TIFR 2014 Problem 21 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.
Also Visit: College Mathematics Program of Cheenta
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 $f=0$.
TIFR 2014 Problem 21 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.
Also Visit: College Mathematics Program of Cheenta
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 ?