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

Prerequisites

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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