Let $X,Y$ be infinite-dimensional Banach spaces and $T:X \rightarrow Y$ be a bounded linear operator. Let $M$ be an infinite-dimensional subspace of $X$ ($M$ is not necessarily closed). Let $N$ be an infinite-dimensional closed subspace of $\overline{M}$. My question is the following:

Question. Is there an infinite-dimensional closed subspace $W$ of $M$ such that $\|T|_{W}\|\leq \|T|_{N}\|$?

  • $\begingroup$ Do you mean "is there ALWAYS..."? $\endgroup$ – erz Mar 11 '18 at 8:33
  • $\begingroup$ Yes. I mean that there is ... $\endgroup$ – Dongyang Chen Mar 11 '18 at 8:34

In general, no. Let $T:X\to Y$ be a bounded linear operator between infinite dimensional Banach spaces, as assumed in the question. Assume further $N:=\ker T$ is a separable subspace of both infinite dimension and infinite co-dimension. Then, there is an infinite dimensional dense subspace $M\subset X$ of $N$ such that $M\cap N=(0)$ and $N\subset \overline{M} $. In this situation, of course, any non-null subspace $W\subset M$ verifies $\|T_{|W}\| >0=\|T_{|N}\|$.

Details. Finding the subspace $M$ requires a little argument. By the assumptions on $N$: There is a sequence $\{u_k\}_k$ such that $\overline{\operatorname{span}}\{u_k\}_k=N$. There is an infinite dimensional subspace $N'$ such that $N'\cap N=(0)$. There is a bounded linearly independent double sequence $\{v_{j,k}\}_{j,k}\subset N'$. Then $\{u_k+2^{-j}v_{j,k}\}_{j,k}$ is a linearly independent family that generates a linear subspace $M$ such that $N\subset \overline{M}$ and $N\cap M=(0)$, as wanted.

  • $\begingroup$ Thanks, Pietro. But I do not understand why $N\cap M=\{0\}$. $\endgroup$ – Dongyang Chen Mar 11 '18 at 13:47
  • $\begingroup$ If $u\in N\cap M$ then $u= \sum_{i,j}c_{i,j} (u_k+2^{-j}v_{i,j})$ apply $T$ and get $0=T\big( \sum_{i,j} 2^{-j}c_{i,j} v_{i,j}\big)$ that is $\sum_{i,j} 2^{-j}c_{i,j} v_{i,j}\in N'\cap N=(0)$ whence $ c_{i,j}=0$ for all $i,j$ and $u=0$ . $\endgroup$ – Pietro Majer Mar 11 '18 at 21:22

When $N=\overline{M}$ clearly we have $$||T_{N}||=||T|_{\overline{M}}||\geqslant||T_{M}||\geqslant ||T|_W||$$ for any closed subspace $W\subseteq M$. Also analogue when $\overline{M}=M$.

When $N\subset \overline{M}$ is a proper subspace then since $N=\overline{N}$ (closed) then it must be the case $N\subseteq M$ since $\overline{M}$ is the smallest closed set containing $M$. Let $W\subseteq M$ be some closed subspace. If $N\cap W=\emptyset$ then either case is possible $||T_N||\geqslant ||T_W||$ or $||T_W||\geqslant ||T_N||$. If $N\cap W\neq\emptyset$ then $N\cap W\subseteq M$ is closed subspace (both $N$ and $W$ are closed subspaces). Then $$||T_{N\cap W}||\leqslant\min\{||T_N||,||T_W||\}\leqslant ||T_N||$$ So $N\cap W$ fullfills the condition whenever it is nonempty and infinite dimensional.

  • $\begingroup$ (Actually, "whenever it is infinite dimensional" as per the OP ) $\endgroup$ – Pietro Majer Mar 11 '18 at 21:32
  • $\begingroup$ correct. thanks for pointing it out. added it. $\endgroup$ – Arian Mar 11 '18 at 22:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.