4

Let $f:[0,\pi] \to \mathbb{R}$ be a twice differentiable function.

Show $f(x) \geq \sin(x) \forall x \in [0,\pi]$ if $f(0)=0$, $f'(0)=1$ and $f''(x)+f(x)\geq 0$ $\forall x \in [0,\pi]$.

I have tried making the taylor series.

Let $g(x) =f(x)-\sin(x)$. Then $g(x) =g(0)+g'(0) x +\frac{1}{2} g''(c) x^2=(f''(c)+\sin(c)) x^2/2 \geq (\sin(c)-f(c))x^2/2$

Thats all I can think of. Please help. Thank in advance.

EDIT: $\sin(x)$ is not unique since $e^x-1$ satisfies the condition too.

EDIT2: Conditions clarified.

k99731
  • 1,298
  • Come on, do not be too pedantic: It states $f''(x)+f(x) \geq 0$ for all $x \in [0, \pi]$. This pretty much implies that $f$ ist at least two times differentiable on the interval. – MooS Feb 08 '15 at 11:26

3 Answers3

1

your $g(x) = f(x) - \sin x$ satisfies $h = g'' + g \ge 0, g(0) = 0, g'(0) = 0.$ the integral representation gives $$g(x) = \int_0^x h(t)\sin (x-t) \, dt $$ that both $h$ and $\sin(x-t)$ are nonnegative for $0 \le x \le \pi$ implies $g(x) \ge 0$ on the same interval.

i can't see a way to extend this interval beyond $\pi.$

abel
  • 29,170
1

Let us denote $g=\left(f-\sin\right)$ and assume that $g<0$ . It is clear that $g\left(0\right)=0$, $g'\left(0\right)=0$

and that $g''\geq-g$ with the hypothesis done on $f$ and $f''$ . Now, let $x>0$ be a non-negative element of $\left[0,\pi\right]$ . By Taylor's formula with integral rest, one has$$0>g\left(x\right)=x^{2}\intop_{t=0}^{1}\left(1-t\right)g''\left(tx\right)dt\geq-x^{2}\intop_{t=0}^{1}\left(1-t\right)g\left(tx\right)dt.$$ But $g\left(tx\right)\leq0$ because $0\leq t\leq1$ , so $-g\left(tx\right)\geq0$ , whence $0>g\left(x\right)\geq0$, that is impossible. So the assumption $g<0$ must be rejected, and then $g\geq0$ .

Nicolas
  • 3,326
0

$g(x)=f(x)-\sin x$ and $g(0)=0$

$g'(x)=f'(x)-\cos x$ and $g'(0)=0$

$g''(x)=f''(x)+\sin x$ and $g(x)+g''(x)\geq 0$ because $f(x)+f''(x)\geq 0$.

Let $x\in [0,\pi]$ be close to $0$ so, if we assume that $g(x)<0$ then due to continuity of $g$ at $x$ we have that $g(x)<0$ for all $x\in I_1$ , where $I_1$ is an interval and $0=\sup I_1$.

Then in $I_1$ we have that $g(x)<0\Rightarrow g''(x)>0\Rightarrow g'$ is increasing and thus $g'(x)\geq g'(0)=0\Rightarrow $ $g$ is increasing and thus $g(x)\geq g(0)=0$. Contradiction.

So $g(x)\geq 0 $ in $I_1$. Write $I_1=(0,a)$ and do the same for a $x$ close to $a$ where you know that there $g(x)\geq 0$.

Use induction to cover the interval $[0,\pi]$.

I could prove it more strictly the close to $0$ thing, but i don't know your education level if you're at high school or something.

Siminore
  • 35,136
Haha
  • 5,648
  • What if $I=(0,b)$, where $b=\lim(a_n)$ and $g(b)=0$ so that the induction cannot continue? We don't have or not yet have $g'(b)=0$. – k99731 Feb 08 '15 at 12:20
  • @k99731,the set $(0,0)=\emptyset$ – Haha Feb 08 '15 at 12:22
  • I dont quite understand what you said. If we have construct $I_1=(0,a_1)$, $I_2=(a_1,a_2)$, ..., s.t. $g(x)\geq 0$ on those intevals, but $\lim(a_n)=b<\pi$, how could be continue the induction? – k99731 Feb 08 '15 at 12:27