# Finding "limit-function" of $f_n=\frac{x}{1+n^2x^2}$, Pointwise/Uniform convergence of ${f_n'}$

I am preparing for my exam and need help with the following tasks:

Let $${f_n}:[-1,1]\to \mathbb{R}$$ be defined as $$f_n=\frac{x}{1+n^2x^2}$$

• Is $${f_n}$$ pointwise/uniformly convergent? Specify the "limit-function". Determine the derivative of this function.

So I know that $$fn$$ is pointwise and uniformly convergent to $$0$$. But what is this "limit-function" supposed to look like? Can I just say $$f(x)= 0$$ ? The derivative would be $$0$$ too... But whats the point in this task then?

• Check if $${f_n'}$$ is pointwise/uniformly convergent. Specify the "limit-function" if there is one.

So $$f_n'= \frac{1-n^2x^2}{(1+n^2x^2)^2}$$

The problem is, that pointwise and uniform convergence are totally confusing me. To see if the sequence is pointwise converging, we show that $$\lim\limits_{n\to\infty} f_n'(0)=1,\lim\limits_{n\to\infty} f_n'(1)=0,\lim\limits_{n\to\infty} f_n'(-1)=0$$ and thus is converging to $$f(x)=0$$ for [$$-1,0$$) ($$0,1$$] and $$f(x)=1$$ for $$x=0$$ Is this correct? Unfortunately I am not able to show if the sequence is uniformly converging. The only thing I have is that $$|f_n'|\leq \frac{1-n^2x^2}{4n^2x^2}=\frac{1}{4n^2x^2}-\frac{1}{4}$$.

Is there anyone who could give me an advice?

(Also this is not a duplicate, since the focus of this question lays on the "limit-function" and the derivative and I couldn't find anything that could solve my problem)

We have that $$\displaystyle f_n'(x) = \frac{1 - n^2x^2}{(1+ n^2x^2)^2}$$ converges pointwise to $$\displaystyle g(x) = \begin{cases} 1, & x= 0 \\0, & x \in [-1,1] \setminus \{0\}\end{cases}$$

Thus,

$$|f'_n(x) - g(x)| = \begin{cases} 0, & x= 0 \\ \left|\frac{1 - n^2x^2}{(1+ n^2x^2)^2}\right|,& x \in [-1,1] \setminus \{0\}\end{cases}$$

For uniform convergence to hold, we must have $$\sup_{x \in [-1,1]}|f_n'(x) - g(x)| \to 0$$ as $$n \to \infty$$

However,

$$\sup_{x \in [-1,1]}|f_n'(x) - g(x)| \geqslant \left|f_n'\left( \frac{1}{\sqrt{2}n}\right) - g\left( \frac{1}{\sqrt{2}n}\right)\right| = \frac{1 - n^2\cdot \left(\frac{1}{\sqrt{2}n} \right)^2 }{\left(1 + n^2\cdot \left(\frac{1}{\sqrt{2}n} \right)^2\right)^2 } \\=\frac{1- \frac{1}{2}}{\left(1 + \frac{1}{2}\right)^2} = \frac{2}{9} \underset{n \to \infty}{\not\to0}$$

• Thanks for your answer. But should this be not $\leq$ instead of $\geq$. I mean lets say $\frac{2}{9}$ would go to $0$, then we could'nt say if $\sup |f'n(x) - g(x)$| $\to$ $0$ or not.... or am I wrong? Mar 25, 2022 at 18:06
• @Analysis_Mark: What I'm showing is that $\sup_{x \in [-1,1]}|f_n'(x) - g(x)| \geqslant 2/9$ for all $n$. Therefore it cannot converge to $0$ as $n \to \infty$. This means the conditions for uniform convergence cannot hold. Convergence is NOT uniform.
– RRL
Mar 25, 2022 at 18:15
• Other approaches to show non-uniform convergence such as discontinuity of the limit function (with $f_n'$ continuous) are fine for this problem. What I am showing you is a very general direct approach to prove non-uniform convergence when even when less direct approaches fail.
– RRL
Mar 25, 2022 at 18:17
• You should always be aware that for $h_n \to h$ uniformly on a set $D$, it is necessary and sufficient that $\lim_{n \to \infty}\sup_{x \in D} |h_n(x) - h(x) | = 0$.
– RRL
Mar 25, 2022 at 18:23

Assume $$f_n' \to g$$ uniformly on $$[-1,1]$$.

