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This is exercise 11, Chapter 6 in Baby Rudin:

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

where by "Schwarz inequality," Rudin means the following version of Holder's inequality:

$$\left\lvert \int_a^b f g \ \mathrm{d} \alpha \right\rvert \leq \left\{ \int_a^b \lvert f \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2} \left\{ \int_a^b \lvert g \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2}. $$

The solution to this question in the solutions manual is:

enter image description here

In the first display on the left-hand side, it should read $||f-h||_2^2$ instead of $||f-h||_2^1$. My question is about the only inequality in the proof. In particular, how is it true that $$\int_a^b |f-g||g-h| \ d\alpha \le \left\{ \int_a^b \lvert f-g \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2} \left\{ \int_a^b \lvert g-h \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2}?$$ This does not looks like the proper usage of the "Schwarz inequality" or am I missing something? Can someone please propose a correct version of the solution?

The solution to the same question here seems to suffer from the same issue, if this is an issue at all, that is.

Ricky_Nelson
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1 Answers1

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Regarding your first question I agree, it seems it should be $\|g-h\|_{2}^{2}$ on the LHS.

For your second question, the Cauchy-Schwarz inequality is a special case of H$\ddot{\text{o}}$lder's inequality in the sense that it is the case that is applicable to inner products. So in the second display you have which explains this the intermediate steps are, \begin{align} \bigg|\int_{a}^{b}fgd\alpha\bigg|\leq\int_{a}^{b}|f||g|d\alpha=\|fg\|_{1}\leq\|f\|_{2}\|g\|_{2}=\bigg(\int_{a}^{b}|f|^{2}d\alpha\bigg)^{1/2}\bigg(\int_{a}^{b}|g|^{2}d\alpha\bigg)^{1/2} \end{align}

The other thing to note here is it seems you are misreading the solution. You are asking how, \begin{align} \int_{a}^{b}|f-g||g-h|d\alpha\leq\|f\|_{2}\|g\|_{2}, \end{align} which is not what is being presented in the proof. In Rudin's proof he is using, \begin{align} \int_{a}^{b}|f-g||g-h|d\alpha\leq\|f-g\|_{2}\|g-h\|_{2}. \end{align}

EDIT: To clarify the last inequality, since $f,g,h$ are all Riemann integrable with respect to $d\alpha$ then $f-g$ and $g-h$ are also Riemann integrable. So consider, \begin{align} \int_{a}^{b}|f-g||g-h|d\alpha=\|(f-g)(g-h)\|_{1}\leq\|f-g\|_{2}\|g-h\|_{2}=\bigg(\int_{a}^{b}|f-g|^{2}d\alpha\bigg)^{1/2}\bigg(\int_{a}^{b}|g-h|^{2}d\alpha\bigg)^{1/2} \end{align}

EDIT2: As Paramanand pointed out, since $f-g$ and $g-h$ are Riemann integrable replace $f$ by $|f-g|$ and $g$ by $|g-h|$ in the given Cauchy-Schwarz inequality to get, \begin{align} \int_{a}^{b}|f-g||g-h|d\alpha=\bigg|\int_{a}^{b}|f-g||g-h|d\alpha\bigg|\leq\bigg(\int_{a}^{b}|f-g|^{2}d\alpha\bigg)^{1/2}\bigg(\int_{a}^{b}|g-h|^{2}d\alpha\bigg)^{1/2} \end{align}

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