Here is Prob. 10, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Let $p$ and $q$ be positive real numbers such that $$ \frac{1}{p} + \frac{1}{q} =1. $$
Prove the following statements.
(a) If $u \geq 0$ and $v \geq 0$, then $$ u v \leq \frac{u^p}{p} + \frac{v^q}{q}. $$ Equality holds if and only if $u^p = v^q$.
(b) If $f \in \mathscr{R}$, $g \in \mathscr{R}$, $f \geq 0$, $g \geq 0$, and $$ \int_a^b f^p \ \mathrm{d} \alpha = 1 = \int_a^b g^q \ \mathrm{d} \alpha, $$ then $$ \int_a^b f g \ \mathrm{d} \alpha \leq 1. $$
(c) If $f$ and $g$ are complex functions in $\mathscr{R} (\alpha) $, then $$ \left\lvert \int_a^b f g \ \mathrm{d} \alpha \right\rvert \leq \left\{ \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right\}^{1/p} \left\{ \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right\}^{1/q}. $$ This is Holder's inequality. When $p = q = 2$ it is usually called the Schwarz inequality. (Note that Theorem 1.35 is a very special case of this. )
(d) Show that Holder's inequality is also true for the "improper" integrals described in Exercises 7 and 8.
Here is Theorem 1.35 in Rudin, 3rd edition:
If $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ are complex numbers, then $$ \left\lvert \sum_{j=1}^n a_j \overline{b_j} \right\rvert^2 \leq \sum_{j=1}^n \left\lvert a_j \right\rvert^2 \sum_{j=1}^n \left\lvert b_j \right\rvert^2. $$
Here are the links to my Math SE posts on Probs. 7 and 8, Chap. 6, in Baby Rudin, 3rd edition:
Prob. 8, Chap. 6, in Baby Rudin: The Integral Test for Convergence of Series
My Attempt:
As $$ \frac{1}{p} + \frac{1}{q} = 1,$$ so $$ q+p = pq, $$ which implies that $$ pq- p - q = 0, $$ and hence $$ pq - p - q + 1 = 1, $$ that is, $$ (p-1) (q-1) = 1. \tag{0} $$ So $$ (q-1)p = (q-1)(p-1 + 1) = (q-1)(p-1) + q - 1 = 1 + q-1 = q. \tag{1} $$ in what follows, we will be using (0) and (1).
Prob. 10 (a)
If $u = 0$ or $v=0$, then $$ 0 = uv \leq \frac{u^p}{p} + \frac{v^q}{q}. $$ So let us suppose that $u$ and $v$ both are positive real numbers. For any fixed $v > 0$, let us define a function $f_v \colon (0, +\infty) \longrightarrow \mathbb{R}$ by the formula $$ f_v(u) \colon= \frac{u^p}{p} + \frac{v^q}{q} - uv \ \mbox{ for all } u \in (0, +\infty). $$ Then $$ f_v^\prime(u) = u^{p-1} - v \ \mbox{ for all } u \in (0, +\infty). $$ So $f_v^\prime$ vanishes only at $$ u = v^{\frac{1}{p-1}} = v^{q-1}. $$ [Refer to (0) above.] In fact, $$ f_v^\prime (u) \ \begin{cases} < 0 \ & \mbox{ if } 0 < u < v^{q-1}, \\ = 0 \ & \mbox{ if } u = v^{q-1}, \\ > 0 \ & \mbox{ if } u > v^{q-1}. \end{cases} $$ Thus $f_v$ is strictly decreasing on $\left( 0, v^{q-1} \right]$, and, $f_v$ is strictly increasing on $\left[ v^{q-1}, +\infty \right)$. This implies that $f_v$ has a relative minimum value at $u = v^{q-1}$, and this relative minimum, being the unique extreme value of $f_v$ within the entire domain of this function, is in fact also the absolute minimum value; moreover, from (1) above we also have $$ \begin{align} f_v \left( v^{q-1} \right) &= \frac{ \left( v^{q-1} \right)^p }{p} + \frac{v^q}{q} - v^{q-1} v \\ &= \frac{ v^{p(q-1)} }{p} + \frac{v^q}{q} - v^q \\ &= \frac{v^q}{p} + \frac{v^q}{q} - v^q \qquad [ \mbox{ using (1) above } ] \\ &= v^q \left( \frac{1}{p} + \frac{1}{q} - 1 \right) \\ &= 0 \end{align} $$ because $$ \frac{1}{p} + \frac{1}{q} =1. $$ Hence $$ f_v(u) \geq 0 \ \mbox{ for all } u \in (0, +\infty); $$ that is, $$ \frac{u^p}{p} + \frac{v^q}{q} - uv \geq 0 \mbox{ for all } u \in (0, +\infty), $$ and so $$ uv \leq \frac{u^p}{p} + \frac{v^q}{q} $$ for all positive real numbers $u$ and $v$.
