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I am little bit confused on this example of point wise convergence.

For $x \in [0,1]$ and $ n >= 2$ define

$f_n(x) = \left\{ \begin{array}{lr} n^2x^2 &,\: if \: 0 <= x <=1/n \:,\\ -n^2(x - 2/n)&,\: if \: 1/n <= x <= 2/n\:,\\ 0&, \: if \: 2/n < x <= 1 \end{array} \right.$

So my book states that given any $ x > 0 $ , let $M = 2/x$. Then for $ n > M$ we have $2/n < 2/m = x$ , so that $f_n(x) = 0 $. So what I don't understand here is that I don't get the intuition at all why is $f_n(x) = 0$ always I can see for example that this works for any only for values $M = 2/x$ but not for everything also I don't understand why is the limit of function is identically zero the book is very vague on this part if someone could I explain this then that would be perfect.

I also don't understand why this doesn't converge uniformly I am trying to develop intuition for uniform convergence so I think this example would help if someone could explain that would be very nice.

1 Answers1

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For fixed $n$, $f_n \not\equiv 0$ as you can always get values that are in the interval $0 < x < \dfrac{1}{n}$.

But, as your book states you can get a bigger $n$ (and all the ones who follows it) such that $f_n(x) = 0$, this would imply the pointwise limit (that is the limit with fixed $x$, $\displaystyle \lim_{n \to \infty} f_n(x))$ is zero, thus the pointwise limit of the sequence of functions $f_n$ is the $0$ function.

As to why it doesnt converges uniformly, that is because for all $n$, you can take $x = \dfrac{1}{n^2}$, then $$\sup_{x \in [0,1]}||f_n(x)|| \geq 1$$ and this would contradict the uniform convergence to $0$.

aram
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