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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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