Let $X$ be a Banach space, $L(X)$ the space of all bounded linear operators on $X$. We say that $A ∈ L(X)$ attains its norm if there exists $x ∈ X$ such that $\|x\| = 1$ and $\|Ax\| = \|A\|$. The following question sounds as a classical one, but possibly is still open.

Question: Does there exist an infinite-dimensional Banach space $X$ such that each $A ∈ L(X)$ attains its norm?

If such spaces exist, they should be very unusual (see related results below). If they do not exist, it would be good to know that in each infinite-dimensional Banach space there is a bounded linear operator which does not attain its norm.

Related results and observations:

(1) R. C. James’s characterization of reflexivity implies that if $X$ is such that each $A\in L(X)$ attains its norm, then $X$ and $L(X)$ are reflexive spaces.

(2) J. R. Holub [2] proved that if $X$ has the approximation property, then the reflexivity of $L(X)$ implies that $X$ is finite dimensional.

(3) N. J. Kalton [3, Theorem 2] proved that $L(X)$ cannot be reflexive for nonseparable $X$.

(4) Hence the only possible candidates for $X$ in the problem are separable reflexive spaces without $1$-complemented infinite-dimensional subspaces having the approximation property.

(5) Some more related results and observations can be found in [1] (see p. 693), [4], and [5].

(6) I published this question in Extracta Math., 20 (2005), 65-66, but never got any information on it.


[1] G. Godefroy and P. Saphar, Duality in spaces of operators and smooth norms on Banach spaces, Illinois J. Math., 32 (1988), no. 4, 672–695.

[2] J. R. Holub, Reflexivity of L(E, F), Proc. Amer. Math. Soc., 39 (1973), 175–177.

[3] N. J. Kalton, Spaces of compact operators, Math. Ann., 208 (1974), 267–278.

[4] V. A. Khatskevich, M. I. Ostrovskii, and V. S. Shulman, Extremal problems for operators in Banach spaces arising in the study of linear operator pencils, Integral Equations and Operator Theory, 51 (2005), 109-119.

[5] M. I. Ostrovskii, Extremal problems for operators in Banach spaces arising in the study of linear operator pencils, II, Integral Equations and Operator Theory, 51 (2005) 553-564

  • 1
    $\begingroup$ By Argyros-Haydon there exists an $\infty$-dimensional Banach space in which every operator is scalar-plus-compact. Is it true that every scalar-plus-compact operator attains its norm? $\endgroup$
    – YCor
    Feb 27, 2016 at 19:21
  • 3
    $\begingroup$ In general this is not true, one can construct examples even in a Hilbert space. Consider $I-K$, where $K$ is an operator of coordinate-wise multiplication of a vector in $\ell_2$ by a positive strictly decreasing sequence in $(0,1)$. $\endgroup$ Feb 27, 2016 at 19:35
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    $\begingroup$ @YCor typical example: diagonal matrix in $\ell^2$ with diagonal entries $1-1/n$. $\endgroup$ Feb 27, 2016 at 19:40
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    $\begingroup$ @YCor: The Argyros-Haydon space has the approximation property, and thus has operators which do not attain their norm by the result of Holub (mentioned in my question). Possibly, people who studied the space will see explicit examples. $\endgroup$ Feb 27, 2016 at 20:20
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    $\begingroup$ @YCor: I should mention that since the Argyros-Haydon space is non-reflexive, it has even one-dimensional operators which do not attain their norms. Possibly one should look at further developments related to the Argyros-Haydon space, such as: Argyros, Spiros A.; Motakis, Pavlos, A reflexive hereditarily indecomposable space with the hereditary invariant subspace property. Proc. Lond. Math. Soc. (3) 108 (2014), no. 6, 1381–1416; front.math.ucdavis.edu/1111.3603 $\endgroup$ Feb 27, 2016 at 21:09


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