The properties given for PN help put each partial sum SN in a form that will make it easier to work with. First of all, in the form of SN above, we change variables to u = t-x and use the fact that PN is even to get $S_N(x) = \int_{-\pi-x}^{\pi-x} f(u+x)P_N(u)du$. We will need the lemma below, which is Lemma 1.3, pg. 44, in the text:
Lemma. Let $F$ be a $2\pi$ periodic function that is integrable on each bounded interval in $\mathbb R$. Then, for any real number $c$, $\int_{-\pi+c}^{\pi+c} F(u)du$ is independent of $c$. In particular, $\int_{-\pi+c}^{\pi+c} F(u)du=\int_{-\pi}^\pi F(u)du$.
Proof: See Lemma 1.3, pg. 44, and exercise 25, pg. 99. $\square$
Fix $x$. Both $P_N(u)$ and $f(x+u)$ are $2\pi$-periodic in $u$, and so by the Lemma with $c=-x$, \[ S_N(x) = \int_{-\pi-x}^{\pi-x}f(u+x)P_N(u)du= \int_{-\pi}^{\pi}f(u+x)P_N(u)du. \] We can change variables in the integral going from $u$ to $-u$. The result is that \[ S_N(x) = -\int_{\pi}^{-\pi} f(x-u)P_N(-u)du = \int_{-\pi}^\pi f(x-u)P_N(u)du. \] (Recall that $P_N$ is even.) Add these two expressions together and divide by 2. This gives us \[ S_N(x) = \int_{-\pi}^{\pi}\frac12 (f(x+u)+f(x-u))P_N(u)du \] The integrand in the result, $(f(x+u)+f(x-u))P_N(u)/2$, is even. Using the fact that an integral of an even function over a symmetric interval is twice the integral of the function over half the interval, we get, after canceling factors of 2, that \[ (\dagger) \quad S_N(x) = \int_0^\pi (f(x+u)+f(x-u))P_N(u)du. \] The next ingredient in our proof is the Riemann-Lebesgue Lemma, which we now state:
Riemann-Lebesgue Lemma. If $f$ is piecewise continuous on an interval $[a,b]$, then $ \lim_{\lambda\to \pm\infty} \int_a^bf(x)\cos(\lambda x)dx = \lim_{\lambda\to \pm\infty} \int_a^bf(x)\sin(\lambda x)dx = \lim_{\lambda\to \pm \infty} \int_a^bf(x)e^{i\lambda x}dx= 0$.
Proof: We will prove the cosine version for the case in which $f$ is a continuously differentiable function on $[a,b]$. First, we integrate by parts; this gives us \[ \int_a^b f(x)\cos(\lambda x)dx = \frac{1}{\lambda}(f(b)\sin(\lambda b) -f(a)\sin(\lambda a)) - \frac{1}{\lambda}\int_a^b f'(x)\sin(\lambda x)dx. \] From this and the triangle inequality ($|A+B|\le |A|+|B|$), follows that \[ (\ddagger)\quad \bigg|\int_a^b f(x)\cos(\lambda x)dx\bigg| \le \frac{1}{|\lambda|}\bigg(|f(b)\sin(\lambda b)|+|f(a)\sin(\lambda a)|+\big|\int_a^b f'(x)\sin(\lambda x)dx\big|\bigg). \] Since $|\sin(x)|\le 1$ for all $x$, we have $|f(b)\sin(\lambda b)|=|f(b)||\sin(\lambda b)|\le |f(b)|$; similarly, $|f(a)\sin(\lambda a)|\le |f(a)|$. From calculus, we recall that for a function $g$, the integral $\int_a^b g(x)dx$ is the area above the $x$-axis minus that below the $x$-axis, but the integral $\int_a^b|g(x)|dx$ is the sum of these two areas. Hence, $\int_a^b|g(x)|dx\ge \big|\int_a^b g(x)dx\big|$. This fact implies that \[ \big|\int_a^b f'(x)\sin(\lambda x)dx\big|\le \int_a^b |f'(x)|\underbrace{|\sin(\lambda x)|}_{\le 1}dx \le \int_a^b|f'(x)|dx. \] Using these inequalities in $(\ddagger)$ results in our final inequality: \[\bigg|\int_a^b f(x)\cos(\lambda x)dx\bigg| \le \frac{1}{|\lambda|}\bigg(|f(b)|+|f(a)|+\int_a^b |f'(x)|dx\bigg)=\frac{C}{|\lambda|}, \] where the constant $C$ does not depend on $\lambda$, so we may apply the "squeeze principle" above. From this, letting $\lambda\to \pm\infty$ yields $\lim_{\lambda \to \infty}\int_a^b f(x)\cos(\lambda x)dx=0$. which proves the lemma for the special case of $f$ being continuously differentiable. To do the full theorem requires more work. $\square$
