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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,

then

$$f(t) \leq c$$ 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!

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2  
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

1 Answer 1

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.

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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

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