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Let $T:X\to X$ be a linear bounded mapping. I have to prove $\|T^n\|\leq \|T\|^n$.

Let $Tx=cx$, where $c>0$. This is a linear mapping. $$T^2 x=T(Tx)=T(cx)=cTx=c^2 x.$$ Hence $\|T^2x\|=c^2\|x\|.$ Similarly, $\|T^n x\|=c^n\|x\|$.

$$\|Tx\|^n=c^n \|x\|^n.$$

Hence, if $\|T^n\|\leq \|T\|^n$, then $c^n\|x\|\leq c^n\|x\|^n\implies \|x\|^{n-1}\geq 1$.

How is this always true? Isn't this dependent on the condition that $\|x\|\geq 1$?

Thanks in advance!

Chris Brooks
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  • Yes. But if this is true for all linear transformations, it should also be true for $Tx= cx$. –  Sep 25 '13 at 17:22
  • Ah, I was not taking that as defining $T$ but rather saying $T$ has eigenvector $x$ with eigenvalue $c$... – Thomas Andrews Sep 25 '13 at 17:27

3 Answers3

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The simple way is via induction: $$\|T(T^{n-1}x)\|\leq \|T\|\cdot \|T^{n-1}x\|\leq \|T\|\cdot \|T^{n-1}\|\cdot\|x\|$$

More generally, we have for bounded operators $T_1,T_2$, that $$\|T_1T_2\|\leq \|T_1\|\cdot\|T_2\|$$

Thomas Andrews
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The way you should be doing this is $\|Tx\| \le C\,\|x\|$ where $C$ is the operator norm of $T$, i.e. $C = \|T\|$. This is what it means for $T$ to be bounded. Given that, we have

$$\|T^2x\| = \|T(Tx)\| \le C\,\|Tx\| \le C^2 \|x\| = \|T\|^2\,\|x\|.$$

Thus, $\|T^2x\| \le \|T\|^2\,\|x\|$, i.e. $\|T^2\| \le \|T\|^2.$ Do you see how to do the inductive argument from here?

Martin Sleziak
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Cameron Williams
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  • By $|T^2|$ I'm assuming you mean $|T^2 x|$. How does $|T^2 x|\leq |T^2||x|$ imply $|T^2x|\leq |T^2|$? –  Sep 25 '13 at 17:19
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By definition, $\|Tx\| \le \|T\| \|x\|$ for all $x$.

Also, if $\|A x\| \le L \|x\|$ for all $x$, then you must have $\|A\| \le L$.

So, $\|T^n x \| = \|T (T^{n-1} x) \| \le \|T \| \|T^{n-1} x \| \le \cdots \le \|T\|^n \|x\|$. That is, $\|T^n x \| \le \|T\|^n \|x\|$ for all $x$.

So, letting $A=T^n$ above, we have $\|T^n \| \le \|T\|^n$.

copper.hat
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