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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.
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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