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


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


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 \(f=0\).