[proofplan] We apply Taylor's theorem with integral remainder to write $f$ as a polynomial part (which $L$ annihilates) plus an integral involving the $(k+1)$-th derivative. Interchanging $L$ with the integral produces the Peano kernel representation. [/proofplan] [step:Apply Taylor's theorem and use $L$'s annihilation property] By [Taylor's theorem with integral remainder](/theorems/189), for $f \in C^{k+1}[a,b]$: \begin{align*} f(x) &= \sum_{j=0}^{k} \frac{f^{(j)}(a)}{j!}(x-a)^j + \frac{1}{k!}\int_a^b (x-\theta)_+^k\,f^{(k+1)}(\theta)\,d\mathcal{L}^1(\theta). \end{align*} Apply $L$ to both sides. Since $L$ annihilates polynomials of degree $\leq k$, the sum vanishes. For the integral term, the hypothesis that $L$ commutes with integration allows interchanging $L$ and $\int$: \begin{align*} L(f) &= \frac{1}{k!}\int_a^b L\!\left((x-\theta)_+^k\right) f^{(k+1)}(\theta)\,d\mathcal{L}^1(\theta) = \frac{1}{k!}\int_a^b K_L(\theta)\,f^{(k+1)}(\theta)\,d\mathcal{L}^1(\theta). \end{align*} [/step]