Derivation into dense ideal of Banach algebras 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 $a\in A$?
 A: 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 $b\in\mathscr F(X)$:
$$tabp = D(ab)p = D(a)bp+aD(b)p = D(a)bp+atbp\\
\implies D(a) = [t,a]$$
