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Let $g(x)$ has continuous derivative,with $g(x)\geq 1,x\in \mathbf{R}$,if $f(x)\in \mathbf{C^{2}}(-\infty,+\infty)$,and $$ f(0)=f'(0)=\int_{-\pi}^{\pi}dy\int_{0}^{\pi}\frac{\cos(nx)-\cos(ny)}{\cos{x}-\cos{y}}dx\qquad n=2k+1,k\in \mathbf{N}^{+} $$ also for and real numbers $x$, $$(g(x)f'(x))'+f(x)\geq 0 $$ show that:$f(\sqrt{2})\geq 0 $.

My try: From a know result(American.Mathematical.Monthly.E3145),we get that $$\int_{0}^{\pi}\frac{\cos{nx}-\cos{ny}}{\cos{x}-\cos{y}}dx=\pi\frac{\sin{ny}}{\sin{y}}\qquad (n=0,1,2\cdots) $$ hence $$f(0)=f'(0)=\pi\int_{-\pi}^{\pi}\frac{\sin{nx}}{\sin{x}}dx=2\pi^2$$ Then I don't know how to deal with the $$g'(x)f'(x)+g(x)f''(x)+f(x)\geq 0$$.....

pxchg1200
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  • Where does this come from? Why state a formula for $f(0)=f'(0)=\ldots $ instead of just saying $f(0)=f'(0)=const$? The spesific value of the constant ($2\pi^2$) have no relevance to what you try to show (you can assume $f(0)=f'(0)=1$). – Winther Jul 01 '14 at 00:46
  • @Winther，this problem is proposed by tian27546,he like propose such tough problem as a homework to us.. – pxchg1200 Jul 01 '14 at 01:18
  • Then why not ask him for the solution? Otherwise, it might be better to condence it to the core: "Given: $g(x)\geq 1$, $f(0)=f’(0)>0$ and $(gf’)’ + f \geq 0$. Show: $f(\sqrt{2}) > 0$" ...unless that integral is meant as a hint?! – Winther Jul 01 '14 at 01:24
  • it was really a hard problem to ask him for solution,maybe he don't like to show the solution... – pxchg1200 Jul 01 '14 at 02:00
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    @pxchg1200 - missed your bounty,but here's an answer.. – Nathaniel Bubis Jul 10 '14 at 03:01

1 Answers1

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First a proof for the case $g(x)=1$:

We have that $f''(x)+f(x)\ge 0$. This can be written as $f''(x)+f(x)=b(x)$ where $b(x)$ is a positive function. Thus:

$$f(x)= k_1 \cos(x)+k_2\sin(x)+\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$ $$\quad\quad\quad\cos(x)\int_0^x -b(s)\sin(s)ds+\sin(x)\int_0^x b(s)\cos(s)ds$$ The fact that $f(0)=f'(0)=2\pi^2$ leads to $k_1 = k_2 = 2\pi^2$. So now, by the mean value theorem, since $b(s)$ does not change sign, there exist $a_1,a_2\in[0,\sqrt{2}]$ such that: $$f(\sqrt{2})=2\pi^2\left(\sin(\sqrt{2})+\cos(\sqrt{2})\right) +\quad\quad\quad\quad\quad\quad\quad\ $$ $$\quad\quad\quad\quad\left(-\cos(\sqrt{2})\sin(a_1)+\sin(\sqrt{2})\cos(a_2)\right) \int_0^\sqrt{2} b(s)ds$$ Thus, we have that: $$f(\sqrt{2}) \ge 2\pi^2 \ge 0$$ Since the minimal value of $\left(-\cos(\sqrt{2})\sin(a_1)+\sin(\sqrt{2})\cos(a_2)\right)$ is zero.

Now for the case $g(x) \ge 1:$

We define $h'(x) = g(x)f'(x)$. Then again by the mean value theorem, for each $x$, $h(x)= g(a_x) f(x)$, so that $h(x) \ge G f(x)$, where $G$ is the minima of $g(x)$ on the the interval under consideration, which we take to be $[0,\sqrt{2}]$. Thus, we have that: $$h''(x)+\frac{1}{G}h(x) \ge h''(x)+ f(x) \ge 0 $$ Thus the function $h(x)/\sqrt{G}$ has the same solution as $f(x)$ in the case $g(x)=1$, and using the same logic: $$h(\sqrt{2})\ge \left(h(0)\cos(\sqrt{2}) + h'(0)\sin(\sqrt{2})\right) \ge$$ $$\left(Gf(0)\cos(\sqrt{2}) + Gf'(0)\sin(\sqrt{2})\right) \ge 2\pi^2 \ge0$$

Note that the actual value of $f(0)=f'(0)$ doesn't make any difference, as long as it's positive. Q.E.D.

Nathaniel Bubis
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