# Prob. 11, Chap. 6, in Baby Rudin: Triangle Inequality for Riemann-Stieltjes Integrals using $L^2$-Norm

Here is Prob. 11, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Let $\alpha$ be a fixed increasing function on $[a, b]$. For $u \in \mathscr{R}(\alpha)$, define $$\lVert u \rVert_2 = \left\{ \int_a^b \lvert u \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2}.$$ Suppose $f, g, h \in \mathscr{R}(\alpha)$, and prove the triangle inequality $$\lVert f-h \rVert_2 \leq \lVert f-g \rVert_2 + \lVert g-h \rVert_2$$ as a consequence of the Schwarz inequality, . . .

My Attempt:

Here is the link to my Math SE post on the Minkowski's inequality for Riemann-Stieltjes integrals:

Minkowski Inequality for Riemann-Stieltjes Integrals

Supposing that $f, g, h$ are complex functions in $\mathscr{R}(\alpha)$ on $[a, b]$, we obtain \begin{align} \lVert f-h \rVert_2 &= \left( \int_a^b \lvert f-h \rvert \ \mathrm{d} \alpha \right)^{1/2} \\ &=\left( \int_a^b \left\lvert \ (f-g) \ + \ (g-h) \ \right\rvert \ \mathrm{d} \alpha \right)^{1/2} \\ &\leq \left( \int_a^b \left\lvert f-g \right\rvert \ \mathrm{d} \alpha \right)^{1/2} + \left( \int_a^b \left\lvert g-h \right\rvert \ \mathrm{d} \alpha \right)^{1/2} \qquad \mbox{ [ using Minkowski's inequality ] } \\ &= \lVert f-g \rVert_2 + \lVert g-h \rVert_2, \end{align} as required.

Is my proof correct and the same as demanded by Rudin?

P.S.:

On $[a, b]$, as $$0 \leq \left\lvert f - h \right\rvert = \left\lvert (f-g) + (g-h) \right\rvert \leq \left\lvert f-g \right\rvert + \left\lvert g-h \right\rvert,$$ so $$\left\lvert f - h \right\rvert^2 \leq \left( \left\lvert f-g \right\rvert + \left\lvert g-h \right\rvert \right)^2 = \left\lvert f-g \right\rvert^2 + 2 \left\lvert f-g \right\rvert \left\lvert g-h \right\rvert + \left\lvert g-h \right\rvert^2.$$ Therefore we have \begin{align} \int_a^b \left\lvert f - h \right\rvert^2 \ \mathrm{d} \alpha &\leq \int_a^b \left( \left\lvert f-g \right\rvert^2 + 2 \left\lvert f-g \right\rvert \left\lvert g-h \right\rvert + \left\lvert g-h \right\rvert^2 \right) \ \mathrm{d} \alpha \\ & \qquad \qquad \mbox{ [ by Theorem 6.12 (b) in Rudin ] } \\ &= \int_a^b \left\lvert f-g \right\rvert^2 \ \mathrm{d} \alpha + 2 \int_a^b \left\lvert f-g \right\rvert \left\lvert g-h \right\rvert \ \mathrm{d} \alpha + \int_a^b \left\lvert g-h \right\rvert^2 \ \mathrm{d} \alpha \\ &\qquad \qquad \mbox{ [ by Theorem 6.12 (a) in Rudin ] } \\ &\leq \int_a^b \left\lvert f-g \right\rvert^2 \ \mathrm{d} \alpha + 2 \left( \int_a^b \lvert f-g \rvert \ \mathrm{d} \alpha \right)^{1/2} \left( \int_a^b \lvert g - h \rvert \ \mathrm{d} \alpha \right)^{1/2} \\ & \qquad + \int_a^b \left\lvert g-h \right\rvert^2 \ \mathrm{d} \alpha \\ & \qquad \mbox{ [ by Holder's inequality for integrals with p=2 ] } \\ &= \left[ \left( \int_a^b \lvert f-g \rvert^2 \ \mathrm{d} \alpha \right)^{1/2} + \left( \int_a^b \lvert g - h \rvert^2 \ \mathrm{d} \alpha \right)^{1/2} \right]^2. \end{align} Thus we have obtained the inequality $$\int_a^b \left\lvert f - h \right\rvert^2 \ \mathrm{d} \alpha \leq \left[ \left( \int_a^b \lvert f-g \rvert^2 \ \mathrm{d} \alpha \right)^{1/2} + \left( \int_a^b \lvert g - h \rvert^2 \ \mathrm{d} \alpha \right)^{1/2} \right]^2. \tag{1}$$

Since all the integrals in (1) are non-negative (and so are their square roots), therefore upon taking the square roots on both sieds of (1), we obtain $$\left( \int_a^b \left\lvert f - h \right\rvert^2 \ \mathrm{d} \alpha \right)^{1/2} \leq \left( \int_a^b \lvert f-g \rvert^2 \ \mathrm{d} \alpha \right)^{1/2} + \left( \int_a^b \lvert g - h \rvert^2 \ \mathrm{d} \alpha \right)^{1/2} ,$$ which is the same as $$\lVert f-h \rVert_2 \leq \lVert f- g \rVert_2 + \lVert g-h \rVert_2,$$ as required.

Here are the links to some relevant Math SE posts of mine:

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 $\int_a^b f_1 d\alpha \leq \int_a^b f_2 d\alpha$

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

Since all the norms are nonnegative, it suffices to prove $$\|f-h\|_2^2 \le (\|f-g\|_2 + \|g-h\|_2)^2.$$ By writing $f-h=f-g+g-h$ and expanding both sides, we reduce the inequality to $$2\langle f-g,g-h\rangle \le 2 \|f-g\|_2 \|g-h\|_2,$$ where $$\langle f-g, g-h\rangle := \int_a^b (f-g)(g-h) \mathop{d\alpha}.$$ The above inequality is simply Cauchy-Schwarz.