Let $f:(-1,1)\to \mathbb{R}$ be a twice differentiable function such that, $f(0)=1$, $f'(x)≤0$, $f(x)≥0$ and $f''(x)≤f(x)$ for all $x≥0$. Prove that, $f'(0)≥-\sqrt2$

Progress: I was able to prove $f'(0)≥-2$. For this I have applied Cauchy MVT for $f'$ and $g$ on $[0,x]$ for any $x\in (0,1)$ where $g: [0,x]\to\mathbb{R}$ is defined by, $g(x)=\sqrt{x+1}$. But, couldn't find another approach to get $-\sqrt{2}$.

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    $\begingroup$ It's usually discouraged to simply post PSQs (problem statement questions) .. otherwise it simply gets closed as off-topic. Showing some work/attempts you have made is commendable. $\endgroup$ – r9m Jun 17 '15 at 19:37
  • $\begingroup$ I was able to prove f'(0)≥-2. But, for the -√2 case. I have applied Cauchy MVT for f' and g on [0,x] for any x€(0,1) where, g: [0,x]--->IR is defined by, g(x)=√(x+1). But, couldnt find another approach. I am totally helpless. $\endgroup$ – user125828 Jun 17 '15 at 19:45
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    $\begingroup$ math.stackexchange.com/questions/1328977/… $\endgroup$ – Jack D'Aurizio Jun 17 '15 at 20:31
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    $\begingroup$ That question has no solution too, but I am writing a solution to this. So please don't mark as duplicate. $\endgroup$ – Landon Carter Jun 17 '15 at 20:45
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    $\begingroup$ @r9m: we can't do that until the other question has an accepted answer. Despite this one, the other question has some efforts from the OP, so it would be better (imho) to answer that one, Landon Carter. $\endgroup$ – Jack D'Aurizio Jun 17 '15 at 20:47

Initially I had posted a solution but it was incorrect. Here's a final solution. Beautiful problem, by the way.

For any $x\in(0,1)$, we can write, for some $c\in(0,x)$ that $f(x)=f(0)+f'(0)x+\dfrac{f''(c)x^2}{2}\leq 1+f'(0)x+\dfrac{f(c)x^2}{2}\leq1 + f'(0)x +\dfrac{x^2}{2}$ since $f(x)\leq1$ for $x\in(0,1)$.

Now using $f''(x)\leq f(x)$ we see that $f''(x)\leq f'(0)x + \dfrac{x^2}{2}+1$. Integrating both sides from $0$ to $x$ we have,

$f'(x)-f'(0)\leq \dfrac{f'(0)x^2}{2}+x+\dfrac{x^3}{6}$ and again integrate from $0$ to $x$ for any $x\in(0,1)$. We get $f'(0)(x+\dfrac{x^3}{6})\geq-\dfrac{x^2}{2}-\dfrac{x^4}{24}+f(x)-f(0)\geq-\dfrac{x^2}{2}-\dfrac{x^4}{24}-1$

Finally we have $$f'(0)\geq \dfrac{-\dfrac{x^2}{2}-\dfrac{x^4}{24}-1}{x+\dfrac{x^3}{6}}$$for any $x\in(0,1)$.

Note that the function on the right side is increasing and achieves its maximum at $x=1$, where its value is $\dfrac{-37}{28}>-\sqrt{2}$.

Hence $f'(0)\geq \sup_{x\in(0,1)}\dfrac{-\dfrac{x^2}{2}-\dfrac{x^4}{24}-1}{x+\dfrac{x^3}{6}}=\dfrac{-37}{28}>-\sqrt{2}$.

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    $\begingroup$ What is the basis for the first equation "$f(x)=f(0)+f'(0)x+\frac12 f''(0)c^2$" for some $c\in (0,x)$? Did you mean $f(x)=f(0)+f'(0)x+\frac12 f''(c)x^2$? $\endgroup$ – Mark Viola Jun 17 '15 at 21:28
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    $\begingroup$ You're welcome. And well done!! +1 $\endgroup$ – Mark Viola Jun 17 '15 at 21:37
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    $\begingroup$ (+1) Well done. I wonder if it is possible to improve the bound up to $$f'(0)\geq -\coth 1=-1.313\ldots,$$ since $f(x)=\cosh x-\coth(1)\sinh x$ should be the extremal solution to our problem ($f(x)\geq 0,f(0)=1$ and $f''(x)=f(x),f(1)=0$). $\endgroup$ – Jack D'Aurizio Jun 17 '15 at 21:56
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    $\begingroup$ Probably we just need to iterate the integration procedure since your bound yet depends on partial sums for the Taylor series of $\cosh$ and $\sinh$. $\endgroup$ – Jack D'Aurizio Jun 17 '15 at 21:58
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    $\begingroup$ @JackD'Aurizio added a simple proof for $f'(0) \ge -\coth 1$ :-) $\endgroup$ – r9m Jun 17 '15 at 23:52

As @JackD'Aurizio has already mentioned in comment to the other answer, a possible lower bound is:

$$f'(0) \ge -\coth 1$$

One way to get it, starting from

$$\begin{align}f''(x) \le f(x) &\implies e^{x}(f'(x)+f''(x)) \le e^{x}(f(x)+f'(x))\\& \implies \frac{d}{dx}(e^x f'(x)) \le \frac{d}{dx}(e^xf(x))\,\,\,\textrm{ [integrating both sides from $0$ to $x$] } \\&\implies f(0) - f'(0) \le e^{x}(f(x)-f'(x)) \\&\implies e^{-2x}(1-f'(0)) \le \frac{d}{dx}(-e^{-x}f(x)) \,\,\,\textrm{ [integrating both sides from $0$ to $x$] }\\&\implies \frac{1}{2}(1-e^{-2x})(1-f'(0)) \le f(0) - e^{-x}f(x)\\& \implies 0 \le f(x) \le e^{x} + (f'(0) - 1)\sinh x\\&\implies -\coth x \le f'(0)\end{align}$$

Supremum of LHS is attained at $x = 1$ for $x \in (0,1)$.

Note that if we were to use $f'(x) \le 0$, that would further give us:

$$\sup_{x \in [0,1)} f(x) = f(0) = 1 \le \sup_{x \in [0,1)} e^{x} + (f'(0) - 1)\sinh x$$

Then I guess we'd have to check if it's an improvement on the previous bound or not.


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