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Let $A$ be a Banach algebra and $I$ be an ideal of $A$. A derivation $D\colon A\to I$ is a linear bounded map, with the following property: $$D(ab)=aD(b)+D(a)b,\qquad a,b\in A.$$ Suppose that $I$ is dense in $A$, and any derivation $D\colon A\to I$ is an inner derivation i.e., there is a $x\in I$ such that for all $a\in A$ we have $D(a)=ax-xa.$ Now could we say that any derivation from $A$ into $A$ is an inner derivation? In other word, for derivation $D\colon A\to A$, is there an element $x\in A$, such that $D(a)=ax-xa$, for all $a\in A$?

Joonas Ilmavirta
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Hamid Shafie Asl
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1 Answers1

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If you ask for proving
Assume that $I$ is a dense ideal of $A$ and that every bounded derivation $D\colon A\to I$ is inner. Then any bounded derivation from $A$ into $A$ is inner.
then what follows is not an answer.
I continue in the hope that it may still be helpful since there's a pretty rich class of examples of concrete Banach algebras which admit inner derivations only.

Let $\mathscr{L}(X)$ denote the algebra of bounded linear operators on a Banach space $X$, and let $\mathscr F(X)$ be the finite-rank operators. If $A$ is a Banach subalgebra in between, i.e. $\mathscr F(X)\subset A\subset\mathscr{L}(X)$, then all bounded derivations on $A$ must be inner.

Concretely, the Schatten ideals $\,\ell^p(H)\,$ come to my mind when reading the OP. These are completions of $\mathscr F(H)$ and they are dense ideals in $\mathscr K(H)$, the compact operators on a (separable) Hilbert space. For $p=1$ one deals with the trace-class $\,\ell^1(H)\,$, and for $p=2$ with the Hilbert-Schmidt operators $\,\ell^2(H)\,$.

Algebra in the sequel is, to my best knowledge, due to I. Kaplansky.

Consider a derivation $\,D\colon A\to A\,$ and let $\,p\in\mathscr F(X)\subset A\,$ be a projection of rank one.
WLOG one may assume that $D(p) = 0$ by the following argument: From $\,D(p) = pD(p) + D(p)p\,$ one concludes $pD(p)p = 0$, and with $y:=[p, D(p)]$ one has $$[p,y] = p(pD(p)-D(p)p) - (pD(p)-D(p)p)p = D(p)$$ hence by defining $\,D'(a):=D(a)-[a,y]\,$ one arrives at $D'(p)=0$.

Now $D(ap)=D(a)p\;\forall a$ shows that $$ap\longmapsto D(a)p$$ is a well-defined linear map on the left-ideal $\,\mathscr{F}(X)p$. The latter is isomorphic to $X$ so that $t\in\mathscr{L}(X)$ exists with $tap=D(a)p$. This $t$ does the job because for any $b\in\mathscr F(X)$: $$tabp = D(ab)p = D(a)bp+aD(b)p = D(a)bp+atbp\\ \implies D(a) = [t,a]$$

Hanno
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  • For Banach algebras between finite rank operators and $B(H)$, an example is $K(H)$, compact operators, and this algebra has outer derivations. E.g. if $S$ is a nonunitary isometry, then $a\mapsto aS-Sa$ is an outer derivation on $K(H)$. I was just wondering, would $F(H)$ in $K(H)$ provide a counterexample? Are there any outer derivations $K(H)\to F(H)$? If not, we have an answer. When you call a derivation on $A\subset L(H)$ inner you don't seem to be referring to $D_a$ with $a\in A$, but rather $D_t$ with $t\in L(H)$, so it is a different problem. – Jonas Meyer Mar 23 '17 at 02:01
  • I posted a question in part inspired by your answer, and citing it: http://math.stackexchange.com/questions/2199319/is-every-bounded-derivation-from-compact-to-finite-rank-operators-inner – Jonas Meyer Mar 23 '17 at 03:54
  • Thanks for putting emphasis on that point, I shall take it away into my few days of sabbatical. My belief is that there are no outer derivations around from $\mathscr F(H)\to\mathscr K(H)$ ... but one never knows. – Hanno Mar 23 '17 at 16:10
  • Bounty awarded for this post's usefulness e.g. in inspiring a new question (and the grace period was going to end soon). I'm not upvoting because that would remove it from the Unanswered list, and because it takes a different meaning of inner than used in the question, and doesn't answer the question. (You mean $\mathscr K(H)\to \mathscr F(H)$ I think.) – Jonas Meyer Mar 23 '17 at 17:29
  • Yes, true, the arrow needs reversal, i.e., $\mathscr F(H)\leftarrow\mathscr K(H)$ ... I'm still (convinced and) seeking for the reason that a derivation can be realised as commutator with a finite rank op. Encouraged also by your bounty, thanks for that! – Hanno Mar 28 '17 at 20:53