Volterra's Function

Write $u = x - a$, $v = b - x$ (so $u + v = b - a$ and $u, v > 0$); inside, $\varphi = u^2 v^2 \sin\frac{1}{u}$.

The endpoint values follow by squeeze: $\lvert u^2 v^2 \sin\frac{1}{u} \rvert \le u^2 v^2 \to 0$ as $x \to a^+$ or $x \to b^-$.

Outside the interval $\varphi \equiv 0$, so $\varphi$ is continuous on $[0,1]$ with $\varphi(a) = \varphi(b) = 0$.

Inside, differentiate directly with the product and chain rules: $\frac{d}{dx}[u^2 v^2] = 2uv^2 - 2u^2v = 2uv(v-u)$ (using $\frac{dv}{dx} = -1$), and $\frac{d}{dx}\sin\frac{1}{u} = -\frac{1}{u^2}\cos\frac{1}{u}$, giving $\varphi'(x) = 2uv(v-u)\sin\frac{1}{u} - v^2\cos\frac{1}{u}$ (the display formula above).

At the endpoints use the difference quotient, not the limit of the formula above.

For $h > 0$, $\varphi(a+h) = h^2(b-a-h)^2\sin\frac{1}{h}$, so $\frac{\varphi(a+h) - \varphi(a)}{h} = h(b-a-h)^2\sin\frac{1}{h}$, squeezed to $0$ by $h(b-a)^2$, so $\varphi'(a) = 0$.

Likewise $\frac{\varphi(b-h) - \varphi(b)}{-h} = -h(b-h-a)^2\sin\frac{1}{\,b-h-a\,} \to 0$, so $\varphi'(b) = 0$.

But $\varphi'$ is not continuous at $a$: at $x_m = a + \frac{1}{2\pi m}$ we have $\sin\frac{1}{x_m - a} = 0$ and $\cos\frac{1}{x_m - a} = 1$, so $\varphi'(x_m) = -(b - x_m)^2 \to -(b-a)^2 \ne 0 = \varphi'(a)$.

The oscillatory term $-(b-x)^2\cos\frac{1}{x-a}$ keeps amplitude $(b-a)^2$ as $x \to a^+$ and never dies. (At the right endpoint $b$, the factor $(b-x)^2 \to 0$ kills the oscillation, so $\varphi'$ is continuous there.)

Hence each bump is differentiable everywhere with value and derivative $0$ at the endpoints, but not $C^1$ — and exactly this failure is what makes $f'$ discontinuous on the Cantor set in Problem 3.

Each $g_k = 2^{-k}\varphi_k$ is differentiable on $[0,1]$ by Problem 1 (endpoints included, derivative $0$).

With $u, v \ge 0$ and $u + v = b_k - a_k \le 1$, AM-GM gives $u^2 v^2 \le \left(\frac{u+v}{2}\right)^4 \le \frac{1}{16}$, so $\lvert g_k \rvert \le 2^{-k}/16$.

Since $\sum 2^{-k}/16 < \infty$, the M-test makes $\sum g_k$ converge uniformly on $[0,1]$ — the limit $f$ is continuous.

The derivatives obey the same treatment: from $\lvert\varphi'_k\rvert \le 2uv\lvert v-u\rvert + v^2$, using $uv \le \left(\frac{u+v}{2}\right)^2 \le \frac{1}{4}$, $\lvert v - u \rvert \le u + v \le 1$, and $v^2 \le 1$, we get $\lvert\varphi'_k(x)\rvert \le \frac{1}{2} + 1 = \frac{3}{2}$ — uniform in $k$ and $x$. Hence $\lvert g'_k \rvert \le 2^{-k}\cdot\frac{3}{2}$, and the M-test makes $\sum g'_k = \sum 2^{-k}\varphi'_k$ converge uniformly as well.

Now the standard theorem: if a series of differentiable functions converges at one point and the series of derivatives converges uniformly, the sum is differentiable and its derivative is the sum of the derivatives. Here the M-test uniform convergence in step 1 satisfies the 'one point convergence' hypothesis — $\sum g_k$ converges uniformly on all of $[0,1]$, hence in particular at a point (e.g. $0$).

