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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 $L^2[0,1]$.

avs
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  • What do you mean by "manifold structure"? – gerw Apr 02 '19 at 05:59
  • @gerw, I mean, whether $A$ is a manifold in $L^2[0, 1]$, just like, say, the sphere $x^2 + y^2 + y^2 = 1$ is a manifold in the Euclidean 3-D space. – avs Apr 02 '19 at 06:01
  • What is your definition of "manifold"? Typically, a manifold is finite dimensional, but $A$ is not. – gerw Apr 02 '19 at 06:02
  • Apologies. I mean, Hilbert manifold. I'll edit the question accordingly. – avs Apr 02 '19 at 06:04
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    If you found Dirac deltas then your curve is not differentiable. A Dirac delta is not an element of $L^2$. https://math.stackexchange.com/a/2198242/8157 – Giuseppe Negro Apr 02 '19 at 20:26
  • @GiuseppeNegro, yes. I am now looking for other curves. – avs Apr 02 '19 at 20:28
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    But are you sure that thing is a Hilbert submanifold of $L^2$? It does not seem obvious to me. Anyway, I have come across some references on Hilbert manifolds sometime (not that I read them). One is Lang, "Fundamentals of differentiable geometry", but it is probably not what you need. I found a more concrete reference in the book of Braess, "Nonlinear approximation theory". – Giuseppe Negro Apr 02 '19 at 20:45
  • @GiuseppeNegro, thank you so much for finding the references. To answer your question, No, I am not sure: I was trying to determine whether it is a Hilbert (or any finite-dimensional) submanifold or not. But it seems doubtful. At any rate, thanks again. If I find out new stuff, I'll add in updates. – avs Apr 02 '19 at 21:44

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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.

Giuseppe Negro
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    Thanks much for following up, Guiseppe. This is only a small part of a problem I am trying to work on. What I am really interested in is finding (which may be extremely difficult and, likely, only partly successful) the orbits of a certain family of vector fields acting on $A$. (Think of $A$ as the state space of a control system, and the family of vector fields as the control space.) So, ideally, it would be to find the sets of $A$ reachable from a given, initial, state in $A$, by these controls. And even $A$ is a simplification of the true set of interest (which arises – avs Apr 03 '19 at 16:58
  • from yet another problem, fundamental to computation theory). – avs Apr 03 '19 at 17:02
  • And sorry I misspelled your name--MSE is not allowing me to edit that comment any more. – avs Apr 03 '19 at 17:05
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    In my opinion is it better to ask another question, more focused on the problem. – Giuseppe Negro Apr 03 '19 at 17:54
  • I agree. I just didn't want to put the labor of my research on crowdsourcing, without being able to give credit appropriately. – avs Apr 03 '19 at 18:00