 Solution

Let $L(f)=f^{\prime }(0)-\mu (f)$. Direct evaluation on $1,x,x^2$ gives $L=0$, so the Peano kernel theorem givesThis representation also follows by applying $L$ to the Taylor expansion of $f$ with integral remainder. The derivative at zero of the truncated quadratic is zero for $t\ge 0$, and its sampled values giveThe Peano kernel is nonnegative, andHence $|L(f)|\le \frac{1}{3}\Vert f^{(3)}{\Vert }_{\infty }$. For $f(x)=x^3/6$, the third derivative is $1$, while $f^{\prime }(0)=0$ and $\mu (f)=-1/3$. Thus equality holds, proving the least constant is $c=1/3$, attained by $f(x)=x^3/6$.