If $f$ is twice differentiable and satisfies the following constraints, prove that $f'(0)\geq-\sqrt 2$.

Question

Let $f$ be a twice differentiable function on the open interval $(-1,1) $such that $f(0)=1$. Suppose $f$ also satisfies $f(x) \ge 0, f'(x) \le 0 $and $f''(x) \le f(x)$, for all $ x\ge 0$. Show that $f'(0) \ge -\sqrt2.$

My attempt $ \rightarrow$ From Taylor's we see that $0 \leq f(x)=f(0)+f'(0)x+\frac{f''(\zeta)x^2}{2} $ for some $\zeta \in (0,x) .$

$f'(x)\leq 0$ and $ f'(x) \le 0 \implies$ $ f''(\zeta) \leq f(\zeta)\leq 1$

Thus $1+f'(0)x+\frac{x^2}{2} \geq 0 $.

I'm stuck upto this,I can see that if the discriminant of the quadratic is less than $ 0$ we're done but how do I conclude that the discriminant would be $0$ ?

Answer 1

Let $u(x) = e^x f(x)$; for $x \ge 0$, $$u'' = e^x(f'' + 2f' + f) \le 2e^x(f + f') = 2u'$$ Thus $(e^{-2x}u')' = e^{-2x}(u'' - 2u') \le 0$, so that $e^{-2x}u' \le u'(0)$. Since $u'(0) = f(0) + f'(0) = 1 + f'(0)$, then $u' \le [1 + f'(0)]e^{2x}$. Integrating from $0$ to $x$ yields $u - f(0) \le \dfrac{1 + f'(0)}{2}(e^{2x} - 1)$, or, $e^xf(x) - 1 \le \dfrac{1 + f'(0)}{2}(e^{2x} - 1)$, using the condition $f(0) = 1$. Rearranging the inequality, $$f'(0) \ge \frac{2(e^x f(x) - 1)}{e^{2x} - 1} - 1$$ By the assumption that $f$ is nonnegative, the right hand side of the inequality is no less than $$\frac{-2}{e^{2x}-1} - 1 = - \frac{e^{2x} + 1}{e^{2x}-1}$$ Therefore $$f'(0) \ge -\frac{e^{2x}+1}{e^{2x}-1} \quad (0 \le x < 1)$$ Taking the limit as $x \to 1^{-}$ results in $$f'(0) \ge -\frac{e^2 + 1}{e^2 - 1} > -\sqrt{2}$$

Answer 2

Another approach to get the lower bound $f'(0)\geq-\sqrt 2$.

By the mean value theorem, for any $x\in(0,1)$, there is $\xi_x\in(0,x)$ such that $f(x)=f(0)+f'(\xi_x)x=1+f'(\xi_x)x$; since $f(x)\geq0$ and $f'(x)\leq 0$, we have $$-\frac1x\leq f'(\xi_x)\leq0, \qquad x\in(0,1). \tag{1}$$ What's more, $f''(x)\leq f(x)$ and $f'(x)\leq 0$ imply that $f''(x)f'(x)\geq f(x)f'(x)$ for all $x\in[0,1)$, and thus the function $F(x):=(f'(x))^2-(f(x))^2$ is increasing in $[0,1)$. So, for all $x\in(0,1)$, using the decreasing of $f$ and $(1)$ we have \begin{align*} (f'(0))^2-1&=F(0)\leq F(\xi_x)=(f'(\xi_x))^2-(f(\xi_x))^2\\ &\leq (f'(\xi_x))^2-(f(x))^2=(f'(\xi_x))^2-(1+f'(\xi_x)x)^2\\ &=(1-x^2)(f'(\xi_x))^2-2xf'(\xi_x)-1\\ &\leq (1-x^2)\left(-\frac1x\right)^2-2x\left(-\frac1x\right)-1\tag{2}\\&=\frac1{x^2}, \end{align*} where $(2)$ follows from the fact that for fixed $x\in(0,1)$ the quardratic function $t\mapsto (1-x^2)t^2-2xt-1$ is decreasing in $t\in(-\infty, 0)$. Therefore, $$(f'(0))^2\leq 1+\frac1{x^2},\qquad \forall x\in(0,1).$$ Since $\lim_{x\to1-}\left(1+\frac1{x^2}\right)=2$, we have $(f'(0))^2\leq2$ and hence $f'(0)\geq-\sqrt2$.