# Is {$fn$} and {$gn$} is uniformly convergent?

Consider two sequence $f_n$ and $g_n$ of functions where $f_n:[0,1] \to \mathbb{R}$ and $g_n: \mathbb{R} \to \mathbb{R}$ which are defined by

$f_n(x) = x^n$ and
$g_n(x) = \begin{cases} cos(x-n)\frac{π}{2} &\mbox{if } x ∈ [n-1,n+1] \\ 0 & \text{otherwise}. \end{cases}$

Then

1) Neither $f_n$ nor $g_n$ is uniformly convergent

2) $f_n$ is not uniformly convergent but $g_n$ is

3) $g_n$ is not uniformly convergent but $f_n$ is

4) Both $f_n$ and $g_n$ are uniformly convergent...

I think both $f_n$ and $g_n$ are pointwise convergent so they will not uniformly convergent. Therefore option 1 is correct that is neither $f_n$ nor $g_n$ is uniformly convergent.

• Uniform convergence implies pointwise convergence but not vice versa. To check for pointwise convergence, just consider each fixed $x$ and send $n \to \infty$. Checking for uniform convergence is a bit harder, but you should show that you can do the pointwise part in your question before we help with the uniform part. – Ian Dec 6 '17 at 15:15
• @lomberlego Try to find the limit of $f_n$ as $n \to \infty$. Then, we can discuss whether it is only the pointwise limit, or it is both pointwise and uniform limit. – ThePortakal Dec 6 '17 at 15:17
1. The pointwise limit of $(f_n)_{n\in\mathbb{N}}$ is the function$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\mathbb{R}\\&x&\mapsto&\begin{cases}0&\text{ if }x<1\\1&\text{ otherwise.}\end{cases}\end{array}$$which is discontinuous. But a sequence of continuous functions cannot converge uniformly to a discontinuous function.
2. The pointwise limit of $(g_n)_{n\in\mathbb{N}}$ is the null function but, for each $n\in\mathbb N$, there is a real number $x$ (namely, $n$) such that $g_n(x)=1$. Therefore, the convergence cannot be uniform.