Smooth basis of $L^2$ with uniformly controlled derivatives Is there a basis $\{f_i\}_{i\in\mathbb{N}}$ of $L^2[0,1]$, where each $f_i$ is smooth and the second derivative of $f_i$ is uniformly bounded in $i$?
Thoughts: I have a vague feeling that the existence of such basis might make it impossible to approximate certain functions in $L^2[0,1]$ with rapid variation. However, I'm not able to give a rigorous proof of this (or give an example of such a basis if it does exist).
 A: No. Suppose $(f_j)$ is an orthonormal basis.
Note first that if $f\in L^2$ then $||f||_2^2=\sum|\langle f, f_j\rangle|^2$, hence $$\lim_{j\to\infty}\langle f,f_j\rangle=0.$$Setting $f(t)=e^{2\pi int}$ gives $$\lim_{j\to\infty}\widehat{f_j}(n)=0\quad(n\in\Bbb Z),$$and now since $\sum_n|\widehat{f_j}(n)|^2=||f_j||_2^2=1$ it follows that for every $N$ we have $$\lim_{j\to\infty}\sum_{|n|>N}|\widehat{f_j}(n)|^2=1.$$This implies that $$\lim_{j\to\infty}\sum_{n\in\Bbb Z}n^4|\widehat{f_j}(n)|^2=\infty.$$
But if $f$ is smooth then $\sum_{n\in\Bbb Z}n^4|\widehat{f_j}(n)|^2=c||f_j''||_2^2$ for some irrelevant constant $c$. So $$||f_j''||_\infty\ge||f_j''||_2\to\infty.$$
In fact the same holds if $(f_j)$ is just an infinite orthonormal  set (ie, it all works without assuming that the linear combinations of the $f_j$ are dense). The only place we used the fact that the $f_j$ span $L^2$ was in $||f||_2^2=\sum|\langle f, f_j\rangle|^2$, and for any orthonormal set we still have $||f||_2^2\ge\sum|\langle f, f_j\rangle|^2$.
(And of course the same argument shows that in fact $||f_j'||_\infty\to\infty$, no need to go to the second derivative..)
