TIFR 2014 Problem 8 Solution -Checking one-one/onto-ness

TIFR 2014 Problem 8 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.

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Problem:True/False

Let (f:\mathbb{R} \to \mathbb{R}) be a continuous function such that (|f(x)-f(y)| \ge |x-y| ), for all (x,y \in \mathbb{R}). Then

A. f is both one-one and onto.

B. f is one-one and may be onto.

C. f is onto but may not be one-one.

D. f is neither one-one nor onto.

Discussion:

Let (f(x)=f(y)) for some (x,y\in \mathbb{R}). Then from the given inequality, we get (0 \ge |x-y| ) which is saying, (|x-y|=0 ) (since modulus can take non-negative values only) and that implies (x=y). So (f) is one-one.

From the inequality, we can see that (f) increases in a steady rate, we want to see whether it is onto or not.

We have (|f(x)-f(0)| \ge |x-0| = |x| ).

If now, (|f(x)| \le M ) then we will end up having (|f(x)-f(0)| \le |f(x)|+|f(0)| \le M+|f(0)| ) which is a contradiction.

So, (|f(x)|) is not bounded.

Already, (f) is continuous and one-one, so f must be increasing or decreasing (strictly). And since (|f|) is not bounded above, we use intermediate value theorem to conclude that (f) must be onto.

Helpdesk

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