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It is well-known that for bounded linear operators $T$ and $S$ on a Banach space $X$, the operator norm satisfies $\|T \circ S\| \leq \|T\| \|S\|$. Equality, however, is not generally achieved. A trivial case where equality holds is when one operator is the identity $I$, as $\|T \circ I\| = \|T\| \|I\| $ for any $ T \in \mathcal{L}(X)$.

  1. General Banach Spaces:
    Are there non-identity operators $ S \in \mathcal{L}(X) $ such that $ \|T \circ S\| = \|T\| \|S\| $ for all $ T \in \mathcal{L}(X) $? If such operators exist, what structural properties must they have (e.g., isometry, invertibility)?

  2. Specific Banach Spaces:
    If such operators are rare in arbitrary Banach spaces, are there examples of Banach spaces $X$ where:

    • A non-trivial subset of $\mathcal{L}(X)$ satisfies this property?
    • The entire space $\mathcal{L}(X)$ universally satisfies $\|T \circ S\| = \|T\| \|S\|$ for all $T, S \in \mathcal{L}(X)$, beyond the trivial one-dimensional case?

This question seeks to characterize operators and spaces where the operator norm composition equality holds universally.

Scottish Questions
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    What did you try? For instance, what happens in the case of a finite dimensional $X$? This is the first case to be considered! Are we working with a Banach space over $\Bbb R$ or over $\Bbb C$ or both cases should be considered? Consider a simpler case, $S$ is diagonalizable, and the diagonal form is $(d_1,d_2,\dots)$ on the diagonal with values in decreasing order of absolute value. For instance $(8, -8,6,-5,3,2,1)$. Take some other operator having w.r.t. the same basis on the diagonal $(0,0,1,1,1,1,1)$. What about $(-8,8)$ only? – dan_fulea Feb 14 '25 at 11:06
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    these may be interesting https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/abs/multiplicative-norms-on-banach-algebras/8CF6DBC6A2B4769C052BCBD9922969AA https://math.stackexchange.com/questions/110443/normed-algebra-with-multiplicative-norm-non-isomorphic-to-r-c-h – Evangelopoulos Foivos Feb 14 '25 at 11:12

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It's a fairly rigid example but in any Banach space $X$ you have the group of surjective isometries $\text{Iso}(X)$, the set of surjective linear operators $S:X\to X$ such that $||Sx||=||x||$ for all $x \in X$. Clearly there is equality in the composition norm in this case since for all linear operators $T$, $$||T \circ S|| = \sup\{||(T(S(x))||: ||x|| \leq 1\}=\sup\{||Tx||:||x||\leq1\}=||T||.$$

If $X$ is a Hilbert space this is exactly the unitary/orthogonal group $U(X)$ or $O(X)$ when $X$ is over the complex or real field, respectively. For $X$ over $\mathbb{C}$, the larger set $G = \{\lambda T | T\in \text{Iso}(X), \lambda\neq 0 \in \mathbb{C}\}$ also has the property you're looking for.

An operator $S$ for which $||T\circ S||=||T||$ always holds need not be an isometry nor invertible. Indeed, the elements in $G-\text{Iso}(X)$ are not isometries. And, there exist non-surjective (and thus not invertible) isometries which satisfy your property. For example, in $\ell^2(\mathbb{N})$ there's the right shift operator which maps the sequence $(x_1, x_2, x_3, \dots)$ to $(0, x_1, x_2, x_3, \dots)$.

Eleanor Blake
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