Completion of the space of bounded linear operators by completing the image. Let $V, W$ be normed vector spaces and $B(V,W)$ be the space of bounded linear operators from $V$ to $W$. In particular, $W$ may be a Banach space or not, i.e. not complete. I have learned that if $W$ is a Banach space, then $B(V,W)$ is also a Banach space with respect to the operator norm. I want to know if the following statement is true.
Claim: Suppose $W$ is a normed space but not a Banach space and also suppose that $B(V,W)$ is not a Banach space.  Since every metric space can be uniquely completed up to isometry, let $W'$ be the completion of $W$ and let $B'(V,W)$ be the completion of $B(V,W)$. Then, $B'(V,W)$ and $B(V,W')$ is isometric.
I believe this is true since $B(V,W')$ is a Banach space as $W'$ is a Banach space, and (although I cannot argue right now) clearly, $B(V,W)\subset B(V,W')$, where by uniqueness, $B'(V,W)=B(V,W')$ up to isometry.
Could someone please confirm if this is true? If so, can someone please elaborate this by pointing out things to check or improve, or even better, proof as well as examples? If not, can someone provide a counterexample?
 A: The conclusion is true in the case of separable inner product spaces. Assume $V$ is a dense proper subspace of a separable Hilbert space $\mathcal{H}_1$ and $W$ a dense proper subspace of a separable Hilbert space $\mathcal{H}_2.$ Let $A\in B(V,\mathcal{H}_2).$ Actually $B(V,\mathcal{H}_2)$ can be identified with $B(\mathcal{H}_1,\mathcal{H}_2).$  Fix an orthonormal basis $\{e_n\}_{n=1}^\infty$ in $ V$ and  an orthonormal basis $\{f_n\}_{n=1}^\infty$ in $ W.$ For every element $e_n$ there exists $k_n$ such that
$$\sum_{k=k_n+1}^\infty |\langle Ae_n, f_k \rangle |^2<2^{-n}\varepsilon^2 $$
Consider the operator $C$ defined on ${\rm lin}\{e_1,e_2,\ldots \}$ by
$$Ce_n=\sum_{k=k_n+1}^\infty \langle Ae_n,f_k\rangle f_k$$ The operator $C$ satisfies
$$\sum_{n=1}^\infty \|Ce_n\|^2=\sum_{n=1}^\infty\sum_{k=k_n+1}^\infty |\langle Ae_n, f_k \rangle |^2 \le \varepsilon^2\sum_{n=1}^\infty 2^{-n}=\varepsilon^2$$
Hence $C$ extends to a Hilbert-Schmidt operator from $\mathcal{H}_1$ to $\mathcal{H}_2$  and $\|C\|\le \varepsilon. $ For $D:=A-C$ we have
$\|A-D\|=\|C\|\le \varepsilon.$ Moreover $D\in B(V,W)$ as $e_n\in V$ and
$$\displaylines{De_n=Ae_n-Ce_n=\sum_{k=1}^\infty\langle Ae_n,f_k\rangle f_k-\sum_{k=k_n+1}^\infty\langle Ae_n,f_k\rangle f_k\\ =\sum_{k=1}^{k_n}\langle Ae_n,f_k\rangle f_k\in W}$$
Hence $B(V,W)$ is dense in $B(V,\mathcal{H}_2).$ According to the comment by @Severin Schraven, this completes the proof.
