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Norm of a Bounded Operator

  1. Mar 22, 2006 #1
    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?
    Last edited: Mar 22, 2006
  2. jcsd
  3. Mar 22, 2006 #2


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    That's supposed to be a supremum.
  4. Mar 22, 2006 #3
    Ah, that makes perfect sense. I should have realized that. Thanks.
  5. Mar 22, 2006 #4


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    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.
  6. Mar 22, 2006 #5
    Is that for a general operator, or have you assumed linearity?
  7. Mar 23, 2006 #6


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    to my knowledge, the word "operator" in ths context of hilbert space always means linear operator.
  8. Mar 27, 2006 #7


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    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.
    Last edited: Mar 27, 2006
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