I know that the set $A$ of all the functions in $L^2[0,1]$ that are bounded by $1$ in the sup norm, $$ A = \{ f : f \in L^2[0,1], ||f||_{\infty} \leq 1\}, $$ has no interior points in $L^2([0,1])$. But, it is a convex and bounded subset of $L^2([0,1])$. Can someone recommend (if it exists) what material I might read to learn to analyze such a set for a Hilbert manifold structure? Many thanks.
UPDATE #1: In an attempt to study the tangent space of $A$, I look at curves in it, parameterized by "time". Denoting by ${\bf 1}$ the indicator function, one such curve is: $$ f[t](x) = {\bf 1}_{[\, t, \; t+0.1 \, ]}(x), \quad 0 \leq t \leq 0.9. $$ An increment (for small $h > 0$, and $t < 0.9 - h$) of this curve is $$ f[t+h](x) - f[t](x), $$ but if on dividing this by $h$ and letting $h \rightarrow 0+$, the resulting expression doesn't tend to any function in $A$; instead it tends to a difference of two Dirac delta functions, which is not in $L^2[0,1]$.
This is not an answer, hence the "community wiki"; rather, it is another question. Why do you want such a structure on $A$?
First, it is unlikely that such structure, if it exists, would be natural. The analogous problem on $\mathbb R^2$ is searching for a manifold structure on the unit square $$ \{ (x, y)\ :\ \max(|x|, |y|)\le 1\}. $$ This cannot be a submanifold of $\mathbb R^2$ because of its corners (we talked about this here, by the way). So, any manifold structure we can put on it will be just an academic exercise, unlikely to be useful.
Second, the set $A$ already possesses a great deal of useful properties. It is closed, bounded and convex, which implies that it is a Chebyshev set, as they say in approximation theory; every element of $L^2(0,1)$ admits exactly one element of best approximation in $A$. Now, the theory of infinite-dimensional manifolds arises precisely to try and study the best approximation problem with respect to non-convex sets, as far as I understand from skimming the book "Nonlinear approximation theory" of Braess. Searching for a manifold structure in this case seems, therefore, rather pointless.