Prove uniform converging for $\{f_n(x)\} = x^n$,

Question

The definition says that:

Suppose $D \subseteq R$.

$\{f_n(x)\}$ uniform converging to $f(x)$ in $D$ if any $\epsilon > 0$ , $\exists N(\epsilon)$, such that for any $n > N(\epsilon)$ and any $x \in D$:

$|f_n(x) - f(x)| < \epsilon$

I need to prove uniform convergence for $\{f_n(x)\} = x^n$, where $0 \leq x \leq \frac{1}{2}.$ Can you please solve this in detail? I couldn't find examples on youtube so I'm here wish you can help me. thanks!

Answer 1

Here's a rough sketch for how you might determine $N(\epsilon)$. Note this is definitely not in the form of a rigorous proof and there are several unjustified statements left for you to think about.

We can show that $f_n(x) \to 0$ as $n \to \infty$ for all $x$ in $D = \lbrack0, 1/2\rbrack$. Given $\epsilon > 0$, you need to determine an $N = N(\epsilon)$ such that $n \geq N$ implies for all $x \in D$ that $|f_n(x) - f(x)| = x^n < \epsilon$. Since $|f_n(x) - f(x)| = x^n$ is maximized in $D$ at $x = 1/2$ (why?), you need $N$ large enough that $(1/2)^N < \epsilon$ (which will in turn make $(1/2)^n < \epsilon$ for $n \geq N$). So $$ N > -\log_2 \epsilon $$ must hold. Hence you can choose $N = \max\{\lceil -\log_2 \epsilon \rceil, 1\}$.