5

One of the first problems in Spivak's Calculus on Manifolds asks you to prove the Cauchy-Schwarz inequality for real integrable functions, namely, that $|\int_{a}^{b}fg|^2 \leq |\int_a^bf^2||\int_a^bg^2|$. Now, the easiest way I see of doing this is to argue that $\int_{a}^{b}(f - \lambda g)^2$, where $\lambda \in \mathbb{R}$, is a quadratic with at most one real solution, so the discriminant must be non-negative. However, he gives a cryptic hint to consider seperately the cases of $\int_{a}^{b}(f - \lambda g)^2 = 0$ and $\int_{a}^{b}(f - \lambda g)^2 > 0$. The second case boils down to, essentially, my solution, save for the fact that arguing there's no real solution is a smidge easier than arguing there's at most one real solution. The first case, however, has always left me vaguely mystified, and any attempt I've taken to use it has always had me run into a brick wall. I would argue someting relating to sets of measure $0$, but this has the unfortunate problem of Spivak not defining such sets until two chapters later. So, does anyone have any idea what Spivak would have done?

Specifically, the question is 1-6. (a) on page 4.

Duncan Ramage
  • 6,928
  • 1
  • 20
  • 38
  • Just to clarify, are $f$, $g$, and $\lambda$ real-valued? Also, if I recall correctly, these are Riemann integrals, not Lebesgue? –  Aug 29 '17 at 01:10
  • Yes, they are real valued and Riemann integrable. Well, he only says "integrable" but he also says at the top of the book that he only expects the student to have taken a "good" calculus course, so I doubt he expects one to be familiar with measure theory. – Duncan Ramage Aug 29 '17 at 01:12
  • 1
    He might want to point out that the inequality holds when $\int (f-\lambda g)^2 = 0$. – Math Lover Aug 29 '17 at 01:19
  • I agree, but I don't see how to get from that to the inequality. – Duncan Ramage Aug 29 '17 at 01:19
  • Can I say $\int (f-\lambda g)^2 = 0 \iff f=\lambda g$? – Math Lover Aug 29 '17 at 01:21
  • Not quite, but you can prove that $\int (f - \lambda g)^2 = 0 \iff f = \lambda g\ \text{a.e.}$ (assuming $f$ and $g$ are integrable). However, if Spivak doesn't define a.e. until later, that doesn't help you. Have you tried proving your equality in case $f$ and $g$ are step functions, then making a limiting argument? –  Aug 29 '17 at 01:23
  • @Bungo Just for clarification (as I have no idea about measure theory etc.), if $(f-\lambda g)^2 \ge 0$ then $\int (f-\lambda g)^2 \ge 0$, where the equality holds when $f- \lambda g =0$. Is this logic right? – Math Lover Aug 29 '17 at 01:31
  • $\int (f - \lambda g)^2 = 0$ doesn't imply $f - \lambda g = 0$ everywhere, only almost everywhere. If Spivak hasn't introduced the concept of almost everywhere yet, then you will have to make a different argument. –  Aug 29 '17 at 01:33
  • Having paged through his book on Calculus, and also quietly assuming that his calculus book is what he imagines a good calculus course to be, it doesn't look like he proves the characterization of Riemann integrable functions; only the weaker "continuity implies integrability". – Duncan Ramage Aug 29 '17 at 01:35
  • 1
    That's why I suggest going back to the definition of Riemann integrability in terms of approximation by step functions. –  Aug 29 '17 at 01:42
  • 1
    Have you considered reading up on the proof of Hölder's inequality? It's actually relatively simple to prove this inequality in a much more general way. – Michael L. Aug 29 '17 at 03:46
  • @DuncanRamage: Are you interested in how to prove C-S for Riemann integrable functions with no appeal to measure (or roots of quadratic equations for that matter) -- this is possible -- or is this more curiosity on how to follow Spivak's suggestion? I have encountered other problems in his books where the hints are misleading in terms of what tools must be used. – RRL Aug 30 '17 at 03:27
  • @RRL Well, originally the latter, but I think answering that involves at least touching on the former, and frankly now that you've posed it, the former seems like a much more interesting question. – Duncan Ramage Aug 30 '17 at 03:48

2 Answers2

4

The Cauchy-Schwartz inequality can be proved using only the basic properties of Riemann integration (no reference to measure $0$), regardless of what Spivak may be hinting. One necessary component is that if $h \geqslant 0$ is Riemann integrable, then $\int_a^b h \geqslant 0$. This follows because any lower sum must be non-negative when $h \geqslant 0$ and, consequently, for any partition $P$ we have

$$0 \leqslant L(P,h) \leqslant \sup_{P'} L(P',h) = \int_a^b h.$$

Thus, for any $\lambda \in \mathbb{R}$ it follows that

$$\tag{*}\lambda^2 \int_a^bg^2 - 2\lambda\int_a^bfg + \int_a^bf^2 = \int_a^b(f - \lambda g)^2 \geqslant 0$$

If $\int_a^b f^2 > 0$ and $\int_a^b g^2 > 0$, we can take $\lambda = \sqrt{\int_a^b f^2}/\sqrt{\int_a^b g^2}$ to find

$$2 \frac{\sqrt{\int_a^b f^2}}{\sqrt{\int_a^b g^2}}\int_a^b fg \leqslant \int_a^bf^2 + \frac{\int_a^b f^2}{\int_a^b g^2}\int_a^b g^2 = 2 \int_a^b f^2,$$

whence, $\int_a^bfg \leqslant \sqrt{\int_a^b f^2}\sqrt{\int_a^b g^2}.$

Similarly taking $\lambda = - \sqrt{\int_a^b f^2}/\sqrt{\int_a^b g^2}$ we can show that $\int_a^bfg \geqslant -\sqrt{\int_a^b f^2}\sqrt{\int_a^b g^2},$ and it follows that

$$\tag{**}\left|\int_a^b fg \right| \leqslant \sqrt{\int_a^b f^2}\sqrt{\int_a^b g^2}$$

It remains to consider the cases where one or both of $\int_a^b f^2$ and $\int_a^b g^2$ equals $0$.

