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October 22, 2017

TIFR 2014 Problem 7 Solution -Limit of Integration


TIFR 2014 Problem 7 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 Real and Complex Analysis by Walter Rudin. This book is very useful for the preparation of TIFR Entrance.

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Problem


Let (f_n(x); n\ge 1) be a sequence of continuous nonnegative functions on ([0,1]) such that

( \lim_{n\to\infty} \int_{0}^{1}f_n(x)dx = 0 )

Which of the following statements is always correct?

A. (f_n \to 0) uniformly of ([0,1])

B. (f_n) may not converge uniformly but converges to (0) point-wise.

C. (f_n) will converge point-wise and the limit may be non-zero.

D. (f_n) is not guaranteed to have a point-wise limit.


Discussion:


We start by a very well known example: (g_n(x)=x^n).

(  \int_{0}^{1}g_n(x)dx= \frac{1}{n+1}\to 0 \int_{0}^{1}f_n(x)dx ) as (n\to \infty).

We know (g_n) does not converge uniformly on ([0,1]) because the limit is (1) at (x=1) and (0) everywhere else so we have a non-continuous limit.

So straight-away A,B are false. Question is now whether at all the sequence has to have a point-wise limit or not.

For this, we take our hint from (g_n) and construct (f_n(x)= \sqrt{n}x^n ).

Then ( \int_{0}^{1}f_n(x)dx= \sqrt{n}\frac{1}{n+1}\to 0 ) as ( n\to \infty).

But look at (f_n(1)= \sqrt{n} ). Therefore, (f_n) does not converge at the point (x=1).

So option (D) i.e., (f_n) is not guaranteed to have a point-wise limit. is true.


Helpdesk

  • What is this topic: Real Analysis
  • What are some of the associated concept: Point wise limit, convergence, Uniformly Continuous
  • Book Suggestions: Real and Complex Analysis by Walter Rudin

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