I have been trying to solve this problem for hours...
At this point, I believe that there is a mistake in the problem, but I assume that I am in the one who is wrong. May you help me?
Here is the problem:
Let $f_n(x)= \frac{nx}{3+nx},\ n\in \mathbb{N},\ \text{and}\ x>0.$
a) Show that the sequence $\{f_n\}_{n\in \mathbb{N}}$ converges uniformly on $[a, \infty)$ for any $a>0$
b) Show that the sequence $\{f_n\}_{n\in \mathbb{N}}$ does not converge uniformly on $\mathbb{R}$.
Here is my perspective:
To prove that the sequence converges uniformly I need to find an $N$ that depends only on epsilon. I am using this definition:
I made a graph with some of the functions:
Here is the graph:
According to the graph, from $[1, \infty)$, the limit of the $f_n$ is $1$, but when $x$ is close to 0 the limit, from the right, is $0$.
Then I am wondering is the a of the problem $[a, \infty)$ must be greater or equal than 1 instead of greater than 0 to get uniform convergence.
When I calculate absolute value of $f_n(x) - f(x),$ I am getting $nx > \frac{3}{\varepsilon} - 3$.
I do not see how to find an N that depends only on epsilon.
May you tell me if my reasoning is correct?
$\endgroup$2 Answers
$\begingroup$You are absolutely correct that the problem is at $x=0.$ However, you're asked to establish uniform convergence on $[a,\infty)$ for $a>0$, so the problem at zero isn't a problem on this interval. Note that for any $a>0,$ $f_n(a)\to 1$ so it's not the case that points close to $x=0$ have $f(x)\to 0$; this only happens at $x=0$ itself. The points closer to zero will go more slowly, though.
As a reflection of this, notice from your plots that convergence to $1$ improves as $x$ gets larger. This actually helps you see that there's uniform convergence. The $N$ that works for $x=a$ will also work for $x>a.$ Now, you just need to convert this into a rigorous proof.
And then to show that it doesn't converge uniformly on $[0,\infty),$ just use the problem you noticed at $x=0.$ The function that you converge to pointwise isn't continuous, and this always means that the convergence isn't uniform (if each $f_n$ is continuous).
$\endgroup$ 7 $\begingroup$For all $n\in\mathbb{N}$, let $$f_n:(0,+\infty)\to\mathbb{R},x\mapsto\frac{nx}{3+nx}$$ The sequence $(f_n)$ is pointwise convergent to $L:x\mapsto 1$.
To decide whether the convergence is uniform or not, consider :
$$\delta_n(x)=\left|L(x)-f_n(x)\right|=1-\frac{nx}{3+nx}=\frac{3}{3+nx}$$
We see that $\delta_n\left(\frac 1n\right)=\frac 34$, hence $$\sup_{x>0}\delta_n(x)\not\to 0$$
and the convergence is not uniform on $(0,+\infty)$.
However, if we fix $a>0$ and consider $x\in[a,+\infty)$ :
$$\delta_n(x)\le\frac{3}{3+na}$$
Hence :
$$\sup_{x\ge a}\delta_n(x)\le\frac{3}{3+na}\to 0$$
which proves that the convergence is uniform on any $[a,+\infty)$, with $a>0$.
$\endgroup$ 2
