Riemann–Lebesgue Lemma

Since $f$ is 1 on $[0, \pi]$ and 0 on $(\pi, 2\pi]$, $b_n = \frac{1}{\pi}\int_0^{\pi} \sin(nx)\,dx$. With $\int \sin(nx)\,dx = -\cos(nx)/n$, we get $b_n = \frac{1}{\pi}\left[-\frac{\cos(nx)}{n}\right]_0^{\pi} = \frac{1 - \cos(n\pi)}{n\pi}$, and $\cos(n\pi) = (-1)^n$ gives $b_n = \frac{1 - (-1)^n}{n\pi}$.

Since $|1 - (-1)^n| \le 2$, $|b_n| \le \frac{2}{n\pi} \to 0$ as $n \to \infty$.

Meanwhile $a_n = \frac{1}{\pi}\int_0^{\pi} \cos(nx)\,dx = \frac{\sin(n\pi)}{n\pi} = 0$ — exactly 0 for every $n \ge 1$.

Hence $a_n \to 0$ and $b_n \to 0$ are both proved.

The scene's step mode reproduces this formula by numerical quadrature: the odd stems ride the $2/(n\pi)$ curve downward while the even stems sit at 0 up to quadrature error. A discontinuous function's coefficients go at rate $1/n$ — slower than for smooth functions (compare the tent mode).

By linearity of the integral, $b_n(s) = \sum_{k=1}^{m} c_k\,b_n(1_{[x_{k-1}, x_k)})$. Each interval indicator contributes $b_n(1_{[x_{k-1}, x_k)}) = \frac{1}{\pi}\int_{x_{k-1}}^{x_k} \sin(nx)\,dx = \frac{\cos(nx_{k-1}) - \cos(nx_k)}{n\pi}$.

Since $|\cos(nx_{k-1}) - \cos(nx_k)| \le 2$, $\left|b_n(1_{[x_{k-1}, x_k)})\right| \le \frac{2}{n\pi}$ — a uniform bound independent of the interval's position.

Hence $|b_n(s)| \le \sum_{k=1}^{m} |c_k|\,\frac{2}{n\pi} = \frac{C}{n}$ (with $C = \frac{2}{\pi}\sum |c_k|$ independent of $n$), which goes to 0 as $n \to \infty$.

The same computation with $\cos$ in place of $\sin$ sends $a_n(s) \to 0$.

The point is finiteness: with only finitely many intervals, adding up the $O(1/n)$ contributions still gives $O(1/n)$. This is the platform the density argument to continuous functions (problem 3 below) stands on.

Lemma (uniform approximation by step functions): if $f$ is continuous on $[0, 2\pi]$ it is uniformly continuous, so for every $\varepsilon > 0$ there is $\delta > 0$ with $|x - y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon$.

Split $[0, 2\pi]$ by a partition $0 = x_0 < x_1 < \cdots < x_m = 2\pi$ (each width ${} < \delta$), with intervals $I_k = [x_{k-1}, x_k)$ for $k < m$ and the last interval $I_m = [x_{m-1}, 2\pi]$ closed at the right endpoint, so the partition covers all of $[0, 2\pi]$ including $x = 2\pi$.

Define the step function $s$ taking the constant value $f(x_{k-1})$ (left endpoint) on each $I_k$ (with $s(2\pi) = f(x_{m-1})$ as well).

For $x \in I_k$, $|x - x_{k-1}| < \delta$, so $|f(x) - s(x)| < \varepsilon$ — this holds at every point of $[0, 2\pi]$, i.e. $\|f - s\|_\infty < \varepsilon$.

Lemma proved.

Fix $\varepsilon > 0$ and pick a step function $s$ with $\|f - s\|_\infty < \varepsilon$ by the lemma.

By the triangle inequality $|b_n(f)| \le |b_n(f - s)| + |b_n(s)|$.

The first term satisfies $|b_n(f - s)| \le \frac{1}{\pi}\int_0^{2\pi} |f - s|\,|\sin(nx)|\,dx \le \frac{1}{\pi}\,\varepsilon \cdot 2\pi = 2\varepsilon$ — independent of $n$.

The second term $|b_n(s)| \to 0$ as $n \to \infty$ by problem 2.

So choose $N$ large enough that $|b_n(s)| < \varepsilon$ for $n \ge N$; for those $n$, $|b_n(f)| < 2\varepsilon + \varepsilon = 3\varepsilon$.

Since $\varepsilon$ is arbitrary, $b_n(f) \to 0$.

The same argument with $|\cos(nx)| \le 1$ gives $a_n(f) \to 0$.

Every step is complete — nothing is left at idea level.