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 be continuous. Suppose
for all
.
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 . That is,
is the indefinite integral of
.
We know from the fundamental theorem of calculus that:
.
So we have for all
.
We will manage this equation as we do in case of differential equations. Except that, we have an inequality here.
for all
.
Multiplying both sides by the inequality remain unchanged. This is because
.
for all
.
Now, .
So we have for all
.
Therefore, by taking integral from o to y and by using fundamental theorem of calculus:
for all
.
i.e, for all
.
Also, note that by definition of (F), (F(0)=0). So we have
for all
.
But, since for all
, we have
 for all
.
So far, we have not used the fact that is a non-negative function. Now we use it. Since
for all
, therefore by monotonicity of the integral,
is an increasing function. This means
for all
.
By the two inequalities obtained above, we get  for all
.
By the fundamental theorem (again!) we get  for all
.
So there is only one such namely the constant function
.
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 be continuous. Suppose
for all
.
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 . That is,
is the indefinite integral of
.
We know from the fundamental theorem of calculus that:
.
So we have for all
.
We will manage this equation as we do in case of differential equations. Except that, we have an inequality here.
for all
.
Multiplying both sides by the inequality remain unchanged. This is because
.
for all
.
Now, .
So we have for all
.
Therefore, by taking integral from o to y and by using fundamental theorem of calculus:
for all
.
i.e, for all
.
Also, note that by definition of (F), (F(0)=0). So we have
for all
.
But, since for all
, we have
 for all
.
So far, we have not used the fact that is a non-negative function. Now we use it. Since
for all
, therefore by monotonicity of the integral,
is an increasing function. This means
for all
.
By the two inequalities obtained above, we get  for all
.
By the fundamental theorem (again!) we get  for all
.
So there is only one such namely the constant function
.
can we apply liebnitz rule at the first step ?