It is well known that if $T:\ell^1\to\ell^1$ is a bounded/continuous map then $$\Vert T\Vert_{op}=\sup\{\Vert Tx\Vert_1\mid x\in\ell^1\text{ and }\Vert x\Vert_1=1\}=\sup\{\Vert Te_n\Vert_1\mid n\ge1\}$$ where $\{e_n\}_{n=1}^\infty$ is the standard basis for the $\ell^p$ spaces.

However the same thing does not hold in $\ell^2$ (or $\ell^p$ for $1<p$) mainly because in the proof, some inequality depends on the $1$-norm of the sequence and it may not even be defined. So is there any "nice" set $U\subseteq\ell^2$ such that $$\Vert T\Vert_{op}=\sup\{\Vert Tx\Vert_2\mid x\in U\}$$



2 Answers 2


A very nice set $U$ at which every bounded operator attains its norm is the unit sphere of $\ell_2$ $$U=\{x\in\ell_2:\|x\|_2=1\}$$ It is a matter of taste whether there is a "nicer" set than this at all in $\ell_2$. For our purposes, there is certainly no strictly smaller set at which every operator attains its norm, because given any point $x_0\in U$, there is a linear functional whose norm is attained precisely at $x_0$ and nowhere else.

  • $\begingroup$ Yes but you could set the same $U$ for every $\ell^p$ and every Banach space, while $U=$"vectors of the canonical basis" only (seems) to work for $\ell^1$. The "there is a linear functional whose norm is attained precisely at $x_0$" does not convince me as the operator norm doesn't (i think) need to be attained at any particular vector but rather be the supremum $\endgroup$ Aug 12, 2019 at 11:09
  • $\begingroup$ I did not claim that the operator norm is always attained at a unit vector. This is in fact not true. What I did is to prove that in $\ell_2$ you cannot take a smaller norming set than $U$, and in order to prove that, it suffices to show that for every point $x\in U$ I can find a linear operator attaining its norm exactly there and nowhere else. I found one - namely, a linear functional. This is not saying that every linear operator attains its norm. $\endgroup$ Aug 12, 2019 at 11:15
  • $\begingroup$ but why can't you have something like a dense subset of $U=\{x\in\ell^2\mid\Vert x\Vert_2=1\}$ so you can approach any $x_0\in U$ and still get the operator norm by taking the $\sup$?? I think I'm not geting all the details that go into your answer, could you explain a bit more so I can accept it? $\endgroup$ Aug 12, 2019 at 11:22

The wikipedia page for the operator norm lists an equivalent definition of the operator norm of $T : V \to W,$ for any normed vector spaces $V\neq 0, W$ as $$\|T\|_{op} = \sup \left\{\|Tx\|_W : x \in V, \|x\|_V = 1\right\}$$

In particular, you can specialize $V = \ell^m, W = \ell^n.$

It seems you are mostly interested in finding a more specialized subset $V_0 \subset V$ such that $\|T\|_{op} = \sup \{\|Tx\|_W : x \in V_0\}.$

We could of course just take any dense subset of $\{x \in V : \|x\|_V = 1\},$ but this would be of arguable usefulness. Instead, the more useful thing to consider (at least for locally convex spaces) is the set of extreme points of the closed unit ball.

For $1 \leq p \leq \infty,$ $\ell^p$ is locally convex, but the set of extreme points of the closed unit ball is only strictly smaller than the boundary of the closed unit ball for $p\in\{1,\infty\}.$ This means that you shouldn't expect to have an analogous theorem for $\ell^p$ for any $1 < p < \infty.$


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