From Kreyszig's Introductory Functional Analysis with Applications:
Prove that the dual space of $\Bbb R^n$ is $\Bbb R^n$
Let $f(x) = \sum \xi_k \gamma_k$, where $\gamma_k = f(e_k)$ for basis vectors $e_k$ in $\Bbb R_n$.
$|f(x)| \le \sum |\xi_k \gamma_k| \le (\sum_k \xi_k^2)^{1/2}(\sum_k \gamma_k^2)^{1/2} = \|x\| (\sum_k \gamma_k^2)^{1/2}$.
Take the supremum over all $x$ of norm $1$ and we obtain
$\|f\| \le (\sum_k \gamma_k^2)^{1/2}$.
Since for $x = (\gamma_1, \dots, \gamma_n)$ we must have equality in Cauchy schwartz, we must have equality in the previous expression.
Can someone explain where these equalities are coming from? I can't see how either of these show equality.
Let $x = (\gamma_1, \dots, \gamma_n)$ and $z:=\frac{x}{||x||}$. Then $||z||=1$ and
$|f(z)|=f(z)=\frac{1}{||x||}f(x)=\frac{1}{||x||}\sum_k \gamma_k^2=\frac{1}{||x||} \cdot ||x||^2=||x||=(\sum_k \gamma_k^2)^{1/2}.$
Thus we have a vector $z$ of norm $1$ such that $|f(z)|=(\sum_k \gamma_k^2)^{1/2}.$