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Let ${f_n}$ be a sequence of continuous functions from $[ 0 , 1 ]$ to $\mathbb{R}$.

Suppose that $f_n(x) \rightarrow 0 $ as $ n \rightarrow \infty$ for each $x \in [0,1]$ and also that, for some constant $ K$, we have $$|\int_{0}^{1} f_n(x) dx|\le K < \infty$$ for all $n \in \mathbb{N}$.

Does $\lim _{n \rightarrow \infty} \int_{0}^{1} f_n(x) dx =0$?

It' a problem from Souza, Silva - Berkeley Problems in Mathematics and the hint there says that we should consider a sequence $f_n$ of functions whose graphs are given by the straight lines through the points: $(0,0), (\frac{1}{2n}, n), (\frac{1}{n}, 0),(1,0)$.

I can see that the pointwise limit of this sequence is $0$, but I don't know how to evaluate the value of the integral.

Could you help me with that?

Thank you.

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You can draw a figure and you see that the integral has the area of triangle: $$\int_{0}^{1} f_n(x) dx=\frac{1}{2}$$ and it's pretty clear that $f_n\to0$

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The integral is just an area. In this case, the area of a triangle with base $\frac{1}{n}$ and height $n$.

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  • $\begingroup$ So the integral will be $\le \frac{1}{2}$, is that right? $\endgroup$
    – Don
    May 3 '13 at 19:22
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    $\begingroup$ It's exactly $\frac{1}{2}$, actually. If all you got was $\leq \frac 1 2$ you wouldn't be able to conclude anything about what happens when $n\to\infty$. $\endgroup$ May 3 '13 at 19:24
  • $\begingroup$ Yes, of course. Thanks. $\endgroup$
    – Don
    May 3 '13 at 19:29
  • $\begingroup$ I have one more question, though. The functions are defined on $[0,1]$, and each of them intersects $x$ axis in $(1,0)$, after intersecting it in $(\frac{1}{n}, 0)$. This means that between these two points the function's values are negative, doesn't it? $\endgroup$
    – Don
    May 3 '13 at 19:37
  • $\begingroup$ I apologize, that was a stupid question. Of course, $f_n =0$ on $[\frac{1}{n}, 1]$ $\endgroup$
    – Don
    May 3 '13 at 21:56

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