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In case of sequence of functions as above, I am confused about pointwise convergence. Especially when x nears zero, could we say that function tends to $0$ as $n$ approaches $\infty$.

My book while discussing uniform convergence of this sequence says that pointwise limits are $0$ for all $x$. Then it says that function attains the maximum value $\frac{1}{2}$ at $x=\frac{1}{n}$. How could this be the case? Shouldn't the pointwise limit at $x=\frac{1}{n}$ be also $\frac{1}{2}$ instead of $0?$

Also if we consider point $x=\frac{1}{2n}$, we find the function attains value of $\frac{2}{5}$ which is again non-zero. We could find infinitely many points in R in neighborhood of $0$ where the pointwise limit will come out to be non-zero contrary to what my book says. Am I correct? Please suggest.

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  • $\begingroup$ Your question is very hard to read. Please use MathJax and use paragraphs. $\endgroup$ – Toby Mak Nov 25 '19 at 12:15
  • $\begingroup$ @ Toby Mak I have edited the question and paragraphed it. Could you please edit it further as I do not how to use MathJax. $\endgroup$ – HARVEER RAWAT Nov 25 '19 at 12:20
  • $\begingroup$ Related : math.stackexchange.com/questions/588976/… $\endgroup$ – Arnaud D. Nov 25 '19 at 15:41
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Pointwise convergence of $f_n$ to $f$ means $f_n(x) \to f(x)$ for every fixed $x$. Here you cannot allow $x$ to depend on $n$. Uniform convergence means $\sup_x |f_n(x)-f(x)| \to 0$ and this demands that $|f_n(x_n)-f(x_n)| \to 0$ for any sequence of points $(x_n)$.

In your example we do have $f_n(x) \to 0$ for every fixed $x$ but $|f_n(\frac 1 n) -0|$does not tend to $0$. Hence the sequence converges pointwise but not uniformly.

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  • $\begingroup$ @ Kabo Murphy What can we say about pointwise convergence when x tends to 0. Couldn't the function approach to some non-zero value then? $\endgroup$ – HARVEER RAWAT Nov 25 '19 at 12:26
  • $\begingroup$ For pointwise convergence you don't vary $x$. You have to consider two cases: $x=0$ and $x$ is a fixed number not equal to $0$. There is no question of letting $x$ vary an approach $0$. @HARVEERRAWAT $\endgroup$ – Kavi Rama Murthy Nov 25 '19 at 12:30
  • $\begingroup$ Shouldn't it be f(x) in your answer instead of f(xn)? $\endgroup$ – HARVEER RAWAT Nov 25 '19 at 12:46
  • $\begingroup$ @HARVEERRAWAT No, when you replace $x$ by $x_n$, $|f_n(x)-f(x)|$ becomes $|f_n(x_n)-f(x_n)|$. $\endgroup$ – Kavi Rama Murthy Nov 25 '19 at 12:51
  • $\begingroup$ If that's the case then you can't replace $f(x_n)$ with 0 as I said before that it could be non-zero when x is dependent on n? $\endgroup$ – HARVEER RAWAT Nov 25 '19 at 12:58
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Let $f_n(x)= \frac{nx}{1+n^2x^2}.$

Pointwise limit means: for each fixed (!) $x \in \mathbb R$, the sequence $(f_n(x))$ is convergent .

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