du Bois-Reymond lemma

Remarks

Let $c = \frac{1}{b - a} \int_a^b f \, dx$ be the mean of $f$, and set $\eta(x) = \int_a^x (f(s) - c) \, ds$. Then $\eta$ is $C^1$, $\eta(a) = \eta(b) = 0$ and $\eta' = f - c$. The hypothesis gives $\int_a^b f (f - c) \, dx = 0$, and $\int_a^b c (f - c) \, dx = 0$ by the choice of $c$. Subtracting, $\int_a^b (f - c)^2 \, dx = 0$, so $f = c$ by continuity.

A common extension: if $f, g$ are continuous and $\int_a^b (g \eta + f \eta') \, dx = 0$ for all such $\eta$, then $f$ is $C^1$ and $f' = g$. To see it, apply the lemma to $f - G$, where $G(x) = \int_a^x g(s) \, ds$, after an integration by parts of $\int_a^b g \eta \, dx$.

Example: for $f(x) = x$ on $[0, 1]$ and $\eta(x) = x(1 - x)$, one finds $\int_0^1 x (1 - 2x) \, dx = -1/6 \ne 0$, as the lemma predicts for a non-constant $f$.

Going deeper

For a student

The extension above gives the Euler–Lagrange equation in integrated form, $L_p(x, y, y') = \int_a^x L_z(s, y, y') \, ds + C$, for a $C^1$ minimiser $y$, without assuming that $y$ has a second derivative. It follows that $x \mapsto L_p(x, y(x), y'(x))$ is $C^1$, which is the start of regularity theory in one variable. The fundamental lemma is the version where $\eta$ itself, instead of $\eta'$, appears in the integral.