# Minkowski Inequality for Riemann-Stieltjes Integrals

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?

• 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! – Alvin Lepik Aug 7 '17 at 5:41

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 $\|\cdot\|_p$.