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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 $$ \begin{align} & \lVert f-h \rVert_2 \\ &= \left( \int_a^b \lvert f-h \rvert^2 \, \mathrm{d} \alpha \right)^{1/2} \\ &= \left( \int_a^b \left\lvert \, (f-g)\, + \, (g-h) \, \right\rvert^2 \ \mathrm{d} \alpha \right)^{1/2} \\ &\leq \left( \int_a^b \left\lvert f-g \right\rvert^2 \, \mathrm{d} \alpha \right)^{1/2} + \left( \int_a^b \left\lvert g-h \right\rvert^2 \, \mathrm{d} \alpha \right)^{1/2} \qquad \mbox{ [ using Minkowski's inequality ] } \\ &= \lVert f-g \rVert_2 + \lVert g-h \rVert_2, \end{align} $$ 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 \begin{align} & \ \ \ \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. \end{align} 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^{2} \ \mathrm{d} \alpha \right)^{1/2} \left( \int_a^b \lvert g - h \rvert^{2} \ \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 \begin{align} & \ \ \ \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)^{\frac12} + \left( \int_a^b \lvert g - h \rvert^2 \, \mathrm{d} \alpha \right)^{\frac12} \right]^2. \tag{1} \end{align}

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 \begin{align} & \ \ \ \left( \int_a^b \left\lvert f - h \right\rvert^2 \, \mathrm{d} \alpha \right)^{\frac12} \\ & \leq \left( \int_a^b \lvert f-g \rvert^2 \, \mathrm{d} \alpha \right)^{\frac12} + \left( \int_a^b \lvert g - h \rvert^2 \, \mathrm{d} \alpha \right)^{\frac12} , \end{align} 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: $\int_a^b \left( f_1 + f_2 \right) d \alpha=\int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha$

https://math.stackexchange.com/questions/2327134/theorem-6-12-a-in-baby-rudin-if-f-in-mathscrr-alpha-on-a-b-then-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

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  • $\begingroup$ In your first approach, it should be $(\int_a^b |f-h|^2 \ d \alpha)^{ \frac {1}{2} }$ instead of $(\int_a^b |f-h| \ d \alpha)^{ \frac {1}{2} }$ $\endgroup$ Commented Dec 16, 2021 at 14:18

2 Answers 2

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As @angryavian pointed out, your first attempt missed Rudin's point. The original problem is as follows:

enter image description here

You have replaced "as in the proof of Theorem 1.37" with three dots in your post and thus made things complicated.

To answer your edited question: Your second attempt in the "P.S." part is correct. "Hölder inequality" in your argument is the same as the "Schwarz inequality", which is explicitly mentioned in the second last phrase of the problem: "as a consequence of the Schwarz inequality".

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Minkowski's inequality is the triangle inequality in this case, so you are using what you are asked to prove in your proof.

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.

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  • $\begingroup$ please have a look at my post now. I've added a P.S. $\endgroup$ Commented Aug 7, 2017 at 12:39

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