Continuity of y^n

1. Oct 6, 2013

Someone2841

Hello. This is an improvement on a previous post, "Continuity of y^2". My original plan was to first prove that y and y^2 were continuous and then prove by induction that y^n was continuous; however, in the process of doing so I think I found a better way. This proof is for rudimentary practice of proof construction and should be taken in context of an undergraduate Real Analysis I course. How does this look?

---Start Proof---

Define $(f_n)_{n \in \mathbb{N}}$ as a sequence of functions where $f_n:\mathbb{R} \to \mathbb{R}$ maps $y \mapsto y^n$. I intend prove pointwise continuity for any $f_n$ by showing that:

$\forall n \in \mathbb{N} \;\; \forall e > 0 \;\; \forall u \;\; \exists \delta: \forall y (|y-u| < \delta \implies |f_n(y) - f_n(u)| = |y^n - u^n| < \epsilon)$​

Choose any $n \in N, e > 0,$ and $u$.

Case 1: $u = 0$
Pick $\delta = \epsilon^{\frac{1}{n}}$ and then any $y$. It is clear that $|y-0 | < \epsilon^{\frac{1}{n}} \implies |y^n - 0| < \epsilon$.​

Case 2: $u \neq 0$

Choose a $\delta < |y|$ and $\delta < \frac{\epsilon}{(2^{n}-1)|u|^{n-1}}.$ A $\delta$ meeting this criteria can be found since both $|y|$ and $\frac{\epsilon}{(2^{n}-1)|u|^{n-1}}$ are positive, and there exists no smallest positive number.

Note the following:
1. $|y^n-u^n| = |y-u| \left |\sum_{i=0}^{n-1} u^{n-1-i}y^i \right |$. This is obvious from evaluating $\frac{y^n-u^n}{y-u}$.
2. $|y-u| < \delta < |y| \implies |u| < 2|y|$
Proof of 2.
$|y-u| < |y|$ so $u - y < |y|$ and $u < |y| + y \leq 2|y|$. At the same time $y-u < |y|$,which implies $-u < |y| - y \leq 2|y|$. Both $u$ and $-u$ are $<2|y|$; therefore $|u|<2|y|$.​
3. $\left |\sum_{i=0}^{n-1} u^{n-1-i}y^i \right | < \sum_{i=0}^{n-1} 2^i|u|^{n-1} = (2^n-1)|u|^{n-1}$. This follows from 2. and the evaluation of the geometric series.

It follows that:
$\begin{array} &|y^n-u^n| &=& |y-u| | \sum_{i=0}^{n-1} u^{n-1-i}y^i | & \text{1.}\\ &<& \delta (2^{n}-1)|u|^{n-1} & \text{By assumption that } |y-u| < \delta \text{ and 3.}\\ &<& \frac{\epsilon}{(2^{n}-1)|u|^{n-1}}(2^{n}-1)|u|^{n-1} & \delta\text{ was chosen for this}\\ &=& \epsilon \end{array}$​

.
QED​
---End Proof---

Last edited: Oct 6, 2013
2. Oct 7, 2013

johnqwertyful

If two functions are continuous, their product is as well.

It's trivial to show identity continuous. Then you can multiply them together n times.

3. Oct 7, 2013

Someone2841

This is true, and of course I knew that; my goal here is to prove the continuity of $y_n$ from the definition of continuity without assumptions concerning the properties of continuous function like such. With that said, I probably should have thought more about that when constructing this proof. Proving that the product of two continuous functions is continuous is probably simpler enough that I could have either 1) provided it as a lemma to the proof or 2) used the same methodology. For example

Proof Sketch:
Trivially, $|y-u| < \delta = \epsilon \implies |y-u| < \epsilon$ and therefore $f(y) = y$ is continuous. If $y^n$ is assumed to be continuous, then it can be shown that $y^{n+1}$ is also continuous by [insert a special case of the product of continuity functions here where $f(y) = y$ and $g(y) = y^n$]. By induction, $f(y) = y^n$ is continuous for all $n \in \mathbb{N}$.​

Does this seem right?