= Solution
Let $L(f)=f'(0)-\mu(f)$. Direct evaluation on $1,x,x^2$ gives $L=0$, so the <Peano kernel theorem> gives
$$
L(f)=\int_0^2K(t)f^{(3)}(t)\,dt,\qquad K(t)=L\left(\frac{(x-t)_+^2}{2}\right).
$$
This 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 $t\ge0$, and its sampled values give
$$
K(t)=-(1-t)_+^2+\frac14(2-t)^2=\begin{cases}t-\frac34t^2,&0\le t\le1,\\\frac14(2-t)^2,&1\le t\le2.\end{cases}
$$
The <Peano kernel> is nonnegative, and
$$
\int_0^2|K(t)|\,dt=\left[\frac{t^2}{2}-\frac{t^3}{4}\right]_0^1+\int_1^2\frac{(2-t)^2}{4}\,dt=\frac14+\frac1{12}=\frac13.
$$
Hence $|L(f)|\le\frac13\|f^{(3)}\|_\infty$. For $f(x)=x^3/6$, the third <derivative> is $1$, while $f'(0)=0$ and $\mu(f)=-1/3$. Thus equality holds, proving \b[the least constant is $c=1/3$, attained by $f(x)=x^3/6$].
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