It is known that the differentiation operator $D$ is not bounded on $C^1([0,1])$ with $L^2$ norm (counterexample: $f(x)=x^n$). Now I am wondering whether there is an infinitely dimensional subspace such that $D$ is indeed bounded on it? For example, $$ X=\{f\in C^{\infty}([0,1])\vert\sup_{n\ge0}\Vert f^{(n)}\Vert_{\infty}<\infty\} $$ (I do not know whether it is true and cannot give a proof). Any help will be greatly appreciated.
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$\begingroup$ To be clear, you want an infinite-dimensional (closed?) subspace of $L^2([0, 1])$, on which the differentiation is bounded in the $L^2$-norm? $\endgroup$– Aleksei KulikovCommented Sep 6 at 11:33
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$\begingroup$ @Aleksei Kulikov Yes, an infinite-dimensional subspace of $L^2([0,1])$ with as few restrictions as possible, but not necessarily closed. $\endgroup$– grahamCommented Sep 6 at 11:38
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$\begingroup$ A negative result, somehow related : mathoverflow.net/questions/376087/… $\endgroup$– Ayman MoussaCommented Sep 6 at 13:28
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I think the answer is no, there is no such infinite-dimensional subspace. Say, we have such a subspace $X$. I will assume that $X$ lies in the Sobolev space $W^{1,2}([0, 1])$, because otherwise it is not even clear what is $Df$, and what is its $L^2$-norm.
The key idea is that the eigenvalues of $D$ on $W^{1, 2}([0, 1])$ tend to infinity (they are $\lambda_n = \pi (n-1)$, $n\in \mathbb{N}$, up to the unimodular factos, if I didn't make any arithmetic mistakes), and therefore by the min-max principle in any $n$-dimensional subspace we can find a vector $f$ for which $||Df||\ge |\lambda_n| ||f||$.
Hence, if our space in infinite-dimensional, we can find such an $f_n$ for any $n$, and since $|\lambda_n| \to \infty$, the derivative operator is not bounded (by bounded I of course here mean that there exists a universal $C > 0$ such that $||Df|| \le C ||f||$).
answered Sep 6 at 11:54
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$\begingroup$ I think technically I should talk about singular values, or eigenvalues of $-D^2$, but the idea still stands. $\endgroup$ Commented Sep 6 at 12:01
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$\begingroup$ An easier but simialr approach: on this subspace $X$, the $L^2$ and $H^1$ norms are equivalent. Since the embedding of $H^1 \to L^2$ is compact, the unit ball of $X$ is compact. $\endgroup$ Commented Sep 6 at 14:34
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$\begingroup$ @GiorgioMetafune this requires explaining why we can assume that $X$ is complete (which is not hard, but still requires a couple of words). Very nice argument regardless! $\endgroup$ Commented Sep 6 at 15:04
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$\begingroup$ I think the problem with this argument is that $iD$ with domain $H^1$ not self-adjoint of $L^2(0,1)$, and we have a one-parameter family $T_{\alpha}$ of self-adjoint restrictions (corresponding to boundary conditions). Your argument would apply to a single $T_{\alpha}$, but it could be that an infinite-dimensional subspace intersects all of the $D(T_{\alpha})$. $\endgroup$ Commented Sep 6 at 15:20
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$\begingroup$ @ChristianRemling by sacrificing two extra dimensions we can assume zero boundary conditions, say. However, the eigenvalues I wrote are (I think) for the Neumann boundary conditions, which, if I was taught right, are the biggest ones and hold for the whole $W^{1, 2}(0, 1)$. $\endgroup$ Commented Sep 6 at 15:31