Density of bounded linear operators space Assume that $D$ is a dense subspace of a Banach space $X$. Can we conclude that the space of bounded linear operators $\mathcal{B}(D,D)$ is dense in $\mathcal{B}(D,X)$?
Thank you in advance
 A: Functional analysis is not my field of expertise, so this answer is incomplete. But since I had an idea, I wanted to share and revisit it later. Maybe it helps someone else in the meantime:
Let $i\colon D\hookrightarrow X$ be the canonical inclusion (which is continuous). To be precise, we have to look at the space $i(\mathcal{B}(D,D))=\{i\circ f|f\in\mathcal{B}(D,D)\}$ instead of $\mathcal{B}(D,D)$ to really make it a subspace of $\mathcal{B}(D,X)$. Since $X$ is a Banach space, so is $\mathcal{B}(D,X)$.
Let $(e_i)_{i\in I}$ be a basis of $D$, then we can expand it to a basis $(e_i)_{i\in J}$ of $X$ for $I\subset J$. Let $T\in\mathcal{B}(D,X)$, then for every $i\in I$, we have sequences $(d_n^{(i)})_{n\in\mathbb{N}}$ in $D$ with $d_n^{(i)}\rightarrow T(e_i)$ since $D$ is dense in $X$. These sequences are bounded since $T$ is bounded, so we can define a sequence $(T_n)_{n\in\mathbb{N}}$ of linear operators by $T_n\colon D\rightarrow D,e_i\mapsto d_n^{(i)}$. Here I assume, that there exists a $N\in\mathbb{N}$, so that for all $n\geq N$, $T_n$ is bounded (and therefore continuous) but haven't been able to prove it yet.
For $d\in D$ we have a unique representation $d=\sum_{i\in I}\lambda_ie_i$ and:
\begin{equation}
T_n(d)
=T_n\left(\sum_{i\in I}\lambda_ie_i\right)
=\sum_{i\in I}\lambda_iT_n(e_i)
=\sum_{i\in I}\lambda_id_n^{(i)}
\rightarrow\sum_{i\in I}\lambda_iT(e_i)
=T\left(\sum_{i\in I}\lambda_ie_i\right)
=T(d),
\end{equation}
which shows $T_n\rightarrow T$ pointwise.
A: I can answer the question for a particular case of separable Hilbert space and a particular dense subspace.
Let $\{e_n\}_{n=1}^\infty $ denote an orthonormal basis in $\mathcal{H}$ and $E={\rm span}\,\{ e_n\,:\, n\ge 1\}.$ Let $A\in B(\mathcal{H}).$ For $\delta>0$ there exists $v_1\in E$ such that $$\|Ae_1-v_1\|< \delta $$ Let $$A_1x=\begin{cases} x & x\perp e_1\\
v_1 & x=e_1
\end{cases} $$
For $x\perp e_1$ we have
$$A_1(x+\lambda e_1)=Ax+\lambda v_1=A(x+\lambda e_1)+\lambda  (v_1-Ae_1)$$
Therefore
$$\|(A-A_1)(x+\lambda e_1)\|=|\lambda|\|Ae_1-v_1\|<\delta |\lambda|=\delta {|\lambda|\over \sqrt{\|x\|^2+|\lambda|^2}}\|x+\lambda e_1\|$$
Hence $\|A-A_1\|\le \delta.$ In a similar way we construct the operator $A_2$ such that $A_2e_2=v_2\in  E,$ $A_2x=A_1x$ for $x\perp e_2$ and $\|A_1-A_2\|\le \delta/2.$ We keep applying that procedure obtaining a sequence of operators $A_n$ such that $A_ne_n=v_n\in E,$ $A_nx=A_{n-1}x$ for $x\perp e_n$ and $\|A_{n-1}-A_n\|\le \delta/2^{n-1}.$ Then the sequence of operators $A_n$ satisfies the Cauchy condition. Let $B=\lim _n A_n.$ By construction $\|A-B\|\le 2\delta$ and $$Be_n =A_ne_n=v_n\in E, \quad n\ge 1$$ Therefore $B(E)\subset E.$
