Answer: The Absolutely Continuous Function That Forgets Its Derivative

Key Idea / Intuition

The key point is that absolute continuity lets us use the Fundamental Theorem of Lebesgue integration: $f(x) = \int_0^x f'(t)\,dt$. Once we have this, the condition $f' = f$ a.e. turns the functional equation $f(x) = \int_0^x f(t)\,dt$ into a Gronwall-type inequality. Gronwall's inequality (or a simple iteration argument) then forces $f \equiv 0$ — the function is "too small to be nonzero."

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Formal Proof / Solution

Step 1: Use absolute continuity.

Since $f$ is absolutely continuous and $f(0)=0$, by the Lebesgue FTC: $f(x) = \int_0^x f'(t)\,dt = \int_0^x f(t)\,dt \quad \text{for all } x \in [0,1].$

Step 2: Bound $|f|$.

Let $M = \sup_{x \in [0,1]} |f(x)|$. Since $f$ is continuous (absolute continuity implies continuity) on a compact set, $M < \infty$. Then for all $x \in [0,1]$: $|f(x)| \leq \int_0^x |f(t)|\,dt \leq M \cdot x.$

Step 3: Iterate the estimate.

Substitute this improved bound back: $|f(x)| \leq \int_0^x M \cdot t \,dt = M \cdot \frac{x^2}{2}.$

Iterate $n$ times: $|f(x)| \leq M \cdot \frac{x^n}{n!}.$

Step 4: Conclude.

For any fixed $x \in [0,1]$, taking $n \to \infty$: $|f(x)| \leq M \cdot \frac{1}{n!} \to 0.$

Therefore $f(x) = 0$ for all $x \in [0,1]$. $\blacksquare$

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Remark (why AC is essential): A function that is merely differentiable a.e. with $f' = f$ a.e. could potentially be pathological — for instance, the Cantor function has $f' = 0$ a.e. but is not identically zero, precisely because it is not absolutely continuous. Absolute continuity is exactly the condition that makes the Lebesgue FTC valid, tying $f$ to its derivative via an integral.