What is the Definition and Equivalence of the Norm of a Bounded Operator?

[Euclid]

I'm having trouble with this for some reason. If [tex]A:\mathcal{H}\to \mathcal{H}[/tex] is a bounded operator between Hilbert spaces, the norm of [tex]A[/tex] is
[tex] ||A|| = \inf\limits_{\psi \neq 0} \frac{||A\psi||}{||\psi||}[/tex].
My trouble is in verifying that [tex]||A||[/tex] is in fact a bound for [tex]A[/tex] in the sense that [tex]||A\psi|| \leq ||A|| ||\psi||[/tex]. I'm actually not even sure if that's true, but I was able to verify this by the definition given here http://en.wikipedia.org/wiki/Operator_norm. I basically just want to make sure the definitions are equivalent. The trouble is that if [tex]\psi\in \mathcal{H}[/tex], then by definition [tex]||A|| \leq \frac{||A\psi||}{||\psi||}[/tex] and this gives the incorrect inequality.
Did I overlook something?

 

[Hurkyl]

[tex] ||A|| = \inf\limits_{\psi \neq 0} \frac{||A\psi||}{||\psi||}[/tex]

That's supposed to be a supremum.

 

[Euclid]

Ah, that makes perfect sense. I should have realized that. Thanks.

 

[mathwonk]

and you can restrict to phi of length one for clarity.

so the definition is just the (largest) radius of the image of the unit sphere.

 

[Euclid]

Is that for a general operator, or have you assumed linearity?

 

[mathwonk]

to my knowledge, the word "operator" in ths context of hilbert space always means linear operator.

 

[benorin]

My trouble is in verifying that [tex]\|A\|[/tex] is in fact a bound for [tex]A[/tex] in the sense that [tex]\|A\psi\| \leq \|A\| \|\psi\|[/tex]

Notably, in Rudin's Real & Comlex Analysis, the norm of [tex]A[/tex] is defined by the above, and by

[tex] \|A\| = \sup\limits_{\psi \neq 0} \frac{\|A\psi\|}{\|\psi\|}[/tex]

and by

[tex] \|A\| = \sup \left\{\|A\psi\| : \|\psi\| =1\right\} [/tex]

(as mathwonk said) and these are equivalent.