# Norm of a Bounded Operator

1. Mar 22, 2006

### 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?

Last edited: Mar 22, 2006
2. Mar 22, 2006

### Hurkyl

Staff Emeritus
That's supposed to be a supremum.

3. Mar 22, 2006

### Euclid

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

4. Mar 22, 2006

### 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.

5. Mar 22, 2006

### Euclid

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

6. Mar 23, 2006

### mathwonk

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

7. Mar 27, 2006

### benorin

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.

Last edited: Mar 27, 2006