Remark 4

Let E be a locally convex space over $\mathbf{C}$. Let us denote by $E_0$ the locally convex space over $\mathbf{R}$ underlying E. In order that E be an approximation space, it is necessary and sufficient that $E_0$ be one. In fact, the condition is necessary; let us prove that it is sufficient. Suppose therefore that $E_0$ is an approximation space. Let C be a precompact subset of E and U a balanced convex neighbourhood of 0 in E. Put $C'= C\cup iC$. There exists a continuous $\mathbf{R}$-linear mapping $u$ of finite rank from $E_0$