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# TIFR 2013 Problem 14 Solution -Uniform Continuous or Not? TIFR 2013 Problem 14 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 program 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 defined by $f(x)=sin(x^3)$. Then f is continuous but not uniformly continuous.

Hint: Can you find a sequence whose terms can get arbitrarily close to each other but the function gives distant values?

## Discussion:

The function sin takes the value 1 at $4n+1$ multiples of $\pi/2$ and it is -1 at $4n-1$ multiples of $\pi/2$ .

Let $x_n=(n\pi+\pi/2)^{1/3}$ That is, the function takes the value +1 when n is even and it is -1 when n is odd.

Now $x_{n+1}-x_n$ $=$ $\frac{( n+1 )\pi+\pi/2-(n\pi+\pi/2)}{\text{terms involving n}}$

This gives that the two terms $x_{n+1}$ and $x_n$ are close to each other. Because, the limiting value of the difference is zero. (So if you give me any positive real number $\delta$ I can find an n such that the difference of two consecutive terms is less than that (\delta\) )

And what happens to $f(x_n)$? It is +1 and -1 for two consecutive terms (or -1 and +1). Therefore, the difference $|f(x_{n+1})-f(x_n)|$ is always 2.

In particular, if I give $\epsilon=1$ then whatever $\delta$ you produce I will select two consecutive terms in the above sequence $x_n$ which has distance less than $\delta$ and the difference of values of $f$ would not be less than 1.

This proves that $f$ is not uniformly continuous.

Remark: $f$ is continuous because it is a composition of two continuous functions ( the sine function applied to the polynomial function $x\to x^3$ ).

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Uniform continuity,
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

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