# Bound on Norm of Power of Element in Normed Algebra

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## 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 | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm x^{k + 1}\) |

$\blacksquare$