I've read in a book of Faries (called: Self-adjoint Opertors) the following statement: Given an inner product, it satisfies the Cauchy-Schwarz inequality, $$|\langle f,g \rangle | \leq \|f \| \|g \|.$$ This means that the norm of g, $\|g\|$, can be computed as the supremum of $|\langle f,g \rangle|$, with $\|f\| \leq 1$. So this is a first question. A second question is: I really don't understand that first question above for one reason. The fact is that I know the definition of sup, but I don't understand it's real significance or even when do I have to use it, that is, how a mathematician think about "pass the supremum". Thanks!


The claim is $$\|g\| = \sup_{f:\|f\| \le 1} |\langle f,g\rangle |.$$

The Cauchy-Schwarz inequality shows that $\|g\|$ us an upper bound for $|\langle f,g\rangle|$ for any $f$ satisfying $\|f\| \le 1$. Thus, $$\|g\| \ge \sup_{f:\|f\| \le 1} |\langle f,g\rangle |.$$ To show the reverse inequality, take $f := g/\|g\|$.

  • $\begingroup$ Thank you very much. Now I see. By your text, I can say that if something (let's say A) is an upper bound for another (say B) then A=sup B. Right? $\endgroup$ – R. Ferreira May 5 '17 at 21:14
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    $\begingroup$ @R.Ferreira No, $\sup B$ is the least upper bound. If $A$ is an upper bound, then all you can say is $A \ge \sup B$ (which is the first step in my proof). To show equality you need to show that it is the least upper bound, which is the second step of my proof. $\endgroup$ – angryavian May 5 '17 at 21:28
  • $\begingroup$ Now I got the idea! Thanks once more! $\endgroup$ – R. Ferreira May 6 '17 at 1:26

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