Prove f(x) >= 0

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Prove f(x) >= 0

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

97 posts

#1 h Mar 26, 2025, 5:47 AM

Y by

Let $f$ be a function that is at least twice differentiable on an open interval containing $[0, 2\pi]$ . Given that
$f(0) = f(2\pi) = f'(0) = f'(2\pi) = 0$ and
$f(x) + f''(x) \geq 0, \quad \forall x \in [0,2\pi].$ Prove that $f(x) \geq 0$ for all $x \in [0,2\pi]$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

768 posts

#2 h Mar 26, 2025, 10:16 AM

Y by

Suppose $f''(x)+f(x)=g(x)$ (i) such that $g:[0,2\pi] \rightarrow [0,\infty)$ is differentiable and whose Laplace Transform $G(s)$ exists. Taking the Laplace Transform of (i) leads to:

$[s^2F(s)-sf(0)-f'(0)] + F(s)=G(s) \Rightarrow (s^2+1)F(s) = G(s) \Rightarrow F(s) = G(s) \cdot \frac{1}{s^2+1}$ (ii);

and the inverse Laplace Transform computes to:

$f(x)=L^{-1}[F(s)] = \int_{0}^{x} \sin(\tau) \cdot g(x-\tau) d\tau$ (iii).

Given that $f(2\pi)=0$ , we obtain $f(2\pi) = \int_{0}^{2\pi} \sin(\tau) \cdot g(x-\tau) d\tau = 0$ . If $\int_{0}^{2\pi} \sin(t) dt = 0$ , then $g(x)=C$ is a non-negative constant function over $[0,2\pi] \Rightarrow \textcolor{red}{f(x)=\int_{0}^{x} C\sin(\tau) d\tau \ge 0}$ for all $x \in [0,2\pi]$ .

This post has been edited 3 times. Last edited by Mathzeus1024, Mar 27, 2025, 10:32 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3 posts

#3 h Mar 26, 2025, 11:20 AM • 1 Y

Y by Mathzeus1024

If we assume that $g(x) = f(x) + f''(x)$ is continuous then
$F(x) \; = \; \int_0^x \sin(x-t)g(t)\,dt$ is such that $F''(x) + F(x) = g(x)$ and $F(0) = F'(0)=0$ . Thus $F(x) = f(x)$ . Since $f(2\pi) = f'(2\pi)=0$ we deduce that
$\int_0^{2\pi} \sin t \,g(t)\,dt \; = \; \int_0^{2\pi} \cos t \,g(t)\,dt \; = \; 0$ and hence
$\int_0^{2\pi} \sin(x-t)g(t)\,dt \; = \; 0 \hspace{2cm} \forall x$ Thus
$f(x) \; =\; \int_0^x \sin(x-t)\,g(t)\,dt \; = \; \int_x^{2\pi} \sin(t-x)\,g(t)\,dt$ Since $g(x) \ge 0$ for all $x$ , the first integral is clearly non-negative when $0 \le x \le \pi$ . while the second integral is clearly non-negative when $\pi \le x \le 2\pi$ . Thus $f(x) \ge 0$ for all $0 \le x \le 2\pi$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

611 posts

#4 h Mar 26, 2025, 12:36 PM • 1 Y

Y by Mathzeus1024

Very nice! In fact you don't need to assume continuity of $g$ if you rewrite your argument using antiderivatives instead of integrals:

Let $x \in [0,\pi]$ and define $F_x(t) = \sin(x-t) f'(t) + \cos(x-t) f(t)$ . $F_x$ is increasing on $[0,x]$ since $F_x'(t) = \sin(x-t) (f''(t) + f(t)) \ge 0$ . This implies that $f(x) = F_x(x) \ge F_x(0) = 0$ . By symmetry the same holds for $x \in [\pi ,2\pi]$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a