1. ## Pointwise convergence

(1) Let $\displaystyle f_n(x)= \frac{nx}{1+nx^2}$.

(a) Find the pointwise limit of $\displaystyle (f_n)$ for all $\displaystyle x \in (0,\infty)$
(b) Is the convergence uniform on $\displaystyle (0,\infty)$?
(c) Is the convergence uniform on $\displaystyle (0,1)$?
(d) Is the convergence uniform on $\displaystyle (1,\infty)$?

(2) Let $\displaystyle g_n(x)= \frac{nx+sin(nx)}{2n}$.

Find the pointwise limit of $\displaystyle (g_n)$ on $\displaystyle \mathbb{R}$. Is this convergence uniform on $\displaystyle [-10,10]$? Is the convergence uniform on all of $\displaystyle \mathbb{R}$?

(3) For each $\displaystyle n \in \mathbb{N}$, find the points on $\displaystyle \mathbb{R}$ where the function $\displaystyle f_n(x) = x/(1+nx^2)$ attains its maximum and minimum values. Use this to prove $\displaystyle (f_n)$ converges uniformly on $\displaystyle \mathbb{R}$. What is the limit function?

(4) For each $\displaystyle n \in \mathbb{N}$, define $\displaystyle f_n$ on $\displaystyle \mathbb{R}$ by
$\displaystyle f_n(x)=\left\{\begin{array}{cc}1,&\mbox{ if } |x|\geq 1/n\\n|x|, & \mbox{ if } |x|<1/n\end{array}\right.$

(a) Find the pointwise limit of $\displaystyle (f_n)$ on $\displaystyle \mathbb{R}$ and decide whether or not the convergence is uniform.

(b)Construct an example of a pointwise limit of continuous functions that converges everywhere on the compact set $\displaystyle [-5,5]$ to a limit function that is unbounded on this set.

If anyone can show me how to do these, I would appreciate it, thanks!

2. Hello !

(1) Let $\displaystyle f_n(x)= \frac{nx}{1+nx^2}$.

(a) Find the pointwise limit of $\displaystyle (f_n)$ for all $\displaystyle x \in (0,\infty)$
That is to say, what is $\displaystyle f(x)=\lim_{n \to \infty} f_n(x)$ ?

Multiply by $\displaystyle \frac{\frac 1n}{\frac 1n}$:

$\displaystyle f(x)=\lim_{n \to \infty} \frac{x}{\frac 1n+x^2}=\frac{x}{x^2}=\frac 1x$

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Definition of uniform convergence :

$\displaystyle \forall \varepsilon >0 ~,~ \exists N_0 \in \mathbb{N} ~,~ \forall n \in \mathbb{N} ~,~ n>N_0 \quad \implies \quad \forall x \in \dots ~,~ sup \ |f_n(x)-f(x)|< \varepsilon$
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$\displaystyle \left|f_n(x)-f(x)\right|=\left|\frac{nx}{1+nx^2}-\frac 1x\right|=\left|\frac{nx^2-(1+nx^2)}{x(1+nx^2)}\right| \boxed{=\frac{1}{x(1+nx^2)}}$

Try to do with that in a first time... Find if the limit is 0 (this is the definition of the $\displaystyle < \varepsilon$ thing) in the given intervals..

3. Thanks. Could someone show me how to do 1b 1c or 1d as an example please? I am still unsure about the rest too.

(b) Is the convergence uniform on $\displaystyle (0,\infty)$?
(c) Is the convergence uniform on $\displaystyle (0,1)$?
(d) Is the convergence uniform on $\displaystyle (1,\infty)$?
So we were left to :

$\displaystyle |f_n(x)-f(x)|=\frac{1}{x(1+nx^2)}$

For $\displaystyle x \in (0, \infty)$, there exists $\displaystyle \alpha >0$ such that :
$\displaystyle |f_n(x)-f(x)|=\frac{1}{x(1+nx^2)} \le \frac{1}{\alpha(1+n\alpha^2)}$

