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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