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.