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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.

Inequalities section

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What does $(f)_n$ mean? –  Davide Giraudo Dec 25 '11 at 11:42
    
I believe it's a sequence of functions: $f_n(x)$. –  MathematicalPhysicist Dec 25 '11 at 12:01
    
$(f)_n = Min(f, n) $ for $f\ge 0$. It's defined at the beginning of section 6 (page 127 in the second edition, softcover editon, Cambridge Mathematical Library) :-) –  user20266 Dec 25 '11 at 12:12
    
:-D thanks. MP. –  MathematicalPhysicist Dec 25 '11 at 13:41
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1 Answer

up vote 3 down vote accepted

The answer to your question is hidden on the pages before the one from which you made your screenshot (sections 188, 189) which tell you when you have equality in Hölder's inequality. See also http://mathworld.wolfram.com/HoeldersInequalities.html.

Let me summarize what is not visible on the screenshot but to anyone having access to the book:

  • the definition $(f)_n: = \min(f, n)$ for $f\ge 0$ (to be found at beginning of section 6)
  • the definition of effectively proportional: two functions $f, g$ are said to be effectively proportional, if there constants $A, B,$ not both zero, such that $Af\equiv Bg, $ where $\equiv$ is equality with the possible exception of a set of measure zero (to be found at the beginning of section 6.9)
  • the statement of (the usual) Hölder's inequality (Theorems 188, 189): if $k>1$ and $1/k+ 1/k'= 1$, then, unless $f^k$ and $g^{k'}$ are effectively proportional, $$ \int fg dx < \left(\int f^k dx\right)^{1/k} \left(\int g^{k'} dx \right)^{1/k'} $$ (what is possibly not made explicit enough is the statement that equality holds if $f^k, g^{k'}$ are effectively proportional)

With this background information the reasoning of the alternative proof for Theorem 191 (which can be seen on the screenshot) should now be easy to comprehend: $g$ is taken to be effectively proportional to $(f)_n^{k-1}$. This leaves a degree of freedom (scaling), so $g$ can be chosen such that $$\int g^{k'} = G$$ Now the line from the screenshot becomes clear, even more obvious when one additional intermediate step is added: $$ \int (f)_n g dx = \left(\int(f)_n^kdx\right)^{1/k}\left(\int g^{k'} dx \right)^{1/k'} = \left(\int(f)_n^kdx\right)^{1/k} G^{1/k'}$$ the first equality being true because $(f)_n^{k-1}$ and $g$ are effectively proportional (which is the same thing as $f^k$ and $g^{k'}$ being effectively proportional).

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I still don't get it, why does $G^{1/k'}$ is not under integral sign, why $\int fg dx = (\int f_n ^k )^{1/k} G^{1/k'}$, I mean G is defined as: $$G(t)=(1-t)^{-2k'}g(t/(1-t))$$?! –  MathematicalPhysicist Dec 25 '11 at 13:58
    
If you've seen my last comment please forget it. $G$ is a fixed constant. Nonetheless you have to use the fact that you have equality in Hölder's inequality when $g$ is effectivley proportional to $(f)_n^{k-1}$. But this leaves you the freedom to scale $g$ so that $\int g^{k^\prime}dx = G$. (The comment about the book being sometimes difficult to read remains true, though :-)). –  user20266 Dec 25 '11 at 15:42
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