In the pic, in the second proof of Thm 191 (the one that starts at the paragraph:"We can prove theorem 191 without appealing to the more difficult theorem 190..."), I don't understand why:
$\int (f_n)g dx = (\int (f_n)^k dx)^{\frac{1}{k}} G^{\frac{1}{k'}}$
I mean from $g$ proportional to $(f_n)^{k-1}$ I know that there's a constatn $c$ s.t $g=c(f_n)^{k-1}$, i.e
$\int (f_n)g dx= \int c (f_n)^k dx$
Anyone care to explain how did they to their equality?
Thanks.