# Bound on Norm of Power of Element in Normed Algebra

## Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$.

Let $x \in A$ and $n \in \N$.

Then:

$\norm {x^n} \le \norm x^n$

## Proof

The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:

$\norm {x^n} \le \norm x^n$

### Basis for the Induction

We have:

$\norm {x^1} = \norm x = \norm x^1$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$\norm {x^k} \le \norm x^k$

from which it is to be shown that:

$\norm {x^{k + 1} } \le \norm x^{k + 1}$

### Induction Step

This is the induction step.

We have:

 $\ds \norm {x^{k + 1} }$ $=$ $\ds \norm {x^k x}$ $\ds$ $\le$ $\ds \norm {x^k} \norm x$ Definition of Algebra Norm $\ds$ $\le$ $\ds \norm x^k \norm x$ using the induction hypothesis $\ds$ $=$ $\ds \norm x^{k + 1}$

$\blacksquare$