Increasing orthogonal functions 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 $L^2([0,1])$? 
 A: 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
$$
 f^{\star}=f-\int_{[0,1]}f, \ g^{\star}=g-\int_{[0,1]}g
 $$
A: 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.
