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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 $0$ ?

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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 $$f'(0) \ge -\frac{e^2 + 1}{e^2 - 1} > -\sqrt{2}$$

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  • $\begingroup$ how do you know $f(f'(0))$ is defined? $\endgroup$ – Soham Dec 18 '14 at 7:01
  • $\begingroup$ I never used $f(f'(0))$. $\endgroup$ – kobe Dec 18 '14 at 7:13
  • $\begingroup$ then how are you using the inequality? $\endgroup$ – Soham Dec 18 '14 at 7:15
  • $\begingroup$ I've made changes so that the argument will be clearer. $\endgroup$ – kobe Dec 18 '14 at 8:06
  • $\begingroup$ Since $f'(0) < 0$ and $-f'(0) \ge x$, $f'(0)[-f'(0)] \le f'(0)x$. $\endgroup$ – ronno Dec 18 '14 at 8:28

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