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Let $H$ be a complex Hilbert space and let $B(H)$ denote the space of bounded linear operators $H \to H$ equipped with operator norm: $$ \lVert T \rVert = \sup\big\{ \lVert Tx \rVert \: : \: \lVert x \rVert \leq 1\big\}. $$ One easily shows that $B(H)$ is not a Hilbert space whenever $\dim(H) > 1$ holds, for it does not satisfy the parallelogram rule. Furthermore, an abstract argument shows that there exists a Hilbert space norm $\lVert\:\cdot\:\rVert_2$ on $B(H)$, but it does not provide us with a very concrete description of such a norm. In the finite-dimensional case one might take the Hilbert–Schmidt norm: this turns $B(H)$ into a Hilbert space and it is known to be submultiplicative. However, in the infinite-dimensional case this does not work, for now the space of Hilbert–Schmidt operators is a proper subspace of $B(H)$. This leads me to the following question:

Question. Is there a submultiplicative Hilbert space norm on $B(H)$ if $H$ is infinite-dimensional, either by abstract reasoning or by concrete example? For the moment I do not care whether this new norm is equivalent to the operator norm.

This is a strengthening of the question Is B(H) a Hilbert space? which did not ask for submultiplicativity.

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The answer is no: such a norm does not exist. First I'll give a short-ish proof; after that I will spell out some of the details for the benefit of those who do not know these already.

Assumption. All spaces (vector spaces, algebras, etc.) under consideration are complex.

We prove something slightly stronger.

Theorem. Let $H$ be an infinite-dimensional Hilbert space, and let $\lVert\:\cdot\:\rVert$ be a norm on $B(H)$ such that for each $a\in B(H)$ the left and right multiplication operators $L_a : b\mapsto ab$, $R_a : b\mapsto ba$ are continuous. Then $B(H)$ is not reflexive with respect to $\lVert\:\cdot\:\rVert$. In particular, $\lVert\:\cdot\:\rVert$ cannot be a Hilbert space norm.

Proof. If $\lVert\:\cdot\:\rVert$ is not complete, then clearly $B(H)$ is not reflexive with respect to this norm. So we assume for the remainder of this proof that $\lVert\:\cdot\:\rVert$ is complete. Now there exists an equivalent norm $\lVert\:\cdot\:\rVert_{\text{Ban}}$ on $B(H)$ turning $B(H)$ into a unital Banach algebra (in other words, we have $\lVert ab\rVert_{\text{Ban}} \leq \lVert a\rVert_{\text{Ban}} \cdot \lVert b\rVert_{\text{Ban}}$ for all $a,b\in B(H)$, and $\lVert \text{id} \rVert_{\text{Ban}} = 1$).

Now choose some orthonormal basis $\{v_i\}_{i\in I}$ of $H$, and let $A \subseteq B(H)$ be the subalgebra of diagonal operators with respect to this basis. Then $A$ is equal to its own commutator, and therefore it is $\lVert\:\cdot\:\rVert_{\text{Ban}}$-closed. This means that $\lVert\:\cdot\:\rVert_{\text{Ban}}$ turns $A$ into a Banach algebra. But $A$ is algebra-isomorphic to $\ell^\infty(I)$, which is unital, commutative and semisimple, so we know that all Banach algebra norms on $A$ are equivalent. In other words, the algebra isomorphism $A \cong \ell^\infty(I)$ is also an isomorphism of Banach spaces¹ $(A,\lVert\:\cdot\:\rVert_{\text{Ban}})\cong(\ell^\infty(I),\lVert\:\cdot\:\rVert_\infty)$. Since $I$ is infinite, we know that $(\ell^\infty(I),\lVert\:\cdot\:\rVert_\infty)$ is not reflexive, so it follows that $(A,\lVert\:\cdot\:\rVert_{\text{Ban}})$ is not reflexive either. Since $A$ is closed in $(B(H),\lVert\:\cdot\:\rVert_{\text{Ban}})$, it follows that $(B(H),\lVert\:\cdot\:\rVert_{\text{Ban}})$ also fails to be reflexive.

Finally, since the norms $\lVert\:\cdot\:\rVert$ and $\lVert\:\cdot\:\rVert_{\text{Ban}}$ are equivalent, it follows that $B(H)$ is not reflexive with respect to $\lVert\:\cdot\:\rVert$.$\quad\blacksquare$

¹: By isomorphism of Banach spaces we mean an invertible bounded linear operator, not necessarily isometric. Strictly speaking, such a map is not an isomorphism of Banach spaces, since the norm is part of the Banach space structure. A better term would be isomorphism of Banachable spaces. See also this discussion on MathOverflow.

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We proceed to fill in some of the details (references given below).

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Fact 1. A reflexive normed space must be complete.

Proof. Dual spaces are complete.

See also: [Rynne&Youngson] Definition 5.38, or [Conway] Definition III.11.2.$\quad\Box$

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Fact 2. Existence of the equivalent Banach algebra norm $\lVert\:\cdot\:\rVert_{\text{Ban}}$.

Proof. See any one of the following sources:

  • [Conway] Exercise VII.1.1;
  • [Kaniuth] Proposition 1.1.1;
  • [Rudin] Theorem 10.2.$\hspace{10cm}\Box$

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Fact 3. $A$ is equal to its own commutator.

