# Prove that the dual space of $R^n$ is $R^n$

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)

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.

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