Is this proof correct?
Prob. 10 (b)
In this proof we will be using Theorem 6.12 (a) and (b) in Baby Rudin. Here are the links to my Math SE posts on these results.
And, we will also be using Theorem 6.13 (a) in Baby Rudin, which is as follows: If $f \in \mathscr{R}(\alpha)$ and $g \in \mathscr{R}(\alpha)$ on $[a, b]$, then $fg \in \mathscr{R}(\alpha)$.
As $f \geq 0$ and $g \geq 0$ on $[a, b]$, so we also have $$ f g \leq \frac{f^p}{p} + \frac{g^q}{q} \tag{2} $$ on $[a, b]$.
As $f \in \mathscr{R}(\alpha)$ and $g \in \mathscr{R}(\alpha)$ on $[a, b]$, so $fg \in \mathscr{R}(\alpha)$ on $[a, b]$, by Theorem 6.13 (a) in Baby Rudin. Moreover, $$ \begin{align} \int_a^b f g \ \mathrm{d} \alpha &\leq \int_a^b \left( \frac{f^p}{p} + \frac{g^q}{q} \right) \ \mathrm{d} \alpha \qquad \mbox{ [ using (2) and Theorem 6.12 (b) in Rudin ] } \\ &= \frac{1}{p} \int_a^b f^p \ \mathrm{d} \alpha + \frac{1}{q} \int_a^b g^q \ \mathrm{d} \alpha \qquad \mbox{ [ using Theorem 6.12 (a) in Rudin ] } \\ &= \frac{1}{p} \cdot 1 + \frac{1}{q} \cdot 1 \qquad \mbox{ [ using our hypothesis ] } \\ &= 1, \qquad \mbox{ [ using the condition on $p$ and $q$ in our hypothesis ] } \end{align} $$
Prob. 10 (c)
In what follows, we will be using Theorems 6.11, 6.12 (a) , and
6.13 in Rudin. Here are the links to some Math SE posts on these
theorems.
Theorems 6.11 and 6.13 from PMA Rudin
Theorem 6.11 of Rudin's Principles of Mathematical Analysis
And, here is the link to my Math SE post on Prob. 2, Chap. 6, in Baby Rudin:
As $f$ and $g$ are complex functions in $\mathscr{R}(\alpha)$ on $[a, b]$, so $\lvert f \rvert$ and $\lvert g \rvert$ are real functions in $\mathscr{R}$ on $[a, b]$, by virtue of Theorem 6.13 (b) in Baby Rudin.
As $\lvert f \rvert \in \mathscr{R}(\alpha)$ and $\lvert g \rvert \in \mathscr{R}(\alpha)$ on $[a, b]$, so $\lvert f \rvert$ and $\lvert g \rvert$ are bounded functions on $[a, b]$. Let us put $$ M \colon= 1 + \max \left\{ \ \sup \{ \ \lvert f(x) \rvert \ \colon \ a \leq x \leq b \ \}, \sup \{ \ \lvert g(x) \rvert \ \colon \ a \leq x \leq b \ \} \ \right\}. $$ Then $M > 0$.
Let $r$ be a real number. If the mapping $y \mapsto y^r$ is continuous on $[0, +\infty)$, then we can conclude from Theorem 6.11 in Rudin that $\lvert f \rvert^r $ and $\lvert g \rvert^r $ are also in $\mathscr{R}$ on $[a, b]$. Thus in particular, $\lvert f \rvert^p $ and $\lvert g \rvert^q $ are in $\mathscr{R}$ on $[a, b]$.
But how to show that the mapping $y \mapsto y^r$ is continuous on $[0, +\infty)$, especially when $r$ is irrational? I would be really grateful for a rigorous and detailed proof of this (preferably using only the tools we have at our disposal after studying Rudin up to Prob. 9, Chap. 6).
We assume that $\lvert f \rvert^p $ and $\lvert g \rvert^q $ are in $\mathscr{R}$ on $[a, b]$.