Since $$f_n'(0)=1$$ for every $$n \in \mathbb N$$, we have $$g(0)=\lim_{n \to \infty} f_n'(0)=1$$. Moreover, $$g(x)=\lim_{n \to \infty} f_n'(x)=0$$ for every $$x \in [-1,1]\setminus \{0\}$$.

Consider $$\varepsilon=\frac{1}{2}$$. By our assumption there is $$N \in \mathbb N$$ so that $$|f_N'(x)-g(x)|<\frac{1}{2}$$ for all $$x \in [-1,1]$$.

Since $$f_N'$$ is continuous at $$0$$ (being a rational function with no singularity at $$0$$) there is $$\delta>0$$ so that $$|f_N'(x)-1|<\frac{1}{2}$$ whenever $$|x|<\delta$$. Now notice that for $$x \in (-\delta, \delta)\setminus \{0\}$$ we have \begin{aligned} \left|f_N'(x)-g(x)\right|&=\left|\left(f_N'(x)-1\right)-\left(g(x)-1\right)\right| \\&\geq \left|\left|f_N'(x)-1\right|-\left|g(x)-1\right|\right| \quad (\text{Reverse Triangle Inequality}) \\&=\left|\left|f_N'(x)-1\right|-1\right| \\&=1-\left|f_N'(x)-1\right|>1-\frac{1}{2}=\frac{1}{2}. \end{aligned} This is a contradiction. Therefore $$\{f_n'\}$$ does not converge uniformly on $$[-1,1]$$.

• Of course, since $f'$ obviously fails to be continuous on $[-1,1]$, this argument could be reduced to one line if we allow ourselves to invoke the Uniform Limit Theorem. Mar 25, 2022 at 3:21
• Thanks for your answer. I just don't understand one thing: Why is $f'(x)-1=-1$. Because the limit function is 0 for $x\neq 0$? But the limit function is 1 for $x=0$. Why are we allowed to discount that? Mar 25, 2022 at 11:39
• My bad I meant to have the fact that $x≠0$ explicitly baked into my argument but it seems tired eyes got the best of me. Moreover, I failed to notice that $f'$ is not even the pointwise limit of $\{f_n'\}$ and so I apologize for the abusively incorrect notation there. I have edited my post to reflect these corrections. Mar 25, 2022 at 14:24

$$f_n'(x)=\frac{1-n^2x^2}{(1+n^2x^2)^2}$$

$$f_n'(0)=1\implies \lim_{n\to \infty}f_n'(0)=1$$

And for $$x\ne0, f_n'(x)= \frac{1-n^2x^2}{(1+n^2x^2)^2}\to 0$$. So the limit function will be:

$$g(x)=\begin{cases}0; x\ne 0\\1; x=0\end{cases}$$

If the convergence $$f_n'(x)\to g(x)$$ were uniform then by continuity of $$f_n'$$'s, $$g(x)$$ should have been continuous but since this is not the case, it follows that the convergence is not uniform.

• Thanks for your answer! What about the first task? Is the limit function 0? But then this task is so senseless...We also have find the derivative of the limit function which would be again 0... Mar 25, 2022 at 11:04
• @Analysis_Mark For the first task, the limit function is indeed the zero function so the derivative of the limit function is again zero function. However, if $f_n(x)\to f(x)$ then there is no reason to believe at the outset that $f_n'(x)\to f'(x)$.
– Koro
Mar 25, 2022 at 11:45
• All right. So the task should probably confuse the student in thinking that $f_n'(x)$ is converging to f'(x) which is not right. Mar 25, 2022 at 11:50
• @Analysis_Mark: To avoid that, I used the symbol $g(x)$. :-)
– Koro
Mar 25, 2022 at 11:56