And when $\displaystyle n \to \infty$, $\displaystyle \frac{1}{\alpha(1+n\alpha^2)} \to 0$, that is to say that $\displaystyle \forall \varepsilon >0 ~,~ \exists N_0 \in \mathbb{N} ~,~ \forall n \in \mathbb{N} ~,~ n>N_0 \implies \forall x \in (0, \infty) ~,~ \frac{1}{\alpha(1+n\alpha^2)}< \varepsilon$

$\displaystyle \implies |f_n(x)-f(x)| \le \frac{1}{\alpha(1+n\alpha^2)}< \varepsilon.$

Which is what we're looking for...
I hope you understand & that it looks correct to you :/
Same thing for the following questions.

(2) Let $\displaystyle g_n(x)= \frac{nx+sin(nx)}{2n}$.

Find the pointwise limit of $\displaystyle (g_n)$ on $\displaystyle \mathbb{R}$.
$\displaystyle g_n(x)=\frac{x+\tfrac{\sin(nx)}{n}}{2}$

$\displaystyle g(x)=\lim_{n \to \infty} g_n(x)=\frac{x+\lim_{n \to \infty} \tfrac{\sin(nx)}{n}}{2}$

$\displaystyle -1 \le \sin(nx) \le 1 \implies -\tfrac 1n \le \tfrac{\sin(nx)}{n} \le \tfrac 1n$

$\displaystyle \lim_{n \to \infty} -\tfrac 1n = \lim_{n \to \infty} \tfrac 1n=0$

Using the squeeze theorem, we get that $\displaystyle \lim_{n \to \infty} \tfrac{\sin(nx)}{n}=0$

Therefore, $\displaystyle g(x)=\frac x2$.

6. Hi
Originally Posted by Moo
So we were left to :

$\displaystyle |f_n(x)-f(x)|=\frac{1}{x(1+nx^2)}$

For $\displaystyle x \in (0, \infty)$, there exists $\displaystyle \alpha >0$ such that :
$\displaystyle |f_n(x)-f(x)|=\frac{1}{x(1+nx^2)} \le \frac{1}{\alpha(1+n\alpha^2)}$

And when $\displaystyle n \to \infty$, $\displaystyle \frac{1}{\alpha(1+n\alpha^2)} \to 0$, that is to say that $\displaystyle \forall \varepsilon >0 ~,~ \exists N_0 \in \mathbb{N} ~,~ \forall n \in \mathbb{N} ~,~ n>N_0 \implies \forall x \in (0, \infty) ~,~ \frac{1}{\alpha(1+n\alpha^2)}< \varepsilon$

$\displaystyle \implies |f_n(x)-f(x)| \le \frac{1}{\alpha(1+n\alpha^2)}< \varepsilon.$

Which is what we're looking for...
I hope you understand & that it looks correct to you :/
Same thing for the following questions.
What you want to show is that $\displaystyle \sup_{x\in (0,\infty)} |f_n(x)-f(x)| \underset{n\to\infty}{\to} 0$ but $\displaystyle \frac{1}{x(1+nx^2)}\underset{x\to 0}{\to} \infty$ hence $\displaystyle \sup_{x\in (0,\infty)} |f_n(x)-f(x)|$ does not exist and the sequence can't converge uniformly on $\displaystyle (0,\infty)$.

7. Originally Posted by flyingsquirrel
Hi

What you want to show is that $\displaystyle \sup_{x\in (0,\infty)} |f_n(x)-f(x)| \underset{n\to\infty}{\to} 0$ but $\displaystyle \frac{1}{x(1+nx^2)}\underset{x\to 0}{\to} \infty$ hence $\displaystyle \sup_{x\in (0,\infty)} |f_n(x)-f(x)|$ does not exist and the sequence can't converge uniformly on $\displaystyle (0,\infty)$.
That's something I don't understand (someone tried to explain it to me...), when do you know if you fix x or not ? If we can find a sup to the function, which is convergent while n tends to infinity, shouldn't it work ?
The conditions of a uniform convergence seem to apply to n, not to x ?

