Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is the following true?

If $$\int_{0}^{x}f(t)\,dt \leq \int_{0}^{x} c \,dt =cx $$ for all $x>0$, $x$ is real number, and $c$ is some fixed constant,


$$f(t) \leq c$$ for all $t>0$?

EDIT: I should said that $f(t)$ is positive function on $t>0$, and $f(t_{1}+t_{2})\geq f(t_{1})+f(t_{2})$, for all $t_{1},t_{2}>0$ if this helps!

share|improve this question
No,it is not true. Actually,The case that f(t)>c in a countable set is available. –  89085731 Mar 29 '12 at 4:00
Just to be clear, that's a Riemann integral right. –  Patrick Mar 29 '12 at 4:01
To say it more detailed, consider the function f(x)=x in [0,1),f(x)=2 at x=1,and c=1 –  89085731 Mar 29 '12 at 4:03
@kuku Since f(x) may not be continuous,the differentiation may not be available. –  89085731 Mar 29 '12 at 4:05
add comment

1 Answer

With the condition provided in the "EDIT" section, the answer is: yes, it is true. Below is the proof.

Since $\int_0^x f(t)dt\leq \int_0^x cdt$, for all $x > 0$, we have $\int_0^x(c - f(t))dt\geq 0$ for all $x > 0$. It will suffice, of course, to prove from the last inequality (and the condition that for all $t_1, t_2 > 0$, $f(t_1 + t_2)\geq f(t_1) + f(t_2)$) that $c - f(x)\geq 0$ for all $x > 0$.

Due to lack of further information, I am assuming that we're dealing with Riemann integrals, and $f$ is Riemann integrable.

Since for all $x, \epsilon > 0$, $f(x + \epsilon) \geq f(x) + f(\epsilon)$, and $f(\epsilon) > 0$ (since $f$ is positive on the positive Real half-line), we know that $f$ is nondecreasing. Now let us assume that for some $x_0 > 0$, $f(x_0) > c$. Say $f(x_0) - c = \delta > 0$. Then for all $x > x_0$, we must have $f(x) - c \geq \delta\implies f(x)\geq c + \delta$. Therefore we obtain, for any $x > x_0$,

$$ \int_{x_0}^x f(t)dt \geq \int_{x_0}^x c + \delta dt = (c + \delta)(x - x_0). $$

On the other hand, say

$$ cx_0 - \int_0^{x_0}f(t)dt = \eta. $$

Then if $x$ is sufficiently large, we have, from above, $(c + \delta)(x - x_0) > \eta$. In other words, for $x$ sufficiently large, we get

$$ \int_0^x f(t)dt > cx, $$

contradicting the hypothesis.

Note that no extra assumption of continuity of $f$ is needed. Although if we assume that $f$ is Riemann integrable, then $f$ has only countably many discontinuities.

The same argument works to establish analogous result in the case of Lebesgue integrability.

share|improve this answer
I have a question.Since $f(2^n)\geq 2^nf(1)$,can such c exist? –  89085731 Mar 29 '12 at 6:53
If we insist that $f(t)$ is strictly positive, then no such $c$ can exist. –  William Mar 29 '12 at 7:00
The conditions $f(t_1)>0$ and $f(t_1+t_2)\ge f(t_1)+f(t_2)$ for $t_1,t_2>0$ can be replaced by the weaker condition that $f$ is monotonically increasing. This makes the problem a bit less trivial. –  robjohn Mar 29 '12 at 8:10
@robjohn: Maybe I'm missing something (it's late and I should sleep), but why would what you suggested introduce complications? Suppose that $f$ is monotonically increasing. Suppose for some $x_0$, $f(x_0) - c \geq \delta > 0$. Then for all $x\geq x_0$, we have $f(x) - c \geq \delta$, and we carry through as before (assuming only that $f$ is integrable on compact intervals, and the domain under consideration is $[a, \infty)$, for any $a\in\mathbb{R}$. –  William Mar 29 '12 at 8:19
@WNY: I wasn't intending complications; I was trying to make the question more interesting. Under the stated conditions, there can be no $c$ so that $\int_{0}^{x}f(t)\,dt \leq cx$ for all $x>0$, so the problem is vacuous. –  robjohn Mar 29 '12 at 8:49
show 1 more comment

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.