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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