I am reading the Manfred Stoll, Introduction to real analysis, proof of 9.5.8 theorem and stuck at some statement.
9.5.8. THEOREM Let $f$ be a continuous function on $[-\pi, \pi]$ with $f(-\pi)=f(\pi)$, and let $f'$ be piecewise continuous on $[-\pi,\pi]$. If $$f(x)=\frac{1}{2}a_0 + \Sigma_{n=1}^{\infty}(a_n \operatorname{cos}nx+b_n\operatorname{sin} nx) , x\in [-\pi,\pi] ,$$ is the Fourier series of $f$, then at each $x\in (-\pi, \pi)$ where $f''(x)$ exists, $$ f'(x)=\Sigma_{n=1}^{\infty}(-na_n\operatorname{sin}nx+nb_n\operatorname{cos}nx).$$
First I propose main question. It is asserted in the proof of 9.5.8. theorem in the Manfred's book.
Q. Let $f$ be a continuous function of $[-\pi , \pi]$ with $f( -\pi) =f(\pi)$, and let $f'$ be piecewise continuous on $[-\pi,\pi]$. Fix $x_0 \in (-\pi , \pi)$ such that $f''(x_0)$ exists. Then since $f'$ is continous at $x_0$, by the Dirichlet's theorem, $f'(x_0)= \lim_{n\to \infty} S_n(f')(x_0)$ ?, Here $S_n(f')(x_0)$ is the $n$th partial sum of the Fourier series of $f'$ at $x_0$. ( C.f. Manfred's book, p.404 )
Here, the Dirichlet's theorem is 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 $$ \lim_{n\to \infty} S_n(f)(x_0)= \frac{1}{2}[f(x_0^{+})+f(x_0^{-})],$$ where $S_n(f)(x_0)$ is the $n$th partial sum of Fourier series of $f$ at $x_0$.
If $f'$ in the above question satisfies the Dirichlet's condition, then since $f'$ is continuous at $x_0$, we get $\lim_{n\to \infty} S_n(f')(x_0)=f'(x_0).$ But I am stuck at showing that $f'$ satisfies the condition $(b)$ in the Dirichlet's theorem..
$\divideontimes$ In Manfred's book, p.420, he argues for next corollary as follows :
9.5.6 COROLLARY Suppose $f$ is a real-valued periodic function on $\mathbb{R}$ of period $2 \pi$. If $f$ and $f'$ are piecewise coninuous on $[-\pi,\pi]$, then $$ \lim_{n\to \infty} S_n(x) = \frac{1}{2}[f(x^{+}) + f(x^{-})]$$ for all $x\in \mathbb{R}$.
Proof. Returning to Dirichlet's theorem suppose $f$ is such that both $f'(x_0^-)$ and $f'(x_0^+)$ exist at the point $x_0$. If $f'$ is piecewise continous on $[-\pi,\pi]$, then there exists $x_1 > x_0$ such that $f$ and $f'$ are continuous on $(x_0 ,x_1)$. If $f$ is not continuous at $x_0$, redefine $f$ at $x_0$ as $f(x_0^+)$. Then $f$(redefined if necessary) is continuous on $[x_0, x_1)$, and thus by Theorem 5.2.11.,
5.2.11 THEOREM Suppose $f:[a,b) \to \mathbb{R}$ is continuous on $[a,b)$ and differentiable on $(a,b)$. If $\lim_{x\to a^{+}}f'(x)$ exists, then $f'_{+}(a)$ ( the right derivative ; his book, p.168 ) exists and $f'_{+}(a) = \lim_{x\to a^{+}}f'(x) =: f'(a^{+})$ ,
$$\lim_{t\to 0^{+}} \frac{f(x_0+t)-f(x_0^{+})}{t}=f'(x_0^{+}).$$
As a consequence, there exists a constant $M$ and $\delta >0$ such that $$ |f(x_0 +t)-f(x_0^+) | \le Mt $$
for all $t, 0<t<\delta$. Similarly, the existence of $f'(x_0^-)$ implies the existence of a constant $M$ and a $\delta>0$ such that $$ |f(x_0-t)-f(x_0^-)| \le Mt $$
for all $t, 0<t<\delta$. Thus $f$ satisfies the hypothesis of Dirichlet's theorem. On combining the above discussion with Theorem 9.5.3 we are done. QED.
Can we try to imitate this proof? A technical issue is that when we substitute the $f'$ in my original question to the argument for 9.5.6 corollary, $f''$ may not be piecewise continuous on $[-\pi, \pi]$. We have just that $f''(x_0)$ exists ( and $f'$ is piecewise continuous on $[-\pi , \pi]$. ) Is it possible to satisfy the Dirichlet's condition (b) by these conditions only?
Can anyone help?
Let me answer for my original question. Let $I:=[-\pi,\pi]$. Since $f'$ is (right) continuous at $x_0 \in \operatorname{int}I:=(-\pi, \pi) $, $f'(x_0)=f'(x_0^+)$ ( $\because$ Manfred's book 4.4.3. theorem ) So,
$$ \lim_{t\to 0} \frac{f'(x_0+t)-f'(x_0^+)}{t} = \lim_{t\to 0} \frac{f'(x_0+t)-f'(x_0)}{t} =: f''(x_0)$$
So, for $\epsilon >0 $, there exists $\delta>0$ such that if $0< |t-0| < \delta$, then $| \frac{f'(x_0+t) - f'(x_0^+)}{t} - f''(x_0) | < \epsilon$ so that $|f'(x_0 +t) - f'(x_0^+)| < |t|(|f''(x_0)| + \epsilon)$. Letting $M:=|f''(x_0)| + \epsilon$, we have
$$ |f'(x_0 +t)-f'(x_0^+) | \le Mt $$
for all $t, 0<t<\delta$.
Similarly, we also can show the existence of a constant $M$ and a $\delta>0$ such that $$ |f'(x_0-t)-f'(x_0^-)| \le Mt $$ for all $t, 0<t<\delta$. Thus $f'$ satisfies the hypothesis of Dirichlet's theorem.
(Correct?)