For the following sequence of functions and its limit function, we can see that $f_n(x)$ is clearly pointwise convergent
$$f_n(x) = x^n\text{ }\forall x\in[0,1]\text{ and }\forall n\in\mathbb N^*\\
f(x) = \begin{cases}0&\text{if } x\in[0,1)\\1&\text{if } x=1\end{cases}$$
However, I was wondering why this is not uniformly convergent. The condition for uniform convergence is:
$$|f_n(x) – f(x)| < \epsilon,\ \ \ \forall x \text{ when } n > N$$
Now most sources present an argument along the lines of: assume that $f_n(x)$ is uniformly convergent and that $0 < x < 1$, this means that $x^n<\epsilon$ whenever $n>N$. Specifically, this would mean $x^{N+1}<\epsilon$ for some fixed $N$. But if we now pick $x$ such that $1 > x > ε^{\frac{1}{N+1}}$, then this would lead to a contradiction, therefore $f_n(x)$ is not uniformly convergent.
However, I was wondering why couldn't we take $n$ to infinity. If $0 < x < 1$, then $\lim_{n\rightarrow \infty} |f_n(x) – f(x)| = 0$ (which is less than $\epsilon$). Now since $|f_n(x) – f(x)|$ will always be $0$ if $n > \infty$, then wouldn't this be uniformly convergent?
Best Answer
You can't just take $n=\infty$. In effect what you have done is to verify pointwise convergence, not uniform convergence.
Uniform convergence fails because, for every $n$, $\sup_{x\in[0,1]} |f_n(x)-f(x)| = 1$, which you can see by continuity.
Alternatively, uniform convergence must fail because each $f_n$ is continuous, $f$ is not continuous, and a uniform limit of continuous functions is continuous.