Prove that $a\leq 4$ if $0\leq ax^4 + bx^2+c\leq 1$ in the interval $[0,1]$

Question

Consider a real polynomial of the form $p(x) = ax^4 + bx^2+c$. We are given that $0\leq p(x)\leq 1$ in the interval $[0,1]$. Show that $a\leq 4$

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Take $x^2=y$. So $p(x)=q(y)$We assumes $a>0$. Now we consider two cases:

1. $x$ such that $q'(x)=0$ lies outside $[0,1]$ and
2. $x$ such that $q'(x)=0$ lies inside $[0,1]$. In the first case, notice that the extreme values of $q(x)$ in that interval must lie on the endpoints. Say in the interval $q'(x)>0$ then $a+b+c\leq 1$ and $c\geq 0$. I'm stuck after here and made no reasonable progress after this. Please help

Answer 1

Since $\,0\leqslant ax^4+bx^2+c\leqslant1\;\;$ for any $\;x\in[0,1]\,,\,$ in particular, for $\,x=0\,,\,$ $x=\dfrac1{\sqrt2}\,,\,$ $x=1\,,\,$ it follows that

$\begin{cases}0\leqslant c\leqslant1\\[3pt]0\leqslant\dfrac14a+\dfrac12b+c\leqslant1\\[3pt]0\leqslant a+b+c\leqslant1\end{cases}$

consequently ,

$\begin{cases}0\leqslant\dfrac12c\leqslant\dfrac12\\[3pt]-1\leqslant-\dfrac14a-\dfrac12b-c\leqslant0\\[3pt]0\leqslant\dfrac12a+\dfrac12b+\dfrac12c\leqslant\dfrac12\end{cases}$

and, by adding the previous inequalities, we get that

$-1\leqslant\dfrac14a\leqslant1\;\;,$

hence ,

$-4\leqslant a\leqslant4\;.$

Answer 2

One can apply $$ \frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(\zeta) $$ (see for example here) to the function $f(x) = ax^2+bx + c$ with $x=1/2$ and $h=1/2$: $$ 2a = f''(\zeta) = \frac{f(1)-2f(1/2)+f(0)}{(1/2)^2} \, . $$ The expression on the right is in the range $[-8, 8]$ and therefore $-4 \le a \le 4$.

The bound is sharp, equality holds for $f(x) = 4(x-1/2)^2$, i.e. for $q(x) = 4(x^2-1/2)^2 = 4x^4 - 4x^2 + 1$.

Answer 3

Another solution:

Suppose for contradiction that $a > 4$. The boundary conditions also tell us that $0\le c\le 1$.

Note that for $q(x) = p(\sqrt x)$ to be nonnegative, it must be the case that $b^2 \le 4ac$ (the discriminant) and so $b \ge -2\sqrt{ac}$.

Then, we have by the boundary condition that

$$1 \ge a + b + c \ge a - 2\sqrt{ac} + c = (\sqrt a - \sqrt c)^2.$$

But, if $a > 4$ then $\sqrt a > 2$ while $0\le \sqrt c \le 1$. This would imply that $(\sqrt a - \sqrt c)^2 > (\sqrt a - 1)^2 > 1$, a contradiction.

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Edit: didn't notice the constrained nonnegativity constraint. To fix the argument, note that if $a > 4$, then $b < 0$. Then, either $b^2 \le 4ac$ (as above) or the vertex $-\frac{b}{2a} > 1$, or equivalently $b < -2a$. In this latter case, $a + b + c < c - a < -3 < 0$ which is also a contradiction.