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I was going through the proof of this question in "Introductory Functional Analysis with Applications" by Kreyszig.

(Apologies for not being able to type the whole of it)

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I got stuck after the line where we took the supremum over all x of norm 1.

How did we arrive at the step of equality in the Cauchy-Schwarz inequality?

Any good explanation would be helpful.

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Just calculate it directly; $$f(x)=\sum_k \xi_k\gamma_k = \sum_k \gamma_k\gamma_k = \sum_k \gamma_k^2=\left(\sum_k \gamma_k^2\right)^{\frac12}\cdot\left(\sum_k \gamma_k^2\right)^{\frac12}=\|x\|\cdot\left(\sum_k \gamma_k^2\right)^{\frac12}$$ since $\|x\|=\left(\sum_k\xi_k^2\right)^{\frac12} = \left(\sum_k\gamma_k^2\right)^{\frac12}$ when the $\xi$ and $\gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.

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  • $\begingroup$ How did you get $\sum_k \xi_k \gamma_k = \sum _k \gamma_k \gamma_k$? $\endgroup$ – Oliver G Aug 28 '19 at 13:38

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