f must be a constant function

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f must be a constant function

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#1 h Dec 8, 2010, 11:42 PM • 2 Y

Y by Adventure10, Mango247

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous and bounded function such that
$x\int_{x}^{x+1}f(t)\, \text{d}t=\int_{0}^{x}f(t)\, \text{d}t,\quad\text{for any}\ x\in\mathbb{R}.$
Prove that $f$ is a constant function.

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#2 h Dec 10, 2010, 9:03 AM • 2 Y

Y by Adventure10, Mango247

$x\int_{x}^{x+1}f(t)\, \text{d}t=\int_{0}^{x}f(t)\text{d}t,\quad\text \ \Longrightarrow\ x\int_{0}^{x+1}f(t)\, \text{d}t=(x+1)\int_{0}^{x}f(t)\, \text{d}t$

Let's define ${\displaystyle g(x)=\frac { \int_{0}^{x}f(t)\, \text{d}t}x},\ g(0)=f(0)$ , so $\forall_x ,g(x+1)=g(x)$ , periodical function.
$f$ is continous so $g$ derivable . $g$ is a periodical function so $g'$ is a periodical function.

${ \int_{0}^{x}f(t)\, \text{d}t}=xg(x) \ \Longrightarrow \ f(x)=xg'(x)+g(x)$
If there is $x_0$ such that $g'(x_0)\neq 0$ , then $f(n+x_0) =(n+x_0)g'(n+x_0)+g(n+x_0)=(n+x_0)g'(x_0)+g(x_0)$ is not bounded when $n\rightarrow \infty$ (contradiction)
Finally we must have $g'(x)=0$ , so $f(x)=g(x)=constant$ (q.e.d)

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#3 h Mar 17, 2025, 3:31 PM

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@below, thank you for pointing out the mistake

This post has been edited 1 time. Last edited by Fibonacci_math, 3 hours ago

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#4 h 5 hours ago

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Fibonacci_math wrote:

Then we have, $0=\int_0^1F(t)\,dt = \int_1^2F(t)\,dt = \int_2^3F(t)\,dt = \cdots$ So, $\int_0^{\infty} F(t)\,dt=0$ .

This is wrong. For example the function $F(x)=\sin (2 \pi x)$ satisfies that condition, but $\int\limits_0^{\infty} F(t) \mathrm{d}t$ does not exist.

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