**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 \(f=0\).

can we apply liebnitz rule at the first step ?