \mathcal{R}[0,\pi]$.

And we are done.

P.s. Is there any other simpler route to show the integrability, e.g., bypassing usage of the above Lemma?

Plantation
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