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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 .


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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 .


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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