Uniform Convergence of $f_n(x)=x^n$

Question

Does $f_n(x)=x^n$ converge uniformly on $[0,1]$?

My instinct is that it does converge to $0$, but I am unsure why. It just seems logical the function would converge to this point.

Answer 1

If $0 \le x<1$ then $f_n(x)=x^n \to 0$ for $n \to \infty$.

If $x=1$ then $f_n(x)=1\to 1$ for $n \to \infty$.

Therefore function $f(x)= \lim_{n \to \infty} f_n(x)$ is not continuous on $[0,1$]. Hence the convergence is not uniform.

Answer 2

If $x=1$ so it converges to $1$. If $x=0$ so it converges to $0$.

Let $0<x<1$ and $\epsilon>0$.

Hence, $x^n<\epsilon$ it's just $n>\frac{\ln\epsilon}{\ln{x}}$ and by the limit definition in the last case $\lim\limits_{n\rightarrow+\infty}x_n=0$.

If $g(x)=0$ for all $0\leq x<1$ and $g(1)=1$ so we proved that $f_n(x)\rightarrow g(x)$.

But (see the Mdoc's post) we have: $$\lim_{n\rightarrow+\infty}\sup\limits_{x\in[0,1]}|x^n-g(x)|=1\neq0$$ Thus, it's not true that $x^n\rightrightarrows g$ on $[0,1]$.

Answer 3

The sequence of functions $\{x^n\}_{n=1}^\infty$ converges pointwise on $[0,1]$ to function $g$ where $g(x)=0$ if $x\in[0,1)$ and $g(1)=1$ if $x=1$.

However, this sequence does not converge uniformly to $g$. To see that this is the case note that the point $\sqrt[n]{1/2}$ is sent by function $x^n$ to 1/2. This implies that if we take for example $\epsilon=1/10$ then we see that for each $n \in \mathbb{N}$ there is $x\in[0,1]$ such that $|f_n(x)-g|>\epsilon$