Prove that $(f_n)$, $f_n =x^n$, $x \in (0,1)$ is not uniformly convergent on $(0,1)$

Question

Question 1. Prove that $f_n:(0,1) \to \mathbb{R}$ is not uniformly convergent on $(0,1)$, where $f_n = x^n , n\in \mathbb{N}$ .

Proof: We need to show that, $\forall \ k \in \mathbb{N} $, $\exists \epsilon_0>0$ and $x_0 \in (0,1)$ such that $|f_n(x_0) - f(x_0)| \geq \epsilon_0$ for some $n_0> k$.

Clearly, $\displaystyle\frac{1}{2^{\frac{1}{p}}} \in (0,1) \ \forall p \in \mathbb{N}$ [ Since $ 2^{1/p}> 1 $ for any $p$]

The sequence converges pointwise to $f(x)=0$.

We choose any $k \in \mathbb{N}$.

Now, $|f_{k+1}\displaystyle(\frac{1}{2^{\frac{1}{k+1}}}) -0 | = |\frac{1}{2}| \geq \frac{1}{2} = \epsilon_0$ for $x_0 =\displaystyle\frac{1}{2^{\frac{1}{k+1}}} \ n_0=k+1 , \epsilon_0 = 1/2 $.

We are done.

Is this valid?

Answer 1

\begin{align} \underset{x\to1}{\lim} f_n(x) = \underset{x\to1}{\lim}x^{n} = 1 \text{ } \forall n\in \mathbb{N} \end{align}

In other words, as $x$ gets arbitrarily close to $1$, $f_n(x)$ also gets arbitrarily close to 1. Hence, \begin{align*} \exists \epsilon> 0 \text{ and } \exists x\in(0,1) \text{ s.t } ||f_n(x)||_{\infty} > \epsilon \text{ }\forall n \in \mathbb{N}. \end{align*}

Which implies that $f_n(x)$ is not uniformly convergent on $(0,1)$.