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I'm having difficulty understanding how to choose the $F(x)$ given $F_n(x)$ in uniform convergence. For reference, I'm speaking about the formula: $||F(x) - F_n(x)|| < \epsilon$ whenever $x \in D, n \geq N$

It seems that everyone just puts the function evaluated at $0$, but I am not sure if that's correct.

The problem I'm working on is showing that $\gamma_n(t) = (\frac{1}{1 + nt}, \frac{t}{n})$ does not converge uniformly on $[0,1]$

I know that I need to find an $N \in \mathbb{N}$ for my $n$ which does not depend on $t$. In this case, I would be needing to show that it doesn't exist. I guess I'm just not sure how I can show that the $N$ doesn't exist.

EDIT: I'm wanting to show that $\gamma_n \rightarrow \gamma$, so does this mean that my $\gamma(t) = (0, 0)$ when $n$ is sufficiently large?

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    Is $\gamma_n(x)$ an interval? – 2017-02-14
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    It's a sequence in $\mathbb{R}^2$ – 2017-02-14
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    If that's the case, then why are we talking about its convergence in $\mathbb{R}$? Do you mean $[0,1]\times [0,1]$? – 2017-02-14
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    I don't see where it's said that we're converging in $\mathbb{R}$? EDIT: $\gamma : [0, 1] \rightarrow \mathbb{R}^2$ – 2017-02-14

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If $F_n$ is going to converge uniformly to anything, then that thing will be the same as the pointwise limit of $F_n$. The reason: for each $x$, $$ |F_n(x)-F(x)|\le \sup_x|F_n(x)-F(x)|=:||F_n-F||, $$ so if $||F_n-F||\to0$, then $F_n(x)\to F(x)$ for every $x$. So your only choice for the uniform limit of $F_n$ is the pointwise limit.

In your example, you should check that the pointwise limit of the $\gamma_n$ is: $$ \gamma(t):=\begin{cases} (1,0)&\text{if $t=0$}\\ (0,0)&\text{otherwise} \end{cases}. $$ Your job is to decide whether the convergence of $\gamma_n$ to $\gamma$ is uniform (we've already established pointwise convergence). Hint: Show each function $\gamma_n$ is continuous. Is the limiting function $\gamma$ continuous?

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    That totally makes sense. So, now we see that our $N$ depends on which $t$ we choose, because when $t = 0$, our $N$ chosen for $t \neq 0$ could and would fail. – 2017-02-14
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Compute $||\gamma_n(1/n)-\gamma(1/n)||$. Then you will see that $||\gamma_n(1/n)-\gamma(1/n)||>2$ for all $n$. This shows that the convergence of $(\gamma_n)$ is not uniform.