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## ISI MStat PSB 2008 Problem 3 | Functional equation

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This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 3 based on Functional equation . Let’s give it a try !!

## Problem– ISI MStat PSB 2008 Problem 3

Let $g$ be a continuous function with $g(1)=1$ such that $g(x+y)=5 g(x) g(y)$ for all $x, y .$ Find $g(x)$.

### Prerequisites

Continuity & Differentiability

Differential equation

Cauchy’s functional equation

## Solution :

We are g is continuous function such that$g(x+y)=5 g(x) g(y)$ for all $x, y$ and g(1)=1.

Now putting x=y=0 , we get $g(0)=5{g(0)}^2 \Rightarrow g(0)=0$ or , $g(0)= \frac{1}{5}$ .

If g(0)=0 , then g(x)=0 for all x but we are given that g(1)=1 . Hence contradiction .

So, $g(0)=\frac{1}{5}$ .

Now , we can write $g'(x)= \lim_{h \to 0} \frac{g(x+h)-g(x)}{h} = \lim_{h \to 0} \frac{5g(x)g(h)-g(x)}{h}$

$= 5g(x) \lim_{h \to 0} \frac{g(h)- \frac{1}{5} }{ h} = 5g(x) \lim_{h \to 0} \frac{g(h)- g(0) }{ h} = 5g(x)g'(0)$ (by definition)

Therefore , $g(x)=5g'(0)g(x)= Kg(x)$ , for some constant k ,say.

Now we will solve the differential equation , let y=g(x) then we have from above

$\frac{dy}{dx} = ky \Rightarrow \frac{dy}{y}=k{dx}$ . Integrating both sides we get ,

$ln(y)=kx+c$ c is integrating constant . So , we get $y=e^{kx+c} \Rightarrow g(x)=e^{kx+c}$

Solve the equation g(0)=1/5 and g(1)=1 to get the values of K and c . Finally we will get , $g(x)=\frac{1}{5} e^{(ln(5)) x} =5^{x-1}$.

But there is a little mistake in this solution .

What’s the mistake ?

Ans- Here we assume that g is differentiable at x=0 , which may not be true .

Correct Solution comes here!

We are given that $g(x+y)=5 g(x) g(y)$ for all $x, y .$ Now taking log both sides we get ,

$log(g(x+y))=log5+log(g(x))+log(g(y)) \Rightarrow log_5 (g(x+y))=1+log_5 (g(x))+log_5 (g(y))$

$\Rightarrow log_5 (g(x+y)) +1= log_5 (g(x))+1+log_5 (g(y)) +1 \Rightarrow \phi(x+y)=\phi(x)+\phi(y)$ , where $\phi(x)=1+log_5 (g(x))$

It’s a cauchy function as $\phi(x)$ is also continuous . Hence , $\phi(x)=cx$ , c is a constant $\Rightarrow 1+log_5 (g(x))=cx \Rightarrow g(x)=5^{cx-1}$.

Now $g(1)=1 \Rightarrow 5^{c-1}=1 \Rightarrow c=1$.

Therefore , $g(x)=5^{x-1}$

## Food For Thought

Let $f:R to R$ be a non-constant , 3 times differentiable function . If $f(1+ \frac{1}{n})=1$ for all integer n then find $f”(1)$ .

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## ISI MStat PSB 2007 Problem 4 | Application of Newton Leibniz theorem

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 4 based on use of Newton Leibniz theorem . Let’s give it a try !!

## Problem– ISI MStat PSB 2007 Problem 4

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a bounded continuous function. Define $g:[0, \infty) \rightarrow \mathbb{R}$ by,
$g(x)=\int_{-x}^{x}(2 x t+1) f(t) dt$
Show that g is differentiable on $(0, \infty)$ and find the derivative of g.

### Prerequisites

Riemann integrability

Continuity

Newton Leibniz theorem

## Solution :

As $f: \mathbb{R} \rightarrow \mathbb{R}$ be a bounded continuous function hence the function

$|\Phi(t)|=|(2xt+1)f(t)|=|2xt+1||f(t)|<(|2xt|+1)M<(2|x|^2+1)M$ , which is finite for a particular x so it’s a riemann integrable function on t.

Now, by fundamental theorem we have g(x)=F(x)-F(-x) , where F is antiderivative of $\Phi(t)$ .

