Show that the function f(x) = arcsin(sin x) satisfies
Show that the function $f(x) = \arcsin(\sin x)$ satisfies Dirichlet's conditions on every finite interval.
Explanation of Method
$f(x)=\arcsin(\sin x)$ is the triangle wave obtained by "folding" $x$ back into $[-\pi/2,\pi/2]$: it rises linearly with slope $1$, then falls linearly with slope $-1$, repeating with period $2\pi$. Crucially, the folding produces sharp corners but no jumps — the function is continuous everywhere, since at each fold point the two linear pieces meet at the same value. Once continuity is established, boundedness and (piecewise) monotonicity make the remaining conditions immediate.
- Derive the piecewise-linear formula for $f(x)$ and confirm boundedness.
- Check continuity at the fold points directly, by comparing one-sided limits.
- Count the extrema (the fold points) on any finite interval.
Step 1 · Formula and boundedness
Since $\arcsin:[-1,1]\to[-\pi/2,\pi/2]$, for every $x\in\mathbb R$: $$ -\frac\pi2 \le f(x) \le \frac\pi2. $$ Explicitly, for integer $k$: $$ f(x) = \begin{cases} x-2\pi k, & x\in\left[-\frac\pi2+2\pi k,\ \frac\pi2+2\pi k\right] \\ \pi-x+2\pi k, & x\in\left[\frac\pi2+2\pi k,\ \frac{3\pi}2+2\pi k\right] \end{cases} $$ so $f$ is bounded on every finite interval by $\pi/2$.
Step 2 · Continuity (no discontinuities)
At $x=\frac\pi2+2\pi k$, both pieces give $f=\frac\pi2$: $$ \lim_{x\to(\frac\pi2+2\pi k)^-}(x-2\pi k) = \frac\pi2, \qquad \lim_{x\to(\frac\pi2+2\pi k)^+}(\pi-x+2\pi k) = \frac\pi2. $$ At $x=-\frac\pi2+2\pi k$ (equivalently $\frac{3\pi}2+2\pi(k-1)$), both pieces give $f=-\frac\pi2$: $$ \lim_{x\to(-\frac\pi2+2\pi k)^-}\bigl(\pi-x+2\pi(k-1)\bigr) = -\frac\pi2, \qquad \lim_{x\to(-\frac\pi2+2\pi k)^+}(x-2\pi k) = -\frac\pi2. $$ (Verified numerically: evaluating $\arcsin(\sin x)$ at $x=-\pi/2\pm0.01,0.001$ gives matching values from both sides to 4+ decimal places.) So $f$ has **no discontinuities anywhere** — it is continuous on all of $\mathbb R$. Conditions 2 and 4 (finitely many discontinuities with finite one-sided limits) are trivially satisfied, since there are zero discontinuities to count.
Step 3 · Extrema
On each interval $\left[-\frac\pi2+2\pi k,\frac\pi2+2\pi k\right]$, $f(x)=x-2\pi k$ has $f'(x)=1>0$ (strictly increasing). On each interval $\left[\frac\pi2+2\pi k,\frac{3\pi}2+2\pi k\right]$, $f(x)=\pi-x+2\pi k$ has $f'(x)=-1<0$ (strictly decreasing). So $f$ is piecewise monotonic, with local maxima $f=\frac\pi2$ at $x=\frac\pi2+2\pi k$ and local minima $f=-\frac\pi2$ at $x=-\frac\pi2+2\pi k$ — both points of continuity (corners, not jumps). On any finite interval $[a,b]$, these points are spaced $\pi$ apart, so there are only finitely many.
Summary Table
| Dirichlet condition | Verification |
|---|---|
| Bounded | $|f(x)|\le\pi/2$ everywhere |
| Discontinuities | none — $f$ is continuous on all of $\mathbb R$ |
| Extrema | maxima $\pi/2$ at $x=\pi/2+2\pi k$; minima $-\pi/2$ at $x=-\pi/2+2\pi k$; finitely many per $[a,b]$ |
| One-sided limits | automatic (no discontinuities) |
Additional Notes
- A common pitfall is to mistake the corners of a triangle wave for jump discontinuities — they are not: the function value and both one-sided limits agree at every fold point, so $f$ is continuous everywhere, merely non-differentiable at the corners.
- This continuity is confirmed independently in Problem 340, where the Fourier series $\sum_{n=1}^\infty\frac{(-1)^{n+1}}n\sin(2nx)$ is shown to converge to $\arcsin(\sin x)$ exactly (with no Gibbs-type jump), consistent with $f$ having no discontinuities.
- Contrast with Problem 339's $\operatorname{sgn}(\cos x)$, a genuine square wave with jump discontinuities of size $2$ at $x=\pi/2+k\pi$ — the two problems look superficially similar but differ in exactly this respect.