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# 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 .

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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 .

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