Maximum cubic function

Question

Let $f(x)=x^{3}+ax^{2}+bx+c$ with a, b, c real.

Show that

$$\frac{1}4 \le \max_{-1 \le x \le 1\hspace{2mm}} |f(x)|=M$$

and find all cases where equality occurs.

Answer 1

Note that $$ \max_{[-1,1]}\,\left|\,x^3-\tfrac34x\,\right|=\tfrac14\tag{1} $$ Considering the symmetry about $0$ of the domain, we have for any $t\in[0,1]$ $$ \max_{\{t,-t\}}\,\left|\,x^3+ax^2+bx+c\,\right|=\left|\,t^3+bt\,\right|+\left|\,at^2+c\,\right|\tag{2} $$ Using $(2)$, it is obvious that $$ M_b=\max_{[-1,1]}\,\left|\,x^3+bx\,\right|\le\max_{[-1,1]}\,\left|\,x^3+ax^2+bx+c\,\right|\tag{3} $$ It is straightforward to compute $$ M_b=\left\{\begin{array}{} 2(-b/3)^{3/2}&\text{if }b\in\left[-3,-\tfrac34\right]\\ |\,1+b\,|&\text{otherwise} \end{array}\right.\tag{4} $$ and $M_b$ reaches a minimum of $\frac14$ only at $b=-\frac34$. For any other value of $b$, $(3)$ says that $$ \max_{[-1,1]}\,\left|\,x^3+ax^2+bx+c\,\right|\ge M_b>\tfrac14\tag{5} $$ Setting $b=-\frac34$ and $t=\frac12$ in $(2)$ yields $$ \max_{[-1,1]}\,\left|\,x^3+ax^2-\tfrac34x+c\,\right|\ge\tfrac14+\left|\,\tfrac14a+c\,\right|\tag{6} $$ and this can be $\frac14$ only if $c=-\frac a4$.

At $|x|=\frac12$, $$ \left|\,x^3+ax^2-\tfrac34x-\tfrac a4\,\right|=\tfrac14\tag{7} $$ However, at $x=\pm\frac12$, the derivative of $x^3+ax^2-\frac34x-\frac a4$ is $\pm a$. Therefore, the maximum of $\left|\,x^3+ax^2-\tfrac34x-\tfrac a4\,\right|$ will be greater than $\frac14$ unless $a=0$.

Thus, $$ \max_{[-1,1]}\,\left|\,x^3+ax^2+bx+c\,\right|\ge\tfrac14\tag{8} $$ where equality holds only for $x^3-\frac34x$.