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

Here is the link to my earlier post here on Math SE on Probs. 10 (a), (b), and (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

Now I'll be attempting Prob. 10 (d).

Here are the links to my posts here on Math SE on Probs. 7 and 8, Chap. 6, in Rudin:

Prob. 7 (a), Chap. 6, in Baby Rudin: If $f$ is integrable on $[c, 1]$ for every $c>0$, then $\int_0^1 f(x) \ \mathrm{d}x =$ . . .

Prob. 7 (b), Chap. 6, in Baby Rudin: Example of a function such that $\lim_{c \to 0+} \int_c^1 f(x) \ \mathrm{d}x$ exists but . . .

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

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

First, suppose that $f$ and $g$ are bounded function on $[a, b]$ such that $f$ and $g$ are Riemann-Stieltjes integrable with respect to a monotonically increasing function $\alpha$ on $[c, b]$ for every $c \in (a, b)$. Then the Holder's inequality gives $$\left\lvert \int_c^b f g \ \mathrm{d} \alpha \right\rvert \leq \left( \int_c^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_c^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q}. \tag{0}$$

Now if $f$ and $g$ are also Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$, then so are $fg$, $\lvert f \rvert^p$, and $\lvert g \rvert^q$, by virtue of Theorem 6.13 in Baby Rudin, 3rd edition.

Theorem 6.13:

If $f \in \mathscr{R}(\alpha)$ and $g \in \mathscr{R}(\alpha)$ on $[a, b]$, then

(a) $fg \in \mathscr{R}(\alpha)$;

(b) $\lvert f \rvert \in \mathscr{R}(\alpha)$ and $\left\lvert \int_a^b f \ \mathrm{d} \alpha \right\rvert \leq \int_a^b \lvert f \rvert \ \mathrm{d} \alpha$.

Now back to the solution

Therefore by (a variation of) Prob. 7 (a), Chap. 6, in Rudin, we have $$\int_a^b fg \ \mathrm{d} \alpha = \lim_{c \to a+} \int_c^b fg \ \mathrm{d} \alpha, \tag{1}$$ $$\int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha = \lim_{c \to a+} \int_c^b \lvert f \rvert^p \ \mathrm{d} \alpha, \tag{2}$$ and $$\int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha = \lim_{c \to a+} \int_c^b \lvert g \rvert^q \ \mathrm{d} \alpha. \tag{3}$$

Therefore, \begin{align} \left\lvert \int_a^b fg \ \mathrm{d} \alpha \right\rvert &= \left\lvert \lim_{c \to a+} \int_c^b fg \ \mathrm{d} \alpha \right\rvert \qquad \mbox{ [ using (1) above ] } \\ &= \lim_{c \to a+} \left\lvert \int_c^b fg \ \mathrm{d} \alpha \right\rvert \qquad \mbox{ [ using the continuity of the map z \to \lvert z \rvert on \mathbb{C} ] } \\ &\leq \lim_{c \to a+} \left[ \left( \int_c^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_c^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} \right] \\ & \qquad \qquad \mbox{ [ using a property of limits along with (0) above ] } \\ &= \lim_{c \to a+} \left( \int_c^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \lim_{c \to a+} \left( \int_c^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} \\ & \qquad \qquad \mbox{ [ using Theorem 4.4 (b) in Rudin ] } \\ &= \left( \lim_{ c \to a+} \int_c^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \lim_{c \to a+} \int_c^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} \\ & \qquad \qquad \mbox{ [ using the continuity of the map y \mapsto y^r on [0, +\infty) for every real r ] } \\ &= \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}. \qquad \mbox{ [ usinb (2) and (3) above ] } \end{align}

Is what I've done so far correct? If so, then is my proof rigorous enough for Rudin? If not, then where have I fallen short?

In the above chain, we have used the result that the mapping $y \mapsto y^r$ is continuous on $[0, +\infty)$, for any given real number $r$.

How to prove this rigorously (and preferably with the help of only what Rudin has discussed so far), especially when $r$ is irrational?