If $\int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha = 0$ or $\int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha = 0$, then we have $\lvert f \rvert^p = 0$ or $\lvert g \rvert^q = 0$ on $[a, b]$, by virtue of Prob. 2, Chap. 6, in Baby Rudin, provided the functions $\lvert f \rvert$ and $\lvert g \rvert$ are continuous on $[a, b]$ (But what if that is not the case? How to proceed then?); this would imply that $f =0$ or $g = 0$ on $[a, b]$, and hence $fg = 0$ on $[a, b]$, and so the Holder's inequality would be trivially satisfied, as both sides of the inequality would be zero.
So let us assume that $\int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \not= 0$ and $\int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \not= 0$. Now let us put $$ \tilde{f} \colon= \frac{ \lvert f \rvert }{ \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} }, \qquad \tilde{g} \colon= \frac{ \lvert g \rvert }{ \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} } . \tag{3} $$
Then $\tilde{f}$ and $\tilde{g}$ both are in $\mathscr{R}(\alpha)$ on $[a, b]$, because of what we have obtained (or assumed) so far together with Theorem 6.12 (a) in Rudin.
Moreover, $$ \begin{align} \int_a^b \tilde{f}^p \ \mathrm{d} \alpha &= \int_a^b \left( \frac{ \lvert f \rvert }{ \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} } \right)^p \ \mathrm{d} \alpha \qquad \mbox{ [ using (3) above ] } \\ &= \int_a^b \frac{ \lvert f \rvert^p }{ \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha } \ \mathrm{d} \alpha \\ &= \frac{ 1 }{ \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha } \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \\ & \qquad \qquad \mbox{ [ using Theorem 6.12 (a) in Rudin since $ \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha$ is constant ] } \\ &= 1. \end{align} $$ Similarly, we can show that $$ \int_a^b \tilde{g}^q \ \mathrm{d} \alpha = 1. $$
Thus we have $\tilde{f} \in \mathscr{R}(\alpha)$, $\tilde{g} \in \mathscr{R}(\alpha)$, $\tilde{f} \geq 0$, $\tilde{g} \geq 0$, and $$ \int_a^b \tilde{f}^p \ \mathrm{d} \alpha = 1 = \int_a^b \tilde{g}^q \ \mathrm{d} \alpha . $$ So by Prob. 10 (b) above we can conclude that $$ \int_a^b \tilde{f} \tilde{g} \ \mathrm{d} \alpha \leq 1. \tag{4}$$ Now using (3) in (4) we obtain $$ \int_a^b \left( \frac{ \lvert f \rvert }{ \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} } \frac{ \lvert g \rvert }{ \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} } \right) \ \mathrm{d} \alpha \leq 1, $$ which is the same as $$ \int_a^b \left( \frac{ \lvert f \rvert \ \lvert g \rvert }{ \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} } \right) \ \mathrm{d} \alpha \leq 1, $$ which simplifies to $$ \frac{1}{ \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} } \int_a^b \lvert f \rvert \ \lvert g \rvert \ \mathrm{d} \alpha \leq 1, $$ by virtue of Theorem 6.12 (a) in Rudin because both the quantities $\left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} $ and $\left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q} $ are constant. The last inequality implies that $$ \int_a^b \lvert f \rvert \ \lvert g \rvert \ \mathrm{d} \alpha \leq \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q}. \tag{5} $$
Finally, as $f \in \mathscr{R}(\alpha)$ and $g \in \mathscr{R}(\alpha)$ on $[a, b]$, so $f g \in \mathscr{R}(\alpha)$ by Theorem 6.13 (a) in Rudin; moreover, $$ \begin{align} \left\lvert \int_a^b f g \ \mathrm{d} \alpha \right\rvert &\leq \int_a^b \lvert f g \rvert \ \mathrm{d} \alpha \qquad \mbox{ [ using Theorem 6.13 (b) in Rudin ] } \\ &= \int_a^b \lvert f \rvert \ \lvert g \rvert \ \mathrm{d} \alpha \\ &\leq \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha \right)^{1/q}. \qquad \mbox{ [ using (5) above ] } \end{align} $$
Is what I've done so far correct? If so, then is my presentation rigorous and lucid enough?
If the above proofs are all correct, then the following questions remain.
Given a real number $r$, especially an irrational one, how to rigorously (and only using the tools Rudin has provided us up to this point in the book) prove that the map $y \mapsto y^r$ is continuous on $[0, +\infty)$?
If both of $\lvert f \rvert$ and $\lvert g \rvert$ are discontinuous at some point(s) of $[a, b]$, then how to establish the Holder's inequality in case one of $ \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha$ and $ \int_a^b \lvert g \rvert^q \ \mathrm{d} \alpha$ is zero?