# $\int_{0}^{x} f(t) dt \ge {f(x)}$ holds or not on $[0,1]$

Let $$f\colon [0,1]\rightarrow [0, \infty )$$ be continuous. Suppose that

$$\int_{0}^{x} f(t)\,\mathrm dt \ge {f(x)} \quad\text{for all }x\in[0,1].$$

Then

$A.$ No such function exists.

$B.$ There are infinitely many such functions.

$C.$ There is only one such function

$D.$ There are exactly two functions.

Now I was thinking, since $\int_{0}^{x} f(t) dt$ is the area under the graph of $f(t)$ from $0$ to $x$ so it can be easily made greater than the value of $f(t)$ at one point, namely $x$. So there will be infinitely many such functions satisfying this condition.

Is my reasoning correct or not?

Thanks.

• Well, this is not a reasoning... Sep 5, 2015 at 16:46
• If $f(t)\not\equiv 0$ satisfy the inequality then so do the function $\lambda f(t)$ for any $\lambda>0$. This rules out D. Sep 5, 2015 at 16:52
• @Winther Unless $f \equiv 0$, so that $\lambda f = f$ for every $\lambda>0$. Sep 5, 2015 at 16:52
• It is known that the derivative of a function cannot be bounded by function itself. Sep 5, 2015 at 18:02
• OP: What about correcting the mess there?
– Did
Sep 6, 2015 at 0:30

Your reasoning has a big hole - the condition is supposed to hold for all $x$, and you talk about just one $x$.

Say $f\le c$. Then the inequality shows that $f(x) \le cx$. Now integrate $cx$ and you see that $f(x)\le c x^2/2$. And then $f(x)\le cx^3/6$. By induction $f(x)\le cx^n/n!$. Let $n\to\infty$ and you see $f(x)=0$. So there's exactly one such function.

Let us modify the problem slightly.

Assume that the function $$f:[0,a]\to[0,\infty)$$ is continuous and that $$\int_{0}^{x}f(t)\, dt \ge f(x) \quad \text{for all } x\in[0,a].$$ Then $f(x) = 0$ for all $x\in[0,a]$.

Proof. We can rewrite the inequality as $$\dfrac{d}{dx}\left(e^{-x}\cdot\int_{0}^{x}f(t)\, dt\right) \le 0 \quad \text{for all } x\in[0,a].$$ Consequently the function $$x\mapsto e^{-x}\cdot\int_{0}^{x}f(t)\, dt$$ is decreasing and it takes its maximum at $x=0$. Thus $$e^{-x}\cdot\int_{0}^{x}f(t)\, dt \le 0 \Leftrightarrow \int_{0}^{x}f(t)\, dt \le 0.$$ However $f$ is non-negative and continuous. We conclude that $f(x) = 0$ for all $x\in[0,a]$.

• This is the nicest approach. Sep 6, 2015 at 15:38
• Yes, it's certainly the nicest approach. Seems curious that $e^x$ is important here, while the terms in the power series for $e^x$ come up in the solution I gave - could be that in some sense they're really the same...(?) Sep 6, 2015 at 15:53
• The function $e^{-x}$ is an integrating factor. The given inequality could be changed to $\displaystyle \int_{0}^{x}f(t)\, dt \ge g(x)f(x) \quad \text{for all } x\in[0,a],$ where $g$ is a positive, continuous function. As before $f(x) = 0$ for all $x\in[0,a]$ will be the only solution.
– JanG
Sep 6, 2015 at 17:35
• @DavidC.Ullrich, I forgot to address my comment above to you.
– JanG
Sep 8, 2015 at 9:57

Let $f$ be such a function. If $0<x< 1$ then by the MWT, there exists $\xi\in (0,x)$ with $$\int_0^xf(t)\,\mathrm dt=xf(\xi)\stackrel{(1)}\le x\int_0^\xi f(t)\,\mathrm dt\stackrel{(2)}\le \int_0^\xi f(t)\,\mathrm dt\stackrel{(3)}\le \int_0^xf(t)\,\mathrm dt.$$ We conclude that $(1)$, $(2)$, and $(3)$ are in fact equalities. The second means that $\int_0^\xi f(t)\,\mathrm dt=0$, the third that $\int_\xi^x f(t)\,\mathrm dt=0$, so that in fact $\int_0^x f(t)\,\mathrm dt=0$. As this holds for all $x\in(0,1)$ and $f$ is continuous, we conclude $f(x)=0$ for all $x\in[0,1]$. So, $C$ - final answer.

• Of course unless I'm missing something this depends on the fact that $x\le 1$; mine would work just as well for $f:[0,2]\to[0,\infty)$. Sep 5, 2015 at 17:05
• @DavidC.Ullrich In fact yours would work imediately for any $[0,n]\to[0,\infty)$ and then per union even for $[0,\infty)\to[0,\infty)$. -Then again, any method that shows that $f$ must be $\equiv 0$ on some $[0,\epsilon]$ can be adjusted to work for $[0,\infty)\to[0,\infty)$: Assume $f$ is nonzero and let $a=\inf\{\,x\in[0,\infty): f(x)\ne 0\,\}$. Then apply the argument to $[a,a+\epsilon)$ ... Sep 5, 2015 at 17:09
• I don't understand your second equality. Should it be inequality (where one uses the property of $f$ given in the problem)? Sep 5, 2015 at 17:53
• i think same as mikep
– R.N
Sep 5, 2015 at 18:25
• @mickep Ack, I missed that - but it turns out to be an equality in the end as well by the squeezing ... Sep 6, 2015 at 9:41