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.

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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