Hence from above we can say that g(x) is differentiable function over x .
Now by Leibniz integral rule we have $g'(x)=(2x^2+1)f(x)+f(-x)(1-2x^2) + \int_{-x}^{x} (2t)f(t) dt$.

## Food For Thought

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. Now, we define $g(x)$ such that $g(x)=f(x) \int_{0}^{x} f(t) d t$
Prove that if g is a non increasing function, then f is identically equal to 0.

Categories

## ISI MStat PSB 2007 Problem 3 | Application of L’hospital Rule

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 3 based on use of L’hospital Rule . Let’s give it a try !!

## Problem– ISI MStat PSB 2007 Problem 3

Let f be a function such that $f(0)=0$ and f has derivatives of all order. Show that $\lim _{h \to 0} \frac{f(h)+f(-h)}{h^{2}}=f”(0)$
where $f”(0)$ is the second derivative of f at 0.

### Prerequisites

Differentiability

Continuity

L’hospital rule

## Solution :

Let L= $\lim _{h \to 0} \frac{f(h)+f(-h)}{h^{2}}$ it’s a $\frac{0}{0}$ form as f(0)=0 .

So , here we can use L’hospital rule as f is differentiable .

We get L= $\lim _{h \to 0} \frac{f'(h)-f'(-h)}{2h} = \lim _{h \to 0} \frac{(f'(h)-f'(0)) -(f'(-h)-f'(0))}{2h}$

= $\lim _{h \to 0} \frac{f'(h)-f'(0)}{2h} + \lim _{k \to 0} \frac{f'(k)-f'(0)}{2k}$ , taking -h=k .

= $\frac{f”(0)}{2} + \frac{f”(0)}{2}$ = $f”(0)$ . Hence done!

## Food For Thought

Let $f:[0,1] \rightarrow[0,1]$ be a continuous function such $f^{(n)} := f ( f ( \cdots ( f(n \text{ times} ))$ and assume that there exists a positive integer m such that $f^{(m)}(x)=x$ for all $x \in[0,1] .$ Prove that $f(x)=x$ for all $x \in[0,1]$

Categories

## ISI MStat PSB 2013 Problem 2 | Application of sandwich Theorem

This is a very beautiful sample problem from ISI MStat PSB 2013 Problem 2 based on use of Sandwich Theorem . Let’s give it a try !!

## Problem– ISI MStat PSB 2013 Problem 2

Let f be a real valued function satisfying $|f(x)-f(a)| \leq C|x-a|^{\gamma}$ for some $\gamma>0$ and $C>0$
(a) If $\gamma=1,$ show that f is continuous at a
(b) If $\gamma>1,$ show that f is differentiable at a

### Prerequisites

Differentiability

Continuity

Limit

Sandwich Theorem

## Solution :

(a) We are given that $|f(x)-f(a)| \leq C|x-a|$ for some $C>0$.

We have to show that f is continuous at x=a . For this it’s enough to show that $\lim_{x\to a} f(x)=f(a)$.

$|f(x)-f(a)| \leq C|x-a| \Rightarrow f(a)-C|x-a| \le f(x) \le f(a) + C|x-a|$

Now taking limit $x \to a$ we have , $\lim_{x\to a} f(a)-C|x-a| \le \lim_{x\to a} f(x) \le \lim_{x\to a} f(a) + C|x-a|$

Using Sandwich theorem we can say that $\lim_{x\to a} f(x) = f(a)$ . Since $\lim_{x\to a} -C|x-a| = \lim_{x\to a} C|x-a|=0$

Hence f is continuous at x=a proved .

(b) Here we have to show that f is differentiable at x=a for this it’s enough to show that the $\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$ exists .

We are given that , $|f(x)-f(a)| \leq C|x-a|^{\gamma}$ for some $\gamma>1$ and $C>0$ ,

which implies $|\frac{f(x)-f(a)}{x-a} | \le C|x-a|^{\gamma -1}$

$\Rightarrow -C|x-a|^{\gamma -1} \le \frac{f(x)-f(a)}{x-a} \le C|x-a|^{\gamma -1}$

Now taking $\lim_{x\to a}$ we get by Sandwich theorem $\lim_{x\to a}\frac{f(x)-f(a)}{x-a} =0$ i.e f'(a)=0 .