8. Originally Posted by Moo
That's something I don't understand (someone tried to explain it to me...), when do you know if you fix x or not ?
I'm not sure to understand what you mean. Here we can't fix $\displaystyle x$ since we work with $\displaystyle \sup_{x\in(0,\infty)}$. What one sometimes does is showing the uniform convergence on every $\displaystyle (\alpha,\infty),\,\alpha>0$ is that what you mean by "fixing" $\displaystyle x$ ?
If we can find a sup to the function, which is convergent while n tends to infinity, shouldn't it work ?
Yes it should.
The conditions of a uniform convergence seem to apply to n, not to x ?
Then you'd rather study $\displaystyle \sup_{n\in\mathbb{N}}|f_n(x)-f(x)|$ ?

9. Originally Posted by flyingsquirrel
is that what you mean by "fixing" $\displaystyle x$ ?
Yes, this is what I meant ^^

Yes it should.
So how come we don't agree on it ?

Then you'd rather study $\displaystyle \sup_{n\in\mathbb{N}}|f_n(x)-f(x)|$ ?
No, it's still $\displaystyle \sup_{x \in \dots}$, but because $\displaystyle x \ge \alpha >0$, $\displaystyle |f_n(x)-f(x)|<|f_n(\alpha)-f(\alpha)|$, because the function is decreasing..
Isn't it a sup of $\displaystyle |f_n(x)-f(x)|$?

10. Moo, The sequence of functions in #1 appears in many texts. There is an extensive discussion of this very problem in Advanced Calculus by Taylor & Mann. Because that is such a classical text, any good mathematics library should have a copy.

That sequence converges uniformly on any closed interval that does not contain 0 and does not converge uniformly on any interval if 0 is in its closure.

11. Hi Plato,

Originally Posted by Plato
Moo, The sequence of functions in #1 appears in many texts. There is an extensive discussion of this very problem in Advanced Calculus by Taylor & Mann. Because that is such a classical text, any good mathematics library should have a copy.

That sequence converges uniformly on any closed interval that does not contain 0 and does not converge uniformly on any interval if 0 is in its closure.
I doubt I can find that easily such a book here
That's what I thought for the convergence, but I didn't see the use of asking for the three intervals... Can you enlighten my blindness here ?
Basically, is there a difference between asking for convergence in $\displaystyle (0, \infty)$ and $\displaystyle [\alpha, \infty[$, $\displaystyle \alpha>0$ ?
I don't really remember if one can be considered as a closed interval :/ That's a part I don't like in maths lol.

Thanks:

12. Originally Posted by Moo
Yes, this is what I meant ^^
As uniform convergence on every $\displaystyle (\alpha,\infty)$, $\displaystyle \alpha>0$ does not imply uniform convergence on $\displaystyle (0,\infty)$, fixing $\displaystyle x$ is useless here.

Fixing $\displaystyle x$ is useful to show the continuity of the limit : if $\displaystyle f_n(x)\underset{n\to\infty}\to f(x)$ for all $\displaystyle x\in(0,\infty)$, if each $\displaystyle f_n$ is continuous on $\displaystyle (0,\infty)$ and if $\displaystyle (f_n)$ converges uniformly on every $\displaystyle (\alpha,\infty)$ (but not necessarily on $\displaystyle (0,\infty)$) we get that $\displaystyle f$ is continuous on every $\displaystyle (\alpha,\infty)$. This does imply its continuity on $\displaystyle (0,\infty)$.

$\displaystyle \begin{cases} \text{continuity on every }(\alpha,\infty),\, \alpha>0 \Rightarrow \text{continuity on }(0,\infty)\\ \text{uniform convergence on every }(\alpha,\infty),\, \alpha>0 \nRightarrow \text{uniform convergence on }(0,\infty) \end{cases}$

13. Yep, now I've understood the whole
Though I guess it's [alpha, infinity[, since (a,b) = ]a,b[
It doesn't change anything since uniform convergence on every $\displaystyle (\alpha,\infty),\,\alpha>0 \Leftrightarrow$ uniform convergence on every $\displaystyle [\alpha,\infty)$.