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What is the maximal $n$ such that there exist functions $f_1, \dots, f_n:[0,1] \to \mathbb{R}$ that are all bounded, non-decreasing, and mutually orthogonal in $L^2([0,1])$?

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Does "non-decreasing" mean strictly increasing or just increasing? –  Potato Jun 8 '12 at 1:21
    
@Potato - I think both cases are interesting. –  Hans Engler Jun 8 '12 at 11:56
    
Orthogonal functions are necessarily linearly independent. So from this exercise follows that the answer is at most two. I think that it's not too difficult to find two such questions. Anyway, I've posted this as a separate question - maybe someone will find a solution which somehow uses the inner-product structure of $L_2$. And if not, we can always post a CW answer linking to the other question. –  Martin Sleziak Jun 8 '12 at 14:00
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@MartinSleziak: But that exercise wants all elements of the subspace to be monotone, whereas here only $f_1,\ldots, f_n$ must be monotone. –  Noah Stein Jun 8 '12 at 14:49
    
D'oh! I feel quite stupid right now. But I'll leave here the above comment - merely as the explanation why the other question is linked. Thanks for noticing my mistake, @Noah. –  Martin Sleziak Jun 8 '12 at 14:51
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2 Answers

up vote 3 down vote accepted

The answer is TWO. (The proof below is just an easy adaption of somebody else’s answer to a related question, here on MO).

Lemma Let $f^{\star}$ and $g^{\star}$ be two nonzero, nondecreasing functions in $L^1([0,1])$, with $\int_{[0,1]}f^{\star}=\int_{[0,1]}g^{\star}=0$. Then $\int_{[0,1]}f^{\star}g^{\star} \gt 0$.

Proof of lemma There is an $a_1\in [0,1]$ such that $f^{\star}(x)\leq 0$ when $x \lt a_1$ and $f^{\star}(x) \geq 0$ when $x\gt a_1$. Similarly, there is an $a_2\in [0,1]$ such that $g^{\star}(x)\leq 0$ when $x \lt a_2$ and $g^{\star}(x) \geq 0$ when $x\gt a_2$. By symmetry, we may assume $a_1 \leq a_2$.

Then $0=\int_{0}^{a_2}f^{\star}(x)dx+\int_{a_2}^{1}f^{\star}(x)dx$, so $\int_{0}^{a_2}f^{ \star}(x)dx \leq 0$. We deduce

$$ \int_{0}^{a_1} |f^{\star}(x)|dx \geq \int_{a_1}^{a_2} |f^{\star}(x)|dx $$

and hence

$$ \int_{0}^{a_1} f^{\star}(x)g^{\star}(x)dx= \int_{0}^{a_1} |f^{\star}(x)||g^{\star}(x)|dx \geq \int_{0}^{a_1} |f^{\star}(x)||g^{\star}(a_1)|dx \geq \int_{a_1}^{a_2} |f^{\star}(x)||g^{\star}(a_1)|dx =\int_{a_1}^{a_2} |f^{\star}(x)||g^{\star}(x)|dx= -\int_{a_1}^{a_2} f^{\star}(x)g^{\star}(x) dx $$

So the integral $\int_{0}^{a_2} f^{\star}(x)g^{\star}(x)dx$ is nonnegative ; on the other hand, $f^{\star}g^{\star}$ is nonnegative on $[a_2,1]$. So $\int_{[0,1]}f^{\star}g^{\star} \geq 0$, and if this inequality is in fact an equality then $|f^{\star}g^{\star}|$ must be zero a.e. on $[0,1]$. In particular, there is a sequence $(x_n)$ tending to $1$ such that $f^{\star}(x_n)g^{\star}(x_n)=0$ for all $n$. By the pigeon-hole principle, there must be infinitely many $n$ such that $u(x_n)=0$, where $u$ is one of $f^{\star}$ or $g^{\star}$. Then $u \leq 0$, but this contradicts the fact that $u$ is nonzero with integral zero. The lemma is proved.

Corollary : Let $f$ and $g$ be two orthogonal, nonzero, nondecreasing functions in $L^1([0,1])$. Then $0 \gt \int_{[0,1]}f \int_{[0,1]}g$ : the integrals of $f$ and $g$ must have different signs.

Proof of corollary : use the above lemma with

$$ f^{\star}=f-\int_{[0,1]}f, \ g^{\star}=g-\int_{[0,1]}g $$

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in strictly increasing case, 2. Take $f,g$ strictly increasing. Suppose $\int f \ge 0, \int g \ge 0$ which can be arranged. Suppose $\int f = 0$, Let $x_0$ be such that $f(x) \le 0, x < x_0, f(x) \ge 0, x > x_0$. Let $c$ be such that $g-c \le 0, x < x_0, g-c \ge 0, x > x_0$. Then $\int fg = \int f(g-c) > 0$ because the integrand is. The same argument shows that $\int fg > \int f\int g$ , a correlation inequality, that dispose of the $\int h \ne 0$ case. I think the non strictly increasing can be done along the same lines, you must have $f(g-c) = 0$, really limiting your choices.

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"Which can be arranged": How? In general I think this answer is missing a lot of details. –  Nate Eldredge Nov 10 '12 at 22:16
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