Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $f:\mathbb{R}_+\to\mathbb{R}_+$ is a positive function which is locally $L^1$ with $\int_0^t{f(s)ds}=at^n$ for all $t\geq0$ then, by differentiating, it follows that $f(t)=a$ if $n=1$ and $f=0=a$ otherwise.

I would like to show that, if $f:\mathbb{R}_+^n\to\mathbb{R}_+$ is a symmetric function which satisfies $$\int_0^t\int_0^{t_1}\cdots \int_0^{t_{n-1}} f(t_1,t_2,\ldots,t_n)dt_n\cdots dt_2dt_1=at^m \quad \text{for all } t\geq0$$ then we must have $$f ~\text{constant if }n=m \quad \text{and} \quad f=0~\text{ if }n\neq m.$$ Does anyone know if this is possible? Perhaps there is even a counterexample?!

share|cite|improve this question
Note that $f$ is locally $L^1$ but not $L^1$. so your first argument is not true. It's possible that $f$ is not constant and we have $f(t)=ant^{n-1}$. – Mohsen Shahriari Jun 11 '15 at 11:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.