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