If $\int_a^b g^2 = 0$ and $\int_a^b f^2 > 0$, then substituting into (*) we obtain

$$\int_a^b f^2 - 2\lambda\int_a^b fg \geqslant 0.$$

This can only be true for every $\lambda \in \mathbb{R}$ if $\int_a^b fg = 0$ and (**) is true with strict equality. Switching the roles of $f$ and $g$ this can be proved when $\int_a^b f^2 = 0$ and $\int_a^b g^2 > 0$, as well.

Finally, if $\int_a^b f^2 = \int_a^b g^2 = 0$, then since $|fg| \leqslant f^2 + g^2$ it follows that

$$\left|\int_a^b fg \right| \leqslant \int_a^b |fg| \leqslant \int_a^b f^2 + \int_a^b g^2 = 0,$$

and again (**) is true with strict equality.

RRL
  • 90,707
0

Consider real functions $f(x)$, $g(x)$ and real $\lambda$ such that $$\int_{a}^{b} \left(f(x) - \lambda g(x)\right)^2 dx = 0.\hspace{50pt}(1)$$ Note that Eq. (1) holds iff $$\lambda^2 \int_{a}^{b} g(x)^2 dx - 2 \lambda \int_{a}^{b} f(x) g(x) dx + \int_{a}^{b} f(x)^2 dx = 0.$$ Also, the equality cannot hold for two distinct values of $\lambda$ as $\int_{a}^{b} \left(f(x) - \lambda g(x)\right)^2 dx \ge 0$.

By multiplying both sides by $\int_{a}^{b} g(x)^2 dx$, we get $$\color{red}{\left(\lambda\int_{a}^{b} g(x)^2 dx\right)^2 - 2\left(\lambda\int_{a}^{b} g(x)^2 dx\right) \left(\int_{a}^{b} f(x) g(x) dx\right)} + \int_{a}^{b} f(x)^2 dx\int_{a}^{b} g(x)^2 dx=0.~~(2)$$

Observe that $\color{red}{\left(\lambda\int_{a}^{b} g(x)^2 dx\right)^2 - 2\left(\lambda\int_{a}^{b} g(x)^2 dx\right) \left(\int_{a}^{b} f(x) g(x) dx\right)} = \left(\lambda\int_{a}^{b} g(x)^2 dx-\int_{a}^{b} f(x) g(x) dx\right)^2 - \left(\int_{a}^{b} f(x) g(x) dx\right)^2.$

Consequently, Eq. (2) can be written as $$\left(\lambda\int_{a}^{b} g(x)^2 dx-\int_{a}^{b} f(x) g(x) dx\right)^2=\left(\int_{a}^{b} f(x) g(x) dx\right)^2 - \int_{a}^{b} f(x)^2 dx\int_{a}^{b} g(x)^2 dx \ge 0.$$ If $\left(\int_{a}^{b} f(x) g(x) dx\right)^2 - \int_{a}^{b} f(x)^2 dx\int_{a}^{b} g(x)^2 dx > 0$ then we can find two distinct values of $\lambda$ satisfying (1), which is impossible. Therefore, $$\left(\int_{a}^{b} f(x) g(x) dx\right)^2 - \int_{a}^{b} f(x)^2 dx\int_{a}^{b} g(x)^2 dx = 0.$$

Math Lover
  • 15,153
  • I do not see how you go from "Note that..." to "Consequently, Eq. (2) can be written as...", could you please elaborate? – Duncan Ramage Aug 29 '17 at 21:26
  • @DuncanRamage Corrected a typo and used red color to assist you. Let me know if it is still not clear. – Math Lover Aug 29 '17 at 21:44
  • The first sentence is false, right? – littleO Aug 29 '17 at 21:55
  • @littleO Which sentence? The one after "Note that ..." – Math Lover Aug 29 '17 at 21:57
  • I mean the sentence, "for any real functions $f(x), g(x)$ and real $\lambda$, $\int_a^b ( f(x) - \lambda g(x))^2 , dx = 0$. This is not true, for example what if $g = 0$ and $f(x) = 1$, then $\int_a^b ( f(x) - \lambda g(x))^2 , dx = b - a$, right? I must be misunderstanding you somehow. – littleO Aug 29 '17 at 23:36
  • @littleO Thanks for pointing that out. I've corrected it now. – Math Lover Aug 30 '17 at 01:14
  • "Also, the equality cannot hold for two distinct values of $\lambda$" - unless $g=0$ a.e. ... –  Aug 30 '17 at 01:30
  • If $g = 0$ a.e. then the theorem we're trying to prove is trivially true as it collapses to $ 0 \leq 0$, so I think we can safely ignore that. – Duncan Ramage Aug 30 '17 at 02:03