# Maximal subspace on which an operator is bounded

Consider the Banach space $X=C[0,1]$ of real continuous function on $[0,1]$ equipped with the supremum norm. Consider the operator $A:D(A)\to X$, $Af=f'$ for each $f\in D(A)=C^1[0,1]$. We can see that the domain of $A$ is dense in $X$. But we know that $A$ is unbounded (non continuous) on its domain. If we take $Y=$ The space of constant functions, then $A$ is a bounded operator on this subspace.

Do we have the existence of a maximal subspace $Y\subset X$ such that $A:Y\to X$ is a bounded operator ?

• Is the assumption that $Y$ is closed part of your definition of subspace? (I'm curious although this might not be relevant to answering the question.) – Jonas Meyer Aug 4 '14 at 15:35
• @JonasMeyer I am interested to what will happen if we assume $Y$ is closed or not. – user50618 Aug 4 '14 at 15:38

The subspace $S_{n}$ spanned by $\{ e^{2\pi i kx}\}_{k=-n}^{n}$ is invariant under $A$, and $A$ is bounded on $S_{n}$ because its a matrix. However, $A$ is not bounded on the union of these nested subspaces. That puts a limit on any Zorn's lemma argument. I suppose you could find a maximal subspace on which $A$ is bounded by a given fixed constant $M$.
• Thanks a lot, that answers the question. Maybe we can have that if $A$ is bounded on a subspace $Y\subset X$, then $Y$ is necessarily finite dimensional ? – user50618 Aug 4 '14 at 21:03
• If a subspace is finite-dimensional and invariant under differentiation $D$, then Jordan canonical form tells you what spans the space: you have some eigenfunctions (exponentials) and some polynomials times exponentials because $(D-\lambda I)^{n}f=0$ gives $D^{n}(e^{-\lambda t}f)=0$. I'm not sure about the infinite-dimensional aspects, even though I think it should be more obvious. – DisintegratingByParts Aug 4 '14 at 21:38
• I wonder what happens if you take a bounded sequence $\{\lambda_{n}\}_{n=1}^{\infty}$ and take the linear space generated by $\{ e^{\lambda_{n}t}p(t) : n \ge 1, \deg(p) \le N \}$ for a fixed $N$. That space is invariant under $D$. Could $D$ be bounded on that space? If it were to be, then the closure of such functions in $C[0,1]$ would have to be differentiable. Hmmmm.... – DisintegratingByParts Aug 4 '14 at 21:52