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

[Euclid]

I'm having trouble with this for some reason. If A:\mathcal{H}\to \mathcal{H} is a bounded operator between Hilbert spaces, the norm of A is
||A|| = \inf\limits_{\psi \neq 0} \frac{||A\psi||}{||\psi||}.
My trouble is in verifying that ||A|| is in fact a bound for A in the sense that ||A\psi|| \leq ||A|| ||\psi||. 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 \psi\in \mathcal{H}, then by definition ||A|| \leq \frac{||A\psi||}{||\psi||} and this gives the incorrect inequality.
Did I overlook something?

 

[Hurkyl]

||A|| = \inf\limits_{\psi \neq 0} \frac{||A\psi||}{||\psi||}

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 \|A\| is in fact a bound for A in the sense that \|A\psi\| \leq \|A\| \|\psi\|

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

\|A\| = \sup\limits_{\psi \neq 0} \frac{\|A\psi\|}{\|\psi\|}

and by

\|A\| = \sup \left\{\|A\psi\| : \|\psi\| =1\right\}

(as mathwonk said) and these are equivalent.