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

$\blacksquare$