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TIFR 2014 Problem 21 Solution - Determining 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


Problem:


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.


Helpdesk

  • What is this topic: Real Analysis
  • What are some of the associated concept: Fundamental theorem of calculus, Increasing Function
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert


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


Problem:


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.


Helpdesk

  • What is this topic: Real Analysis
  • What are some of the associated concept: Fundamental theorem of calculus, Increasing Function
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

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