Since , $\lim_{x\to a} C|x-a|^{\gamma -1} = \lim_{x\to a} -C|x-a|^{\gamma -1} = 0$ , for $\gamma >1$.

Hence f is differentiable at x=a proved .

## Food For Thought

$f : R \to R$ be such that $|f(x)-f(a)| \le k|x-y|$ for some $k \in (0,1)$ and all $x,y \in R$ . Show that f must have a unique fixed point .

Categories

## ISI MStat PSB 2014 Problem 2 | Properties of a Function

This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 2 based on the use and properties of a function . Let’s give it a try !!

## Problem– ISI MStat PSB 2014 Problem 2

Let $a_{1}<a_{2}<\cdots<a_{m}$ and $b_{1}<b_{2}<\cdots<b_{n}$ be real numbers such
that $\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right| \text { for all } x \in \mathbb{R}$
Show that $m=n$ and $a_{j}=b_{j}$ for $1 \leq j \leq n$

### Prerequisites

Differentiability

Mod function

continuity

## Solution :

Let , $\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right|=f(x) \text { for all } x \in \mathbb{R}$

Then , $f(x)=\sum_{i=1}^{m}\left|a_{i}-x\right|$ is not differentiable at $x=a_1,a_2, \cdots , a_m$ —(1)

As we know the function $|x-a_i|$ is not differentiable at $x=a_i$ .

Again we have , $f(x) = \sum_{j=1}^{n}\left|b_{j}-x\right|$ it also not differentiable at $x= b_1,b_2, \cdots , b_n$ —-(2)

Hence from (1) we get f has m non-differentiable points and from (2) we get f has n non-differentiable points , which is possible only when m and n are equal .

And also the points where f is not differentiable must be same in both (1) and (2) .

As we have the restriction that $a_{1}<a_{2}<\cdots<a_{m}$ and $b_{1}<b_{2}<\cdots<b_{n}$ .

So , we have $a_{j}=b_{j}$ for $1 \leq j \leq n$ .

## Food For Thought

$a<b \in \mathbb{R} .$ Let $f:[a, b] \rightarrow[a, b]$ be a continuous and differentiable on (a,b) . Suppose that $\left|f^{\prime}(x)\right| \leq \alpha<1$ for all $x \in(a, b)$ for some $\alpha .$ Then prove that there exists unique $x \in[a, b]$ such that $f(x)=x$

Categories

## ISI MStat 2018 PSA Problem 7 | Continuous Function

This is a problem from ISI MStat 2018 PSA Problem 7 based on Continuity.

## Continuous Function – ISI MStat Year 2018 PSA Question 7

Let $f$ be a function defined from $(0, \infty)$ to $\mathbb{R}$ such that
$\lim _{x \rightarrow \infty} f(x)=1$ and $f(x+1)=f(x)$ for all x
Then $f$ is

• (A) continuous and bounded.
• (B) continuous but not necessarily bounded.
• (C) bounded but not necessarily continuous.
• (D) neither necessarily continuous nor necessarily bounded.

### Key Concepts

Epsilon-Delta definition of limit

Continuity

Bounded function

ISI MStat 2018 PSA Problem 7

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

Try to use the epsilon-delta definition of limit and the property that $f(x+1)=f(x)$ for all x.

$\lim_{x \to \infty} f(x) = 1$ $\Rightarrow \exists M > 0$ such that $x > M \Rightarrow$ $|f(x) – 1| < \epsilon$. Now $f(x) = f(x+1)$ for all $x \in \mathbb{R} \Rightarrow f(x) = f(x+n)$ for every $n \in \mathbb{Z}$.

Let’s try to use this .

Given any $y \in \mathbb{R}$ we can select a suitable $n \in \mathbb{Z}$ such that $y+n > M$. Then $|f(y+n) – 1| < \epsilon$. But $f(y+n) = f(y)$. Hence , $|f(y) – 1| < \epsilon$. Hence , for all $y \in \mathbb{R},$ we have $|f(y) – 1| < \epsilon$. Since , $\epsilon > 0$ is arbitrary , we must have $f(y) = 1$ for all $y \in \mathbb{R}.$

So $f$ is continuous and bounded

Categories

# Let us take a warm up quiz

[h5p id="16"]

# Understand the problem

Â There exist a non-negative continuous function $f: [0,1] \longrightarrow \mathbb{R}$ such that $\int_{0}^{1} f^{n} dx \longrightarrow 2$ as $\longrightarrow \infty$ (a) TRUE (b) FALSE

##### Source of the problem

TIFR PROBLEM 10

##### Topic
Continuous FunctionÂ
EASY
##### Suggested Book
REAL ANALYSIS BY S.K MAPA

Do you really need a hint? Try it first!

