I was given the very first inf definition, and have to prove that 3rd is the same as the given one. What I tried is taking sup on both sides of llAvll =< llAll llvll, but failed. Any help would be greatly appreciated.
Thanks
I was given the very first inf definition, and have to prove that 3rd is the same as the given one. What I tried is taking sup on both sides of llAvll =< llAll llvll, but failed. Any help would be greatly appreciated.
Thanks
Let $\displaystyle \|A\|_0=\sup\nolimits_{\|v\|=1}\|Av\|$ and let $\displaystyle \|A\|_1=\inf\{c:\|Av\|\leq c\|v\|\textrm{ for all }v\in V\}$.
Consider any $\displaystyle c$ such that $\displaystyle \|Av\|\leq c\|v\|$ for all $\displaystyle v\in V$ and choose any $\displaystyle u\in V$ with $\displaystyle \|u\|=1$.
Then $\displaystyle \|Au\|\leq c\|u\|=c$ and so $\displaystyle \sup\nolimits_{\|u\|=1}\|Au\|\leq c$, i.e. $\displaystyle \|A\|_0\leq c$. Taking the inf over all such $\displaystyle c$, we deduce that $\displaystyle \|A\|_0\leq \|A\|_1$.
If $\displaystyle v\in V$ is non-zero, let $\displaystyle u=v/\|v\|$. Then $\displaystyle \|u\|=1$ and so $\displaystyle \|Au\|\leq\|A\|_0$.
It follows by linearity that $\displaystyle \|Av\|\leq\|A\|_0\|v\|$ for all non-zero $\displaystyle v\in V$ and therefore for all $\displaystyle v\in V$, since it is clearly true when $\displaystyle v=0$.
Hence $\displaystyle \|Av\|\leq c\|v\|$ for all $\displaystyle v\in V$ when $\displaystyle c=\|A\|_0$. By the definition of $\displaystyle \|A\|_1$, we see that $\displaystyle \|A\|_1\leq\|A\|_0$.
Thus we have $\displaystyle \|A\|_0=\|A\|_1$.