Proof. Let \[S_n(x) = \frac {a_0}{2} + \sum ^n_{k = 1} \left [a_k\cos kx + b_k\sin kx\right ]\] where the \(a_k\) and \(b_k\) are the Fourier coefficients. Then \begin {align*} \int _{-\pi }^{\pi }f(x)\, S_n(x)\, dx & = \frac {a_0}{2}\int _{-\pi }^{\pi }f(x)\, dx + \sum ^n_{k = 1}\left [a_k\int _{-\pi }^{\pi } f(x)\, \cos kx\, dx + b_k \int _{-\pi }^{\pi }f(x)\, \sin kx \, dx\right ]\\ & = \frac {\pi a_0^2}{2} + \pi \sum _{k = 1}^n(a^2_k + b_k^2). \end {align*}

It can also be shown that \[\int _{-\pi }^{\pi }S_n(x)\, dx = \frac {\pi \, a^2_0}{2} + \pi \, \sum _{k = 1}^n(a_k^2 + b_k^2).\] Thus \begin {align*} \int _{-\pi }^{\pi }[f(x) - S_n(x)]^2\, dx & = \int _{-\pi }^{\pi }f^2(x)\, dx - 2\int _{-\pi }^{\pi } f(x)\, S_n(x)\, dx + \int _{-\pi }^{\pi } S^2_n(x)\, dx\\ & = \int _{-\pi }^{\pi } f^2(x)\, dx - \left [\frac {\pi \, a^2_0}{2} + \pi \sum ^n_{k = 1}(a^2_k + b_k^2)\right ] \geq 0. \end {align*}

Thus \[\frac {\pi \, a_0^2}{2} + \pi \, \sum _{k = 1}^n(a^2_k + b_k^2) \leq \int ^{\pi }_{-\pi } f^2 (x)\, dx.\] Since \(f(x)\) is integrable, so is \(f^2(x)\). Hence, \(\int _{-\pi }^{\pi }f^2(x)\, dx\,\) exists, and thus \(\sum ^n_{k = 1} (a^2_k + b^2_k)\,\) is a bounded and monotone increasing sequence of partial sums hence, (the corresponding series) \(\sum _{k = 1}^{\infty }(a_k^2+b_k^2)\hspace {0.6cm}\text {convergence}\). By property of convergence of series \[\lim _{k\rightarrow \infty }\left (a^2_k + b^2_k\right ) = 0.\] Hence, \(\lim _{k\rightarrow \infty } a_k = 0\) and \(\lim _{k\rightarrow \infty } b_k = 0\). If \(\lim _{k\rightarrow \infty } a^2_k = 0\), then \(\lim _{k\rightarrow \infty } a_k = 0\). □

Proof. We have that \[\Phi (x)\, \sin \left [\left (n + \frac {1}{2}\right )x\right ] = \left [\Phi (x)\, \cos \frac {x}{2}\right ] \sin nx + \left [\Phi (x)\sin \frac {x}{2}\right ] \cos nx.\] Let \(\Phi _2(x) = \Phi (x)\, \cos \frac {x}{2},\,\, \hspace {0.2cm} \Phi _1(x) = \Phi (x)\, \sin \frac {x}{2}\). Then \(\Phi _1(x)\) and \(\Phi _2(x)\) are Riemann integrable on \([-\pi , \pi ]\). By Lemma 5.0.5 \[\lim _{n\rightarrow \infty } \int _{-\pi }^{\pi } \Phi _1(x)\, \cos nx\, dx = 0 = \lim _{n\rightarrow \infty }\int _{-\pi }^{\pi } \Phi _2(x)\, \sin nx\, dx.\] Adding them up, we get the result. □

Proof. Define the functions \(h_1(x)\) and \(h_2(x)\) by \[h_1(x) = \begin {cases} 0, & 0\leq x \leq \pi \\ \Phi (x), & -\pi \leq x < 0\\ \end {cases}\]

\[h_2(x) = \begin {cases} \Phi (x), & 0 \leq x \leq \pi \\ 0, & -\pi \leq x < 0 \end {cases}\] \(h_1(x)\) and \(h_2(x)\) are Riemann integrable. Hence, by Corollary 5.4.2 \begin {align*} \lim _{n\rightarrow \infty }\int _{-\pi }^0\Phi (x)\, \sin \left [\left (n + \frac {1}{2}\right )x\right ]\, dx = \int _{-\pi }^{\pi } h_1(x)\, \sin \left [\left (n + \frac {1}{2}\right )x\right ]\, dx & = 0\\\\ \lim _{n\rightarrow \infty }\int ^{\pi }_0\Phi (x)\, \sin \left [\left (n + \frac {1}{2}\right )x\right ]\, dx = \int _{-\pi }^{\pi } h_2(x)\, \sin \left [\left (n + \frac {1}{2}\right )x\right ]\, dx & = 0. \end {align*} □

Proof. Let \[G_n(u) = \frac {1}{2} + \sum ^n_{k = 1}\cos ku\] multiply both sides by \(2\sin \frac {u}{2}\) and using the identity \[2\sin \frac {u}{2}\cos ku = \sin \left [\left (\frac {k + 1}{2}\right )u\right ] - \sin \left [\left (k - \frac {1}{2}\right )u\right ].\]

\[2\sin \frac {u}{2}\, G_n(u) = \sin \frac {u}{2} + \sum ^n_{k = 1} \left [\sin \left (k + \frac {1}{2}\right )u - \sin \left [\left (k - \frac {1}{2}\right )u\right ]\right ] = \sin \left [\left (n + \frac {1}{2}\right )u\right ].\] Dividing, get \[G_n(u) = \frac {\sin \left [\left (n + \frac {1}{2}\right )u\right ]}{2\sin \left (\frac {u}{2}\right )}\, , \hspace {0.4cm} \text {if}\hspace {0.3cm} \sin \frac {u}{2}\neq 0.\] □