Rather I want to say that it is a comment on the question that here $f^{n}$ does not mean $f \circ f \circ . . . . . . \circ f$ ($n$ times). Here $f^{n}= f \bullet f . . . . . \bullet f$ ($n$ times). Now do you want to think again with this disclosure?
Make two cases Case 1:Â  $0 [Observe that $f$ is non negative function] Case 2:Â  $f(x) > 1$ for some $x \in [0,1]$ Now prove for each case that $\int_{0}^{1} f^{n} dx \nrightarrow 2$ as $n \to \infty$. So, by the last statement you have guessed the validity of the statement . It is a false statement!! In the next two cases, we will basically prove two cases.
Case 1:Â  If $f(x) \leq 1\ \forall\ x \in [0,1]$ then $f^{n}(x) \leq 1\ \forall\ x \in [0,1]$ So, $\int_{0}^{1} f^{n} dx \leq\ \int_{0}^{1} 1 dx = 1$ $\implies \lim_{n \to \infty} \int_{0}^{1} f^{n} dx \leq 1$ So, $\int_{0}^{1} f^{n} dx \nrightarrow 2$ as $n \to \infty$
Case 2:Â  Suppose $f(x) > 1$ for someÂ  $y \in [0,1]$ Now, as $f(x)$ is continuous function We have $f(x) > 1+\epsilon\ \forall\ x \in (y - \delta, y + \delta)$ for some $\epsilon , \delta > 0$ $\implies f^{n}(x) > (1+ \epsilon)^{n}$ then we have $\int_{0}^{1} f^{n}\ dx > \int_{y - \delta} ^{y + \delta} (1+ \epsilon)^{n}\ dx$ [If $y \in [0,1]]$ or $\int_{1 - \delta}^{1} (1+ \epsilon)^{n}\ dx$ [If $y=1$] or $\int_{0}^{0 + \delta} (1+ \epsilon)^{n}\ dx$ [If $y=0$] In either case , $\int_{0}^{1} f^{n}\ dx > (1 + \epsilon)^{n}\ \delta \to \infty$   So,the statement is false.

# Similar Problems

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## ISI MStat PSB 2009 Problem 6 | abNormal MLE of Normal

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 6. It is based on the idea of Restricted Maximum Likelihood Estimators, and Mean Squared Errors. Give it a Try it !

## ISI MStat PSB 2009 Problem 3 | Gamma is not abNormal

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## ISI MStat PSB 2009 Problem 1 | Nilpotent Matrices

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# College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuingÂ who wish to rediscover the world of mathematics.

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## TIFR 2015 Problem 2 Solution -Image of continuous function

TIFR 2015 Problem 2 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: \mathbb{R} \to \mathbb{R} ) be a continuous function. Which of the following can not be the image of ((0,1]) under (f)?

A. {0}

B. ((0,1))

C. ([0,1))

D. ([0,1])

## Discussion:

If f is the constant function constantly mapping to 0, which is continuous, then the image set is {0}.

Suppose that (f((0,1])=(0,1)) . Then (f((0,1))=(0,1)- {f(1)} ). Now since (f(1)\in (0,1) ) the set ( (0,1)- {f(1)} ) is not connected. But ((0,1)) is connected, and we know that continuous image of a connected set is connected. This gives a contradiction. So ((0,1)) can not be the image of ((0,1]) under f.

Define (f(x)=1-x).  Then (f((0,1])= [0,1)).

Define (f(x)=0) for (x\in [0,\frac{1}{2}] ) and (f(x)= 2(x-\frac{1}{2}) ) for (x\in [\frac{1}{2} ,1 ] ). (f) is continuous on  ((0,\frac{1}{2}] ) and ( [\frac{1}{2} ,1 ] ) and (f) agrees on the common points, by pasting lemma (f) is continuous on ( [0,1] ) . And image of ((0,1] ) is ([0,1]).

