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# Limit of Integration (TIFR 2014 problem 7)

Question:

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 is true.

October 22, 2017