Peano kernel theorem

Remarks

Proof idea: write $f$ by Taylor's theorem with the integral remainder at $a$, as the Taylor polynomial of degree $k$ plus $\frac{1}{k!} \int_a^b (x - \theta)_+^k f^{(k+1)}(\theta)\,d\theta$. The functional kills the polynomial part. The special form of $L$ allows $L$ to be moved inside the integral over $\theta$, which gives the formula.

If $K$ does not change sign on $[a, b]$, the mean value theorem for integrals gives $L(f) = \frac{f^{(k+1)}(\xi)}{(k+1)!}\, L\big(x^{k+1}\big)$ for some $\xi \in [a, b]$. Also $\abs{L(f)} \le \norm{K}_{L^1} \, \norm{f^{(k+1)}}_\infty$ in all cases.

Example: the error of the trapezoidal rule, $L(f) = \int_a^b f - \frac{b - a}{2} \big( f(a) + f(b) \big)$, with $k = 1$. Here $K(\theta) = \int_\theta^b (x - \theta)\,dx - \frac{b - a}{2} (b - \theta) = -\frac12 (\theta - a)(b - \theta)$, which is $\le 0$. Since $\int_a^b K = -\frac{(b - a)^3}{12}$, one gets $L(f) = -\frac{(b - a)^3}{12} f''(\xi)$ for some $\xi$.

The hypothesis $L(p) = 0$ for degree at most $k$ is needed: for $L(f) = f(0)$ on $[0, 1]$ and $k = 0$, the formula would give $f(0) = \int_0^1 K f'$, which fails for $f = 1$.

Going deeper

For a student

The theorem turns the question "how large is the error of this rule" into a computation with one explicit function $K$. It is the standard way to derive the error terms of Newton–Cotes rules, of Gaussian rules and of interpolation and differentiation formulas. When $K$ keeps one sign, the error has the familiar form "constant times $f^{(k+1)}(\xi)$", and the constant is found by testing the rule on $x^{k+1}$. When $K$ changes sign, one still has the bound by $\norm{K}_{L^1} \norm{f^{(k+1)}}_\infty$, which is sharp over the class of functions with bounded $(k+1)$-st derivative. The degree of exactness of a rule fixes the largest $k$ for which the theorem applies.