Uniform convergence of functions and intervals

Question

We define $f_n:\mathbb{R}\to\mathbb{R}$ by $f_n(x)=\dfrac{x}{1+nx^2}$ for each $n\ge 1$.

I compute that $f(x):= \displaystyle\lim_{n\to \infty}f_n(x) = 0$ for each $x\in\mathbb{R}$.

Now, I want to know in which intervals $I\subseteq \mathbb{R}$ the convergence is uniform.

Any hint? Thanks.

Use this technique. – Mhenni Benghorbal Jun 08 '13 at 03:43 — Mhenni Benghorbal, Jun 08 '13 at 03:43
@user73564: You are very welcome. – Mhenni Benghorbal Jun 08 '13 at 04:35 — Mhenni Benghorbal, Jun 08 '13 at 04:35

score 11 · Accepted Answer · edited Apr 13 '17 at 12:20

11

If you use the technique, then the max over $x\in(-\infty,\infty)$ of the function is achieved at $x=\frac{1}{\sqrt{n}}$ and it equals $\frac{1}{2\sqrt{n}}$. So, we have

$$ \sup| f_n(x)-f(x) |= \sup \Big| \dfrac{x}{1+nx^2} \Big|= \frac{1}{2\sqrt{n}} < \epsilon.$$

This shows the sequence converges uniformly over $\mathbb{R}$.

Added: Here is the plot of the function for $n=1,2,3$.

enter image description here

edited Apr 13 '17 at 12:20

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1

answered Jun 08 '13 at 04:10

Mhenni Benghorbal

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Could you please tell me how did you plot this function? – Srijan Jun 08 '13 at 04:25
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This is great, thank you very much. – user73564 Jun 08 '13 at 04:33
@user73564: You are very welcome. – Mhenni Benghorbal Jun 08 '13 at 04:34
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@srijan: I used Maple. Do you want the command? – Mhenni Benghorbal Jun 08 '13 at 04:34
@srijan: Here is the command " plot( {x/(1+1^2x^2),x/(1+2^2x^2), x/(1+3^2*x^2) }, x=-6..6 ); " – Mhenni Benghorbal Jun 08 '13 at 04:51
@MhenniBenghorbal Thanks dear I noted down your command. :) – Srijan Jun 08 '13 at 04:56
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@srijan: Thank you very much. I really appreciate it. – Mhenni Benghorbal Jun 08 '13 at 04:59
@MhenniBenghorbal You are welcome sir. :) – Srijan Jun 08 '13 at 05:00
This supremum that we are calculating , this is of which set ? Set of all functions possible for all values of $n$ over the interval ? And Suppose we admit that $f_n$ is maximum for a certain value of $x$ of $n$. What does this mean ? Does this mean that at each point of the interval , the value attains its maximum value for this particular n and hence on the entire interval ? Also , in different questions , when we solve for the derivative and equate it to zero , how do we make sure that the points that we are getting lie in the Interval concerned.Sorry for bothering. – llecxe Apr 15 '21 at 11:27

score 0 · Answer 2 · answered Jun 08 '13 at 04:18

The sequence converges uniformly on $I$ if for every $\epsilon >0, \exists N$ such that $\forall x \in I $ and $\forall n \geq N$, we have $|f_n(x) - f(x)| < \epsilon$

For the interval $|x| \leq 1$, we see that $x(1+nx) = x +nx^2 \leq 1 + nx^2$, so that $\frac{x}{1+nx^2} \leq \frac{x}{x(1+nx)} = \frac{1}{1+nx}$. So, to make this smaller than $\epsilon$, choose $N_1 = (\frac{1}{\epsilon}-1)$.

For the interval $|x| >1$, we have that $|\frac{x}{1+nx^2}| \leq |\frac{x}{nx^2}| =|\frac{1}{nx}| < \frac{1}{n}$. So, choose $N_2 > \frac{1}{\epsilon}$.

Comparing $N_1, N_2$, we see that we can choose $N = N_2$. Convergence is uniform on $\mathbb{R}$.

Uniform convergence of functions and intervals

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