Prove that $f_n$ with $f_n :[0,a]\rightarrow\mathbb{R},a\in \mathbb{R}, f_n(x)=x^n$ is continous $\forall n \in \mathbb{N}$

Question

I know how to prove it for $n=2$ can I somehow inductively conclude the Statement $\forall n\in\mathbb{N}$?

Because $\forall c,b \in[0,a]:L|c-b|\geq |c^2-b^2|= |(c+b)(c-b)|\iff L\geq |c+b|\iff L\geq 2a $

Answer 1

Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_n\times f_1$ and the product of two continuous functions is again a continuous function.

Answer 2

Using the binomial theorem yields:

$$|(c+h)^n-c^n| \le \sum_{i=1}^n {n \choose i} |c|^{n-i}|h|^i \xrightarrow{h \to 0}$$

because $n \in \mathbb{N}$ is fixed. Hence $x\mapsto x^n$ is continuous at $c$.

Answer 3

For any $x,y\in [0,a]$ we have \begin{eqnarray} |x^n-y^n|&=&|x-y|\sum_{k=0}^{n-1}x^{n-1-k}y^k\\ &\le&|x-y|\sum_{k=0}^{n-1}x^{n-1-k}y^k\\ &\le&|x-y|\sum_{k=0}^{n-1}a^{n-1}\\ &=&na^{n-1}|x-y| \end{eqnarray}

Hence $f_n$ is continuous.