# Justify with Proper Reason which among these is correct?

Let $f:[0,1]\to[0,\infty)$ be a continuous function. Suppose

$\displaystyle \int_0^x f(t) \ dt\ge f(x) \ \$ for all $x\in [0,1]$

Then

A. No Such function exists.

B.There are infinitely many such functions.

C. There is exactly one such function.

D.There is two such functions.

• Hint: In particular, the inequality holds at the point $y$ where $f$ has its maximum value. – David Mitra Dec 7 '14 at 16:33

Pick $b\in (0,1]$ arbitrary. By the Mean Value Theorem there exists $c\in (0,b)$ such that $$\frac{\int_0^b f(t)\,dt - \int_0^0 f(t)\,dt}{b-0} = f(c)$$ so that $$\int_0^b f(x)\,dx = bf(c).$$ We conclude $f(x) = f(c)$ on $[0,1]$, whatever $f(c)$ is. But since $b<1$ we see $bf(c)\le f(c)$ for any $f(c)\ge 0$. Thus $f(c) = 0$, so $f(x) = 0$. The answer is (C).

Because $f$ is continuous, you can write $$\frac{d}{dx}\int_{0}^{x}f(t)\,dt - \int_{0}^{x}f(t)\,dt \le 0 \\ \frac{d}{dx}\left[e^{-x}\int_{0}^{x}f(t)\,dt\right] \le 0 \\$$ So the function $h(x)=e^{-x}\int_{0}^{x}f(t)\,dt$ is a non-increasing function on $[0,1]$, which means $h(x) \le 0$ for all $x \in [0,1]$ because $h(0) = 0$. That means $\int_{0}^{x}f(t)\,dt \le 0$ and, therefore, $$f(x) \le \int_{0}^{x}f(t)\,dt \le 0.$$ By assumption $f \ge 0$. Therefore $f(x)=0$ for $x \in [0,1]$ must hold. And you can see that such $f$ works. So there's one solution: $f\equiv 0$. My answer is (C).

The only function possible is $f(x)=0$.

Take a sequence $(a_i)$ convergent to $0$. Say the sequence is $1,1/2,1/2^2,1/2^3,\dots$

At each point $a$, if you have $\int_0^a{f(t) dt}\geq f(a)$, that would imply that the area covered between the graph and $x$-axis from $0$ to $a$ is $\geq$ the area covered between $x$-axis and the constant function $y=f(a)$ from $0$ to $1$.

As $a$ approaches $0$, for the above condition to be true, $f(x)$ would have to approach infinity as $x\to 0$. Try to draw this on a sheet of paper.

• Not getting....please explain. – user151456 Dec 7 '14 at 16:30