TIFR 2015 Problem 2 Solution is concluded.

## Chatuspathi

• What is this topic: Real Analysis
• What are some of the associated concept: Continuous Function, Metric Space
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

Categories

## TIFR 2014 Problem 2 Solution -Continuous Bounded Function

TIFR 2014 Problem 2 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:\mathbb{R}\to \mathbb{R}) be a continuous bounded function. Then

A. f has to be uniformly continuous

B. there exists an (x\in \mathbb{R}) such that (f(x)=x)

C. f cannot be increasing

D. (\lim_{x\to \infty}f(x)) exists.

## Discussion:

Define (f(x)=sin(x^2)). Then (f) is bounded, continuous.

Take (x_1=\sqrt{n\pi+\pi/2}), (x_2=\sqrt{n\pi}).

Then (|x_1-x_2|=\frac{\pi/2}{\sqrt{n\pi+\pi/2}+\sqrt{n\pi}}<\frac{1}{\sqrt{n\pi}}).

Since (\frac{1}{\sqrt{n\pi}}\to 0), given (\delta >0), we can have

(|x_1-x_2|<\delta) for large values of (n).

But (|f(x_1)-f(x_2)|=1). So given (\epsilon =1/2) we can never find a (\delta >0) for which (|x_1-x_2|<\delta) would imply that (|f(x_1)-f(x_2)|<\epsilon).

So this (f) is not uniformly continuous. So (A) is false.

Also from this example, since (f) does not have limit as (x\to \infty), we conclude that (D) is false.

Increasing does not mean strictly increasing. So, you are allowed to take a constant function as an example of increasing function. And it is bounded,continuous. This disproves (C).

Now we are left only with (C). Since we disproved all of the others, and one option is correct, (C) has to be true. We inspect the proof just to be sure.

Since (f) is bounded, there exists (M>0) such that (|f(x)|<M).

That means (-M<f(x)<M) for all (x\in \mathbb{R}).

Look at (g(x)=f(x)-x). Finding fixed point of (f) is same as finding a zero of (g).

We know that (f) is continuous, therefore (g) is continuous. We would like to find two points where (g) takes values with opposite signs. Now

(g(M)=f(M)-M<0) and (g(-M)=f(-M)-(-M)=f(-M)+M>0).

Therefore, (g) must cut the x-axis. Here we are using the intermediate value theorem for continuous functions.

This proves (B).

## Helpdesk

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

Categories

## TIFR 2013 Problem 28 Solution -Bijective continuous non-homeomorphism

TIFR 2013 Problem 28 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:True/False?

Let $f:X\to Y$ be a continuous map between metric spaces. If $f$ is a bijection, then its inverse is also continuous.

## Hint:

No need to go into weird spaces. Start with familiar spaces, find out counterexamples.

## Discussion:

One way to search for counterexamples is finding spaces that can not be homeomorphic but surely has a one way continuous function. For example, start with a not-connected space and have its image connected. Then you can’t go back, because connected sets will go to a connected set only (under continuous functions). Or similar thoughts can be made about compact sets.

To illustrate, I will give one such example.

Consider $I=[0,1]$. And $A=[-0.5, 0]$ $B=(1,2]$. Translation by $0.5$ takes $A$ to $[0,0.5]$ and translating $B$ by $-0.5$ gives $(0.5,1]$. Therefore, we get $I$ as the whole image of $f$ where $f(x)=x+0.5$, $x\in A$ and $f(x)=x-0.5$, $x\in B$. This $f$ is a continuous function from $A \cup B$ to $I$. This can be checked by sequence criteria, or pasting lemma as well. Note that for pasting lemma we need $A$ and $B$ be closed. But they truly are closed because here we are considering the space $A \cup B$ as a subspace of $\mathbb{R}$. That means, $B$ can be written as a closed set in $\mathbb{R}$ intersection the space $A \cup B$.

Now that we know $f$ is continuous, f is bijective is easy to check. And further, by our starting point, we know that inverse function can not be continuous since our domain space $A \cup B$ is not connected (or compact) while I is.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuity, Bijective, Continuity of inverse function
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert