 Solution

Let $L(f)=\mathcal{T}[f]-f^{\prime }(1/3)$. Direct substitution shows $L(1)=L(x)=L(x^2)=0$, so the functional annihilates all quadratic polynomials. Taylor's integral-remainder formula isApplying $L$ and interchanging the finite evaluations and derivative with the continuous integral gives the Peano kernel representation $L(f)={\int }_0^{1}K(t)f^{\prime \prime \prime }(t)dt$, whereAll three expressions are nonnegative; the middle quadratic has positive leading coefficient and discriminant $-1/9$. Therefore the best possible norm bound has constant ${\int }_0^{1}K(t)dt$. Evaluate that integral particularly simply by using $f(x)=x^3/6$, for which $f^{\prime \prime \prime }=1$:It follows thatThe same cubic attains equality, so the constant is sharp, not merely sufficient. This is the sharp Peano bound for an interior three-point first derivative.