# Prob. 10 (d), Chap. 6, in Baby Rudin: Holder Inequality for Improper Integrals With Infinite Limits

Here is the link to my Math SE post on Probs. 10 (a) through (c), Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Probs. 10 (a), (b), and (c), Chap. 6, in Baby Rudin: Holder's Inequality for Integrals

Here is the link to my first Math SE post on Prob. 10 (d), Chap. 6, in Baby Rudin:

Prob. 10 (d), Chap. 6, in Baby Rudin: Holder's Inequality for Improper Integrals

And, here is the link to my Math SE post on Prob. 8, Chap. 6, in Baby Rudin:

Prob. 8, Chap. 6, in Baby Rudin: The Integral Test for Convergence of Series

Here I'll be attempting a proof of the Holder's inequality for the improper integrals defined in Prob. 8, Chap. 6, in Rudin.

My Attempt:

Let $p$ and $q$ be positive real numbers such that $1/p + 1/q = 1$.

Suppose that $f$ and $g$ are complex functions which are Riemann-Stieltjes integrable with respect to a monotonically increasing function $\alpha$ on $[a, b]$ for every $b > a$, where $a$ is a fixed real number. Then the holder's inequality gives $$\left\lvert \int_a^b f g \ \mathrm{d} \alpha \right\rvert \leq \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q}. \tag{0}$$

Suppose that the integrals $\int_a^\infty f g \ \mathrm{d} \alpha$, $\int_a^\infty \lvert f \rvert^p \ \mathrm{d} \alpha$, and $\int_a^\infty \lvert g \rvert^q \ \mathrm{d} \alpha$ all converge.

Then by the definition in Prob. 8, Chap. 6, in Rudin, we have $$\int_a^\infty f g \ \mathrm{d} \alpha = \lim_{b \to \infty} \int_a^b f g \ \mathrm{d} \alpha, \tag{1}$$ $$\int_a^\infty \lvert f \rvert^p \ \mathrm{d} \alpha = \lim_{b \to \infty} \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha , \tag{2}$$ and $$\int_a^\infty \lvert g \rvert^q \ \mathrm{d} \alpha = \lim_{b \to \infty} \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha . \tag{3}$$

So,
\begin{align} \left\lvert \int_a^\infty f g \ \mathrm{d} \alpha \right\rvert &= \left\lvert \lim_{b \to \infty} \int_a^b f g \ \mathrm{d} \alpha \right\rvert \qquad \mbox{ [ by (1) above ] } \\ &= \lim_{b \to \infty} \left\lvert \int_a^b f g \ \mathrm{d} \alpha \right\rvert \qquad \mbox{ [ using the continuity of the map t \mapsto \lvert t \rvert ] } \\ &\leq \lim_{b \to \infty} \left[ \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} \right] \\ & \qquad \qquad \mbox{ [ using (0) and a property of the limits ] } \\ &= \lim_{b \to \infty} \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \lim_{b \to \infty} \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} \qquad \mbox{ [ by Theorem 4.4 (b) in Rudin ] } \\ &= \left( \lim_{b \to \infty} \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \lim_{b \to \infty} \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} \\ & \ \ \ \mbox{ [ using the continuity of the map y \mapsto y^r for y \geq 0, r being a given real number ] } \\ &= \left( \int_a^\infty \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^\infty \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q}. \qquad \mbox{ [ by (2) and (3) above ] } \end{align}

Is this proof correct? If so, then is it rigorous enough too for Rudin?

If incorrect, then where?

How to show that the map $y \to y^r$ for $y \geq 0$ is continuous, for any given real number $r$, especially when $r$ is irrational?

• $y^r=\exp (r \ln y)$ – Marko Karbevski Aug 7 '17 at 14:43
• @MarkoKarbevski but so far Rudin has not discussed these functions in a systematic and rigorous manner. – Saaqib Mahmood Aug 7 '17 at 17:05