# Uniform convergence of sequence of functions $\frac{nx}{(1+n^2x^2)}$ for real x

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.

• Your question is very hard to read. Please use MathJax and use paragraphs. – Toby Mak Nov 25 '19 at 12:15
• @ Toby Mak I have edited the question and paragraphed it. Could you please edit it further as I do not how to use MathJax. – HARVEER RAWAT Nov 25 '19 at 12:20
• – Arnaud D. Nov 25 '19 at 15:41

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.

• @ 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? – HARVEER RAWAT Nov 25 '19 at 12:26
• 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 – Kavi Rama Murthy Nov 25 '19 at 12:30
• Shouldn't it be f(x) in your answer instead of f(xn)? – HARVEER RAWAT Nov 25 '19 at 12:46
• @HARVEERRAWAT No, when you replace $x$ by $x_n$, $|f_n(x)-f(x)|$ becomes $|f_n(x_n)-f(x_n)|$. – Kavi Rama Murthy Nov 25 '19 at 12:51
• 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? – HARVEER RAWAT Nov 25 '19 at 12:58

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 .