Space of linear operators complete, but target space not.

Question

Let $X$, $Y$ be normed spaces and denote by $L(X,Y)$ the space of bounded linear operators from $X$ to $Y$. We know that $L(X,Y)$ is complete if $Y$ is. Could you provide me with an example where $L(X,Y)$ is complete, but $Y$ is not?

Answer 1

No, because the converse is also true (provided $X$ is non-trivial, i.e. $X \neq \{0\}$).

Indeed, let $\{y_n\}$ be a Cauchy sequence in $Y$. Let $x_0 \in X \setminus \{0\}$. By the Hahn-Banach Theorem, there is a linear functional $L \colon X \to K$, where $K = \mathbb{R}, \mathbb{C},...$ is the field of scalars, such that $L(x_0) = 1.$

Define $T_n \colon X \to Y$ via $x \mapsto L(x)y_n$. One can easily check that $T_n$ is linear and continuous, indeed $$\|T_n\|_{L(X,Y)} = \sup_{x \neq 0}\frac{L(x)y_n}{\|x\|} = \|L\|_{L(X,Y)}\|y_n\|_{Y} < \infty.$$

Similarly, $\|T_n - T_m\|_{L(X,Y)} \le \|L\|_{L(X,Y)}\|y_n - y_m\|_{Y}$, showing that $\{T_n\}$ is a Cauchy sequence in the Banach space $L(X,Y)$. Then there exists $T \colon X \to Y$ linear and continuous such that $T_n \to T$ in $L(X,Y)$. Then in particular $T_n(x) \to T(x)$ in $Y$ and hence $$y_n = L(x_0)y_n = T_n(x_0) \to T(x_0).$$ This shows that $\{y_n\}$ is convergent (to $T(x_0)$).