Smallest constant $A$ such that $f(x) \leq Ax^2$, given $f(0)=f'(0)=0.$ The function $f(x)$ satisfies:
$$ xf''(x)+3x[f'(x)]^2=1-e^{-x}$$
for all real $x$. (Do not attempt to solve this differential equation).
Find the Smallest constant $A$ such that $f(x) \leq Ax^2$, given $f(0)=f'(0)=0.$
This is a problem from Calculus Volume 1 by Tom Apostol. There are 3 questions for this exercise. I am struggling with the third. Perhaps knowing the first two may help answer the last.
First: If $f$ has an extremum at point $c$ with c$\neq0$ show it has a minimum.
$\quad$ Because $f'(c)=0$ we end up with $f''(c)=\frac{1}c (1-e^{-c})\gt0$, regardless if $c\gt0$ or $c \lt0$.
Second: If $f$ has an extremum at $0$ is it a maximum or a minimum?
$\quad$ Solving for $f''(x)$ results in two terms. The second term clearly goes to zero leaving $\frac{1-e^{-x}}x$ but applying L'hopital we have
$\quad$ $lim_{x\to0}\frac{1-e^{-x}}x=1\gt0$ hence it's a minimum.
Third: if $f(0)=f'(0)=0$, find the smallest constant $A$ such that $f(x)\leq Ax^2$ for all $x\geq0$.
The answer at the back gives $A=1/2$. I have no idea how to prove this?
My only argument is that since at $0$ they have the same value and first derivative then $(Ax^2)''>1$ which yields $A=1/2$ but this doesn't prove anything other that $A$ must be greater than $1/2$, but it might not be enough. A sounder argument is needed.
Any help is greatly appreciated!
 A: First note that the first derivative $f'$ is always nonnegative for $x \ge 0$. [Indeed, you are given $f'(0)=0$, and you already observed that $f"(0)=1$. So for $x$ small enough and positive, $f'(x)$ is positive, and in particular, $f(u)=0$ for all $u \in (0,x]$. Suppose that there is another $y>0$ such that $f'(y)<0$. Then $y>x$, and by the Intermediate Value Theorem, there must be a $z \in (x,y)$ such that both $f'(z)=0$ and $f'$ is decreasing at $z$, which would imply $f''(z) < 0$. But, wouldn't that give $z$ a local minimum of $f$?]
Next note that $$f''(0)=1,$$ and $$f''(x) \le 1 \quad  \forall x \ge 0.$$ [Indeed, from the fact that the first derivative $f'$ of $f$ is always nonnegative, from the equation
$$f''(x) = (1-e^{-x})/x-3f'(x)^2,$$
it follows for all nonnegative $x$ that indeed, $f''(x) \le (1-e^{-x})/x \le 1,$ with $f''(0)=1$ because $f'(0)=0$ and $\lim_{x \rightarrow 0^+} (1-e^{-x})/x = 1$.]
So let $$g_A(x) = Ax^2-f(x).$$
Then, for all $A \ge 1/2$:
$$g''_A(x) = (2A-f''(x))$$ $$ \ \ge \ 2A-1 \ \ge \ 0 \quad \forall x \ge 0,$$ and so it follows that $g'_A(x)$ is nondecreasing with $x$ for $x>0$. As $g'_A(0)=0$ however and $g'_A(x)$ is nondecreasing for $x>0$, it follows that $g'_A(x)$ is nonnegative for $x>0$. As $g'_A(x)$ is nonnegative for all $x>0$, it follows that $g_A(x)$ is nondecreasing with $x$ for $x>0$. As $g_A(0)=0$, it follws that $g_A(x)$ is nonnegative for all $x>0$. Thus, as $g_A(x)$ is ninnegative for all $x>0$, it follows that $Ax^2 \ge f(x)$ for all $x>0$ and $A \ge 1/2$. So from this the following is true: For all $A \ge 1/2$, the inequality $Ax^2 \ge f(x)$ is satisfied for all $x>0$.
However, for all $A<1/2$, then $$g''_A(0) = 2A-f''(x)$$ $$\ = \ 2A - 1 \ < \ 0.$$
Can you use this to show that $g'_A(x)$ must be negative for all positive $x$ sufficiently small for all such $A$, and then conclude that $g_A(x)$ must be negative for all positive $x$ sufficiently small for all such $A$. Then that there is an $x>0$ such that $g_A(x)<0$ means the following: There is an $x>0$ such that $Ax^2 < f(x)$ for all such $A < 1/2$....
This all gives you what you want.
A: Let us set $g(x)=f'(x)$ and write
$$ g'(x) + 3 g(x)^2 = \frac{1-e^{-x}}{x}. \tag{1}$$
The RHS is an entire function and over $\mathbb{R}^+$ it is a positive, decreasing and convex function.
$g(0)=0$ implies $g'(0)=1$ and by setting $g(x)=x h(x)$ we get
$$ x h'(x)+h(x)+3x^2 h(x)^2 = \frac{1-e^{-x}}{x},\qquad h(0)=1. \tag{2}$$
At the stationary points of $h$ we have $h'(x)=0$, and there
$$ h(x) + 3(x h(x))^2 = \frac{1-e^{-x}}{x}. $$
If $x>0$ the RHS is $<1$ and $(3xh(x))^2\geq 0$, so $h(x)$ is less than one. This gives $h(x)<1$ for any $x>0$ and
$$ \sup_{x > 0}\frac{f(x)}{x^2} = \frac{1}{2}\tag{3} $$
as wanted.
