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Let $f:(-1,1) \to \mathbb R$ be a twice differentiable function such that $f(0)=1$ $f(x) \ge 0 , f'(x) \le 0 , f''(x) \le f(x) , \forall x \in [0,1)$ , then how to prove that $f'(0) \ge -2$ ? I am totally stuck , Please help . Thanks in advance .

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assume $f'(0) = c$,

$f(1) = 1 + \int_0^1 f'(t) \ge 0$,

$\int_0^1 f'(t)\ge -1$.

then

$\int_0^1 (c + \int_0^t f''(s)ds )dt \ge -1$

i.e.

$c \ge -1 -\int_0^1 \int_0^t f''(s)ds dt$

and $f''\le f \le 1$,

then $\int_0^t f''(s)ds \le t$, $\int_0^1 tdt = 1/2$, then

$c\ge -1 - 1/2 = -\frac{3}{2}\ge -2$.

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  • $\begingroup$ Can you avoid integration please $\endgroup$ – user228168 May 12 '15 at 4:51
  • $\begingroup$ and $f(1)$ is not defined ... $\endgroup$ – user228168 May 12 '15 at 5:02
  • $\begingroup$ @SaunDev you can extend definition by limit. $\endgroup$ – Yimin May 12 '15 at 5:11
  • $\begingroup$ @SaunDev i wonder if integration is avoided, how do you connect $f'$ and $f$ $\endgroup$ – Yimin May 12 '15 at 5:13
  • $\begingroup$ I think I can avoid integration , I will write an answer if I think I'm correct ... it follows your idea +1 $\endgroup$ – user228168 May 12 '15 at 5:16
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For every $n >1 , \exists r_n \in (0,1-\dfrac 1n)$, such that

$f(1-1/n)-f(0)=\dfrac{n-1}n f'(r_n)=f(1-1/n)-1 \ge -1$ ( as $1-1/n \ge 0$ ) . So

$f'(r_n) \ge \dfrac n{1-n}$ . Now $f'(r_n)-f'(0)=r_n f''(m_n)$ , where $0<m_n<r_n<1 , \forall n > 1$ , so

writing $c:=f'(0)$ ,

$c+r_nf''(m_n)=f'(r_n) \ge \dfrac n{1-n}$ , now $f(m_n) \ge f''(m_n)$ , also $f'(x) \le 0$ in $[0,1)$ so $f$

decreases in $[0,1)$ , so $1=f(0)\ge f(m_n) \ge f''(m_n)$ ; as $1>r_n >0$ , so $r_n \ge r_n f''(m_n)$ and

$1+c \ge c+r_n \ge c+r_nf''(m_n) \ge \dfrac n{1-n} , \forall n > 1$ so taking limit $n \to \infty $ , $1+c \ge -1$

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