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

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. 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

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 $f$ on $\mathbb{R}$ by $$f(u) \colon= \frac{u^p}{p} + \frac{v^q}{q} - uv.$$ Then $$f^\prime(u) = u^{p-1} - v.$$ So $f^\prime$ vanishes only at $u = v^{\frac{1}{p-1}} = v^{q-1}$, by (0) above. In fact, $$f^\prime (u) \ \begin{cases} < 0 \ & \mbox{ if } u < v^{q-1}, \\ = 0 \ & \mbox{ if } u = v^{q-1}, \\ > 0 \ & \mbox{ if } u > v^{q-1}. \end{cases}$$ Thus $f$ is strictly decreasing on $\left(- \infty, v^{q-1} \right]$ and hence also on $\left[0, v^{q-1} \right]$; and, $f$ is strictly increasing on $\left[ v^{q-1}, +\infty \right)$; thus $f$ has a relative minimum at $u = v^{q-1}$, and this relative minimum, being the unique extreme value of $f$, is in fact the absolute minimum; moreover, from (1) above we also have $$f \left( v^{q-1} \right) = \frac{ \left( v^{q-1} \right)^p }{p} + \frac{v^q}{q} - v^{q-1} v = \frac{v^q}{p} + \frac{v^q}{q} - v^q = 0$$ because $1/p+1/q=1$. Hence $f(u) \geq 0$ for all real $u$ and hence for all $u > 0$; that is, $$\frac{u^p}{p} + \frac{v^q}{q} - uv \geq 0,$$ and so $$uv \leq \frac{u^p}{p} + \frac{v^q}{q}$$ for all positive real numbers $u$ and $v$.

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.

Theorem 6.12 (a) in Baby Rudin: $\int_a^b \left( f_1 + f_2 \right) d \alpha=\int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha$

Theorem 6.12 (a) in Baby Rudin: If $f\in\mathscr{R}(\alpha)$ on $[a,b]$, then $cf\in\mathscr{R}(\alpha)$ for every constant $c$

Theorem 6.12 (b) in Baby Rudin: If $f_1 \leq f_2$ on $[a, b]$, then $\int_a^b f_1 d\alpha \leq \int_a^b f_2 d\alpha$

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:

Prob. 2, Chap. 6, in Baby Rudin: If $f\geq 0$ and continuous on $[a,b]$ with $\int_a^bf(x)\ \mathrm{d}x=0$, then $f=0$

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.

1. 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 $[0, +\infty)$?

2. 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?

• Isn't this a copy past. How can we spend half an hour to read . – hamam_Abdallah Aug 6 '17 at 14:08
• @Salahamam_Fatima yes, you're right, but you can review as much of this post as you can and then comment on what you've found it like. – Saaqib Mahmood Aug 6 '17 at 14:08
• Has this post of mine been down-voted because of its length? If so, then what are my options, given that another post of mine has been similarly down-voted because I'd tried to break one single problem into several posts? – Saaqib Mahmood Aug 6 '17 at 14:11