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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 $f=0$.

**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

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 $f=0$.

**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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