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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 $\|T|_{W}\|\leq \|T|_{N}\|$?

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  • $\begingroup$ Do you mean "is there ALWAYS..."? $\endgroup$
    – erz
    Commented Mar 11, 2018 at 8:33
  • $\begingroup$ Yes. I mean that there is ... $\endgroup$
    – Dongyang Chen
    Commented Mar 11, 2018 at 8:34

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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.

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  • $\begingroup$ Thanks, Pietro. But I do not understand why $N\cap M=\{0\}$. $\endgroup$
    – Dongyang Chen
    Commented Mar 11, 2018 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
    Commented Mar 11, 2018 at 21:22
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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.

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  • $\begingroup$ (Actually, "whenever it is infinite dimensional" as per the OP ) $\endgroup$
    – Pietro Majer
    Commented Mar 11, 2018 at 21:32
  • $\begingroup$ correct. thanks for pointing it out. added it. $\endgroup$
    – Arian
    Commented Mar 11, 2018 at 22:46

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