For interior points, apply the theorem on every closed subinterval $[0, 1-\delta]$ and $[\delta, 1]$ inside $(0,1)$. For the endpoints $0, 1$, apply the same M-test bound directly to difference quotients: the uniform bound on the tail $\sum_{k>K} 2^{-k}\varphi'_k$ also dominates the tail of the difference quotients, so the one-sided derivatives at the endpoints are $f'(0) = \sum_k 2^{-k}\varphi'_k(0)$ and $f'(1) = \sum_k 2^{-k}\varphi'_k(1)$.

Hence $f$ is differentiable at every point of $[0,1]$ (one-sided at the endpoints) with $f'(x) = \sum_k 2^{-k}\varphi'_k(x)$, bounded by $\lvert f'(x) \rvert \le \sum_k 2^{-k}\cdot\frac{3}{2} = \frac{3}{2}$. Numerically: $\sup\lvert f' \rvert \approx 0.0307$ on the stage-4 grid.

Corollary: at any $c$ in the Cantor set $C$, every $\varphi_k$ and $\varphi'_k$ vanishes (as $c$ lies outside all complementary intervals), so $f(c) = 0$ and $f'(c) = 0$. That is, $f'$ is identically $0$ on $C$ — yet, as Problem 3 shows, discontinuous at every point of $C$.

Lemma: every neighborhood of $c \in C$ contains a whole complementary interval. Let $I_n$ be the stage-$n$ surviving closed interval containing $c$; then $\lvert I_n \rvert \to 0$, and at stage $n$ the middle chunk removed from $I_{n-1}$ is a complementary interval $J_n = (a_k, b_k)$ of length $4^{-n}$.

For $n$ with $\lvert I_{n-1} \rvert < \delta$, all of $J_n \subset I_{n-1}$ lies within $\delta$ of $c$.

So $C$ has empty interior (every open interval eventually has its middle removed).

On such an interval $(a, b) = (a_k, b_k)$, $f' = 2^{-k}\varphi'_k$ (every other term is $0$). With $x_m = a + \frac{1}{\pi m}$, we have $\sin\frac{1}{x_m - a} = \sin(\pi m) = 0$ and $\cos\frac{1}{x_m - a} = (-1)^m$, so $\varphi'_k(x_m) = -(b-x_m)^2(-1)^m$ and $f'(x_m) = 2^{-k}(b-x_m)^2(-1)^{m+1} \to \pm 2^{-k}(b-a)^2$ (sign alternating with the parity of $m$).

But Problem 2's corollary gives $f'(c) = 0$. Since $x_m \to a$ and $a$ is arbitrarily close to $c$, there are points arbitrarily close to $c$ where $f'$ takes values both above $+2^{-k}(b-a)^2/2$ and below $-2^{-k}(b-a)^2/2$ — it cannot converge to $f'(c) = 0$.

Hence $f'$ is discontinuous at $c$, and $c \in C$ was arbitrary: $f'$ is discontinuous at every point of $C$.

Numerically: on the first interval $(0.375, 0.625)$, a dense grid near $x \to 0.375^+$ shows $\max f' - \min f' \approx 0.0625 = 2\cdot\frac12\cdot(0.25)^2$.

Finally we cite Lebesgue's criterion as a theorem (its proof is measure theory, so we keep it at idea level — see the measure/lebesgue scenes): a bounded function on $[a,b]$ is Riemann integrable iff it is continuous almost everywhere, i.e. its discontinuity set has Lebesgue measure $0$.

Here $f'$ is bounded (Problem 2), its discontinuity set contains $C$, and $C$ is closed with measure $1/2$ (the complement's total length is $\lim_{N\to\infty}\frac{1-2^{-N}}{2} = \frac12$). A discontinuity set of positive measure means $f'$ is not Riemann integrable.

The punchline: a function can be differentiable everywhere with bounded derivative, yet its derivative need not be Riemann integrable.