Proof. Since $A$ is commutative, we have $A \subseteq A'$. Now let $b\in A'$ be given. We show that $b$ must be diagonal with respect to $\{v_i\}_{i\in I}$. To that end, choose some $i_0 \in I$ and let $a\in A$ denote the orthogonal projection onto $v_{i_0}$ (this is the diagonal operator given by $v_{i_0} \mapsto v_{i_0}$ and $v_j \mapsto 0$ for all $j \neq i_0$). Then we have $bv_{i_0} = bav_{i_0} = abv_{i_0}$, hence $bv_{i_0} \in \text{im}(a) = \text{span}(v_{i_0})$. We see that every $v_i$ is an eigenvector of $b$, so indeed $b$ is diagonal with respect to $\{v_i\}_{i\in I}$.$\quad\Box$

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Fact 4. A commutator in a Banach algebra is closed.

Proof. Let $X$ be a Banach algebra and $S \subseteq X$ a subset. Then we have $$ S' = \bigcap_{s\in S} \: \ker(L_s - R_s), $$ which is closed since it is an intersection of closed sets. (Here $L_s : r \mapsto sr$ and $R_s : r \mapsto rs$ denote the left and right multiplication operators. We have $L_s,R_s \in B(X)$, so that $\ker(L_s - R_s)$ is closed.)$\quad\Box$

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Fact 5. For any index set $I$, the algebra $\ell^\infty(I)$ is semisimple.

Proof. It admits a $C^*$-algebra structure, and commutative $C^*$-algebras are semisimple. (Recall that a commutative Banach algebra $X$ is semisimple if and only if the Gelfand representation $X\to C_0(\Omega(X))$ is injective.)$\quad\Box$

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Fact 6. In a unital, commutative and semisimple algebra, all Banach algebra norms are equivalent.

Proof. See any one of the following sources:

  • [Conway] Exercise VII.8.15;
  • [Kaniuth] Corollary 2.1.11;
  • [Pedersen] Exercise 4.2.13;
  • [Rudin] Corollary 11.10.$\hspace{10cm}\Box$

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Fact 7. A closed subspace of a reflexive Banach space is again reflexive.

Proof. See any one of the following sources:

  • [Conway] Exercise III.11.4;
  • [Pedersen] Exercise 2.4.8;
  • [Rudin] Exercise 4.1(d);
  • [Rynne&Youngson] Theorem 5.44.$\hspace{8cm}\Box$

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Fact 8. If $\lVert\:\cdot\:\rVert_1$ and $\lVert\:\cdot\:\rVert_2$ are equivalent norms on a vector space $X$, then $(X,\lVert\:\cdot\:\rVert_1)$ is reflexive if and only if $(X,\lVert\:\cdot\:\rVert_2)$ is reflexive. Consequently, if $T : X \to Y$ is an invertible bounded linear operator on normed spaces $X,Y$, then $X$ is reflexive if and only if $Y$ is reflexive.

Proof. Since $\lVert\:\cdot\:\rVert_1$ and $\lVert\:\cdot\:\rVert_2$ induce the same topology, the dual spaces $(X,\lVert\:\cdot\:\rVert_1)^*$ and $(X,\lVert\:\cdot\:\rVert_1)^*$ are not only isomorphic, but actually the same vector space. It is easily seen that the corresponding dual norms are equivalent. Taking this one step further, it is clear that the natural maps $J_1 : (X,\lVert\:\cdot\:\rVert_1) \to (X,\lVert\:\cdot\:\rVert_1)^{**}$ and $J_2 : (X,\lVert\:\cdot\:\rVert_2) \to (X,\lVert\:\cdot\:\rVert_2)^{**}$ are equal. Therefore $J_1$ is surjective if and only if $J_2$ is surjective.

Now let $X$ and $Y$ be normed spaces and let $T : X \to Y$ be an invertible bounded linear operator. We let $\lVert\:\cdot\:\rVert_1$ denote the original norm on $X$ and $\lVert\:\cdot\:\rVert_2$ the norm on $X$ obtained from $Y$ via $T$, so that $T$ is an isometric isomorphism $(X,\lVert\:\cdot\:\rVert_2)\cong Y$. Clearly $(X,\lVert\:\cdot\:\rVert_2)$ is reflexive if and only if $Y$ is reflexive, since they are isometrically isomorphic (and thus, in every sense, the same Banach space). Now it follows from the above that $(X,\lVert\:\cdot\:\rVert_1)$ is reflexive if and only if $Y$ is reflexive.

For an alternative proof, see [Rynne&Youngson] Corollary 5.56.$\quad\Box$

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Fact 9. For any infinite set $I$, the space $\ell^\infty(I)$ is irreflexive.

Proof. Choose some countably infinite subset $J \subseteq I$, then $\ell^\infty(J)$ is easily seen to be a closed subspace of $\ell^\infty(I)$ that is isometrically isomorphic to the sequence space $\ell^\infty(\mathbb{N})$. The latter is known to be irreflexive, so we see that $\ell^\infty(J)$, and therefore $\ell^\infty(I)$, are irreflexive as well (using Fact 7).$\quad\Box$

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References:

[Conway]: John B. Conway, A Course in Functional Analysis (1985), Springer Graduate Texts in Mathematics 96.

[Kaniuth]: Eberhard Kaniuth, A Course in Commutative Banach Algebras (2009), Springer Graduate Texts in Mathematics 246.

[Pedersen]: Gerd K. Pedersen, Analysis Now, Revised Printing (1995), Springer Graduate Texts in Mathematics 118.

[Rudin]: Walter Rudin, Functional Analysis, Second Edition (1991), McGraw–Hill.

[Rynne&Youngson]: Bryan P. Rynne & Martin A. Youngson, Linear Functional Analysis, Second Edition (2008), Springer Undergraduate Mathematics Series.

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