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Based on Prob. 10, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition, we have the following result.

Suppose $p$ is a real number such that $p \geq 1$, and suppose that $f$ and $g$ are complex functions which are Riemann-Stieltjes integrable with respect to a monotonically increasing function $\alpha$ on an interval $[a, b]$. Then $$ \left( \int_a^b \lvert f + g \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \leq \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} + \left( \int_a^b \lvert g \rvert^p \ \mathrm{d} \alpha \right)^{1/p}. \tag{0} $$

Am I right?

Here are the links to my Math SE post on Theorem 6.12 (a) and (b) and Prob. 10 (a), (b), and (c) in Rudin:

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

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

My Proof:

First, we suppose that $p=1$. Then since $$ \lvert f+g \rvert \leq \lvert f \rvert + \lvert g \rvert $$ on $[a, b]$, therefore we have $$ \int_a^b \lvert f+g \rvert \ \mathrm{d} \alpha \leq \int_a^b \lvert f \rvert \ \mathrm{d} \alpha + \int_a^b \lvert g \rvert \ \mathrm{d} \alpha, $$ by Theorem 6.12 (b) in Rudin. This is (0) with $p = 1$.

Now suppose that $p > 1$.

If $ \int_a^b \lvert f+g \rvert^p \ \mathrm{d} \alpha = 0$, then the LHS of (0) equals zero, and since the RHS of (0) is non-negative (by Theorem 6.12 (b) in Rudin), therefore (0) holds.

So we suppose that $ \int_a^b \lvert f+g \rvert^p \ \mathrm{d} \alpha \neq 0$. Then we in fact have $$ \int_a^b \lvert f+g \rvert^p \ \mathrm{d} \alpha > 0, \tag{1} $$ by Theorem 6.12 (b) in Rudin.

Let $q$ be a positive real number such that $1/p + 1/q = 1$. Then $$ p+q = pq, $$ and so $$ pq - p - q + 1 = 1,$$ that is $$ (p-1)(q-1) = 1;$$ therefore $$ (p-1)q = (p-1)(q-1+1) = (p-1)(q-1) + (p-1) = 1 + (p-1) = p. \tag{2} $$

So, $$ \begin{align} \int_a^b \lvert f+g \rvert^p \ \mathrm{d} \alpha &= \int_a^b \lvert f + g \rvert \, \lvert f+g \rvert^{p-1} \ \mathrm{d} \alpha \qquad \mbox{ [ note that $p > 1 $ ] } \\ &\leq \int_a^b \left( \lvert f \rvert + \lvert g \rvert \right) \lvert f+g \rvert^{p-1} \ \mathrm{d} \alpha \qquad \mbox{ [ using Theorem 6.12 (b) in Rudin ] } \\ &= \int_a^b \lvert f \rvert \, \lvert f+g \rvert^{p-1} \ \mathrm{d} \alpha + \int_a^b \lvert g \rvert \, \lvert f+g \rvert^{p-1} \ \mathrm{d} \alpha \\ & \qquad \qquad \mbox{ [ using Theorem 6.12 (a) in Rudin ] } \\ &\leq \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \left\lvert \lvert f+g \rvert^{p-1} \right\rvert^q \ \mathrm{d} \alpha \right)^{1/q} \\ & \qquad + \left( \int_a^b \lvert g \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \left\lvert \lvert f+g \rvert^{p-1} \right\rvert^q \ \mathrm{d} \alpha \right)^{1/q} \\ & \qquad \mbox{ [ by Holder's inequality for integrals, or Prob. 10 (c), Chap. 6, in Rudin ] } \\ &= \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \lvert f+g \rvert^{(p-1)q} \ \mathrm{d} \alpha \right)^{1/q} \\ & \qquad + \left( \int_a^b \lvert g \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \left( \int_a^b \lvert f+g \rvert^{(p-1)q} \ \mathrm{d} \alpha \right)^{1/q} \\ &= \left[ \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} + \left( \int_a^b \lvert g \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \right] \left( \int_a^b \lvert f+g \rvert^{(p-1)q} \ \mathrm{d} \alpha \right)^{1/q} \\ &= \left[ \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} + \left( \int_a^b \lvert g \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \right] \left( \int_a^b \lvert f+g \rvert^{p} \ \mathrm{d} \alpha \right)^{1/q}. \\ & \qquad \qquad \mbox{ [ using (2) above ] } \end{align} $$

Thus we have shown that $$ \int_a^b \lvert f+g \rvert^p \ \mathrm{d} \alpha \leq \left[ \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} + \left( \int_a^b \lvert g \rvert^p \ \mathrm{d} \alpha \right)^{1/p} \right] \left( \int_a^b \lvert f+g \rvert^{p} \ \mathrm{d} \alpha \right)^{1/q}. \tag{3} $$ And as $$ \int_a^b \lvert f+g \rvert^p \ \mathrm{d} \alpha > 0, $$ by (1) above, so $$ \left( \int_a^b \lvert f+g \rvert^{p} \ \mathrm{d} \alpha \right)^{1/q} > 0 $$ as well; dividing (3) out by this integral we obtain $$ \frac{\int_a^b \lvert f+g \rvert^p \ \mathrm{d} \alpha }{\left( \int_a^b \lvert f+g \rvert^{p} \ \mathrm{d} \alpha \right)^{1/q} } \leq \left( \int_a^b \lvert f \rvert^p \ \mathrm{d} \alpha \right)^{1/p} + \left( \int_a^b \lvert g \rvert^p \ \mathrm{d} \alpha \right)^{1/p}. $$ which is the same as (0) above, because $1 - 1/q = 1/p$.

Is my proof correct? If so, then is it as rigorous as Rudin demands? If not, then where are the issues?

Is my presentation clear enough for a not-so-sharp student who is taking their very first course in analysis?

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  • $\begingroup$ I have read it three times now. I haven't checked the proofs of the facts you have used [I know them and I know they're valid]. Kudos to you for a very detailed proof. It's very undergraduate friendly! $\endgroup$ – Alvin Lepik Aug 7 '17 at 5:41
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To answer your question directly: yes, your answer is logically correct and it is very clear.


If you want the proof to be slightly less lengthy, the simple algebra in (2) could have been written in just one line: the equality $1/p+1/q=1$ implies that $$ (p-1)q=(p-1)\cdot\frac{1}{1-1/p}=p. $$

Also, if you want to save your labor to type/write the integral sign, use the norm notation $\|\cdot\|_p$.

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