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Question:

Let $$f:[0,1]\to [0,\infty )$$ be continuous. Suppose $$\int_{0}^{x} f(t)dt \ge f(x)$$ for all $$x\in [0,1]$$.

Then

A. no such function exists

B. there are infinitely many such functions

C. there is only one such function

D. there are exactly two such functions

Discussion:

Basically, the question is to find out how many such functions can exist.

Let $$F(x)=\int_{0}^{x} f(t)dt$$. That is, $$F(x)$$ is the indefinite integral of $$f$$.

We know from the fundamental theorem of calculus that:

$$F'(x)=f(x)$$ .

So we have $$F'(x) \ge F(x)$$ for all $$x\in [0,1]$$.

We will manage this equation as we do in case of differential equations. Except that, we have an inequality here.

$$F'(x)-F(x) \ge 0$$ for all $$x\in [0,1]$$.

Multiplying both sides by $$e^{-x}$$ the inequality remain unchanged. This is because $$e^{-x} >0$$.

$$e^{-x}F'(x)-e^{-x}F(x) \ge 0$$ for all $$x\in [0,1]$$.

Now, $$(e^{-x}F(x))’ = e^{-x}F(x) -e^{-x}F'(x)$$.

So we have $$(e^{-x}F(x))’ \le 0$$ for all $$x\in [0,1]$$.

Therefore, by taking integral from o to y and by using fundamental theorem of calculus:

$$e^{-y}F(y)-e^{0}F(0) \le 0$$ for all $$y\in [0,1]$$.

i.e, $$e^{-x}F(x) \le F(0)$$ for all $$x\in [0,1]$$.

Also, note that by definition of $$F$$, $$F(0)=0$$. So we have

$$e^{-x}F(x) \le 0$$ for all $$x\in [0,1]$$.

But, since $$e^{-x} >0$$ for all $$x\in [0,1]$$, we have $$F'(x) \le 0$$  for all $$x\in [0,1]$$.

So far, we have not used the fact that $$f$$ is a non-negative function. Now we use it. Since $$f(x) \ge 0$$ for all $$x\in [0,1]$$, therefore by monotonicity of the integral, $$F$$ is an increasing function. This means $$F'(x) \ge 0$$ for all $$x\in [0,1]$$.

By the two inequalities obtained above, we get $$F'(x) = 0$$  for all $$x\in [0,1]$$.

By the fundamental theorem (again!) we get $$f(x)=F'(x)=0$$  for all $$x\in [0,1]$$.

So there is only one such $$f$$ namely the constant function $$f=0$$.