Proof of Fejer’s Theorem
Theorem 1.5
(i) If \(f: \mathbb{T} \to \mathbb{C}\) is Riemann integrable then, if \(f\) is continuous at \(t\),
\[\sigma_{n}(f,t) = \sum_{r=-n}^{n} \frac{n+1-|r|}{n+1} \hat{f}(r) \exp irt \to f(t)\](ii) If \(f:\mathbb{T} \to \mathbb{C}\) is continuous then
\[\sigma_{n}(f,t) = \sum_{r=-n}^{n} \frac{n+1-|r|}{n+1} \hat{f}(r) \exp irt \to f(t)\]uniformly on \(\mathbb{T}\).
We start with the following remark. Suppose \(f: \mathbb{T} \to \mathbb{C}\) is Riemann integrable. Then
\[\begin{align} \sigma_{n}(f,t) &= \sum_{r=-n}^{n} \frac{n+1-|r|}{n+1} \hat{f}(r) \exp irt\\ &= \sum_{r=-n}^{n} \frac{n+1-|r|}{n+1}\frac{1}{2\pi} \bigg(\int_{\mathbb{T}} f(x)\exp(-irx)dx\bigg) \exp irt\\ &= \frac{1}{2\pi}\int_{\mathbb{T}} f(x)\sum_{r=-n}^{n}\frac{n+1-|r|}{n+1} \exp(ir(t-x))dx\\ &= \frac{1}{2\pi} \int_{\mathbb{T}} f(x)K_n(t-x)dx \end{align}\]where
\[K_n(s) = \sum_{r=-n}^{n} \frac{n+1 - |r|}{n+1} \exp irs\]Further, making the substitution \(y = t-x\), we have
\[\begin{align} \sigma_{n}(f,t) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x)K_n(t-x)dx\\ &= -\frac{1}{2\pi} \int_{t+\pi}^{t-\pi} f(t-y)K_n(y)dy\\ &= \frac{1}{2\pi} \int_{t-\pi}^{t+\pi}f(t-y)K_n(y)dy\\ &= \frac{1}{2\pi}\int_{\mathbb{T}}f(t-y)K_n(y)dy \end{align}\]We are therefore led to examine the structure of \(K_n\) in some detail.
Lemma 2.1.
\[\begin{align} K_n(s) &= \frac{1}{n+1} \Bigg( \frac{\sin \frac{(n+1)s}{2}}{\sin \frac{s}{2}} \Bigg)^2 \quad \quad [s \ne 0]\\ K_n(0) &= n+1 \end{align}\]Proof.
If \(s \ne 0\) then
\[\begin{align} &\sum_{r=-n}^{n} (n+1-|r|) \exp irs\\ =& \bigg(\sum_{k=0}^{n} \exp i (k - \frac{n}{2})s\bigg)^2 \end{align}\]\[\begin{align} &= \bigg( \exp - \dfrac{ins}{2} \sum_{k=0}^{n}\exp iks \bigg)^2\\ &= \bigg( \exp \bigg( -\dfrac{ins}{2} \bigg) \dfrac{1-\exp i(n+1)s}{1 - \exp is} \bigg)^2\\ &= \Bigg(\dfrac{\exp \bigg( - \dfrac{i(n+1)s}{2} \bigg) - \exp\bigg( \dfrac{i(n+1)s}{2} \bigg)}{\exp -\dfrac{is}{2} - \exp\dfrac{is}{2}} \Bigg)^2\\ &= \Bigg( \dfrac{\sin \dfrac{(n+1)s}{2}}{\sin \dfrac{s}{2}} \Bigg)^2 \end{align}\]Considering the number of situations of drawing two cards individually from two decks, the sum of two number gives \(x\).
If \(s = 0\) then \(K_n(0) = n+1\) by direct computation.
$\square$