 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