Pointwise Convergence Theorem. If f is a 2π-periodic piecewise continuous function that has a right-hand derivative $f'(x+)$ and a left hand derivative $f'(x-)$ at $x$, then \[ \lim_{N\to \infty}S_N(x) = \left\{ \begin{array}{cl} f(x) & \text{if }f \ \text{is continuous at }x.\\ \tfrac12 (f(x+)+f(x-)) & \text{if }f\ \text{has a jump discontinuity at } x. \end{array} \right. \] Proof: The trick here is to put the the error EN(x) =SN(x) - f(x) or EN(x) =SN(x) - (f(x+)+f(x-))/2 in the right form. Since in the continuous case one just has f(x+)=f(x-)=f(x), we can do both cases using $E_N(x) = S_N(x) - \frac12 (f(x+)+f(x-))$. From the third property of PN listed above, $\int_0^\pi P(u)du = \frac12$, so \[ E_N(x)= S_N(x) - \frac12 (f(x+)+f(x-)) = S_N(x)-\int_0^\pi (f(x+)+f(x-)) P(u)du.\] Now, using the final form of $S_N$ given in $(\dagger$), we obtain \[E_N(x) = \int_0^\pi \big(f(x+u) - f(x+) + f(x-u) - f(x-)\big)P_N(u)du.\] Finally, applying property 4 of $P_N(u)$, we can put the error in the form \[ \begin{aligned} E_N(x) =& \int_0^\pi \bigg(\frac{f(x+u) - f(x+)}{2\pi \sin(u/2)} - \frac{f(x-)-f(x-u) }{2\pi \sin(u/2)}\bigg) \sin((N+\tfrac12)u)du \\ =& \int_0^\pi F(u) \sin((N+\tfrac12)u)du, \end{aligned} \] where $F(u) := \frac{f(x+u) - f(x+)}{2\pi \sin(u/2)} - \frac{f(x-)-f(x-u) }{2\pi \sin(u/2)}$. Since x is fixed, we can ignore the dependence of the expression above on $x$. L'Hospital's rule and the fact that $f$ has right and left derivatives at x imply that $\lim_{u \downarrow 0} F(u) = (f'(x+) - f'(x-))/\pi$, so $F$ has a right hand limit at u = 0. This and the piecewise continuity of $f$ and $1/\sin(u/2)$ on the half-open interval $(0,\pi]$ mean that $F(u)$ is piecewise continuous on $[0,\pi]$. From the Riemann Lebesgue Lemma we have that $\lim_{N\to \infty}E_N(x) = 0.$ $\ \square$
Examples. Our first example is for $f(x) = \left\{ \begin{array}{cl} 1 & 0 \le x \le \pi, \\ -1 & -\pi < x< 0 \end{array} \right.$. We did the Fourier series for $f$ in class, but we will repeat the calculation here. The function $f$ is odd, so all of the $a_n$'s are $0$. We also have \[b_n=\frac{2}{\pi}\int_0^\pi \sin(nx)dx= \frac{2}{\pi}\bigg(-\frac{\cos(nx)}{n} \bigg|_0^\pi\bigg)= \frac{2}{n\pi}(1-\cos(n\pi))=\frac{2}{n\pi}(1-(-1)^n).\] For even $n$, $b_n=0$, while for odd $n$, $b_n=\frac{4}{n \pi}$. Letting $n=2k-1$, we have that the Fourier series for $f$ satisfies \[ \sum_{n\ \text{odd}} \frac{4}{n \pi}\sin(nx) = \sum_{k=1}^\infty \frac{4}{(2k-1) \pi}\sin\big((2k-1)x\big) = \left\{ \begin{array}{cl} f(x) = 1, & \text{for } \ 0< x < \pi \\ \frac12 \big(f(0+)+ f(0-)\big) = 0,& \text{for } x=0 \\ f(x) = -1, & \text{for } -\pi < x< 0. \end{array} \right. \] As an illustration of what this means, let's set $x = \pi/2 $. For this value of $ x $, we have $1 = \sum_{k=1}^\infty \frac{4}{(2k-1) \pi}\sin\big((2k-1)\pi/2\big)$. Since $\sin\big(\frac{(2k-1)\pi}{2}\big)= (-1)^{k+1}$, we have, after multiplying both sides by $\pi/4$, \[ \frac{\pi}{4} = \sum_{k=1}^\infty \frac{4(-1)^{k+1}}{2k-1} = 1 - \frac13 + \frac17 - \frac19 + \cdots, \] which is a famous series that was used to compute $\pi$. Since the series has terms that alternate and decrease in absolute value, the error made truncating the series after $K$ terms is $\frac{1}{2K+1}$. Thus to compute $\pi/4$ to within $0.01$, we would need to take $K\ge 50$.
For our second example, we will use the series for $x^2$, which was derived in problem 1 in chapter 1. \[x^2= \frac{\pi^2}{3} + \sum_{n=1}^\infty \frac{4(-1)^n}{n^2}\cos(nx)\] which holds for all $x\in [-\pi,\pi]$. We want to find $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$. To do this, set $x=0$ in the equation above: \[ 0= \frac{\pi^2}{3} + \sum_{n=1}^\infty\frac{4(-1)^n}{n^2}= \frac{\pi^2}{3} + \sum_{n=1}^\infty\frac{4(-1)^n}{n^2}= \frac{\pi^2}{3} - \sum_{n=1}^\infty\frac{4(-1)^{n+1}}{n^2}, \] where we used $(-1)^n = - (-1)^{n+1}$. Solving for the series that we want gives us \[ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}. \]