Is it impossible to pack too much information in the space of sufficiently smooth functions? The basic version of my question is: am I right that there doesn't exist an infinite sequence of mutually orthogonal functions $f_i\colon[0,1]\to\mathbb R$ which are sufficiently smooth, e.g. Lipschitz-continuous with the same constant? My loose intuition is that the existence of such an infinite sequence would require packing too much information in the space of those functions which should not be possible.
The full version is: for a uniform random variable $x\sim U[0,1]$ and a sequence of sufficiently smooth functions $f_i(x)$ for $i=1,\dots$, under what conditions does it follow that whenever $Var(f_i(x))\ge a>0$, $\frac{1}{n^2}\sum_{i,j=1}^n\left|Cov(f_i(x),f_j(x))\right|$ cannot converge to zero as $n\to\infty$. The intuition again is that one can't have covariances too close to zero for all but a small fraction of the pairs of these smooth functions. (Showing this for a general multivariate $x$ would be even better. Or limiting to monotone functions, such that all covariances are non-negative, could be good enough, too.)
Thanks for any thoughts!
 A: (everywhere below $\|\cdot\|$ denotes $L_2$ norm)
If you, similar to full version, require that $\|f_i\| \geq a > 0$, then it's not possible.
For simplicity, assume that Lipschitz constant of every function is at most $1$ and $a < 1$. Then if function changes sign, it's absolute value is at most $1$ in every point. And all but at most two functions have to change sign (because two everywhere positive or two everywhere negative functions are not orthogonal). Discarding them, we now have infinite sequence of functions each taking absolute values of at most $1$.
Let $h(x) = \lceil x / \varepsilon\rceil \cdot \varepsilon$ - rounding with precision of $\varepsilon$. Let $g_i(x) = h(f_i(h(x)))$ - simple function that takes values $n\varepsilon$ that represents "rounding" of $f$ both for argument and value.
Note that $\|g_i - f_i\| < 3\varepsilon$ (because of Lipschitz restriction) and $|g_i(x)| < 1 + 2 \varepsilon$.
Take $\varepsilon = a / 8$. There are finitely many possible $g_i$, so we have $g_i = g_j$ for some $i, j$.
Now, we have
$$2 (f_i, f_j) = \|f_i\|^2 + \|f_j\|^2 - \|f_i - f_j\|^2\geq\\
2a^2 - \|f_i - g_i + g_i - g_j + g_j - f_j\|^2 \geq \\
2a^2 - (\|f_i - g_i\| + \|g_i - g_j\| + \|g_j - f_j\|)^2 \geq \\
2a^2 - (3a / 8 + 0 + 3a / 8)^2 > 22/16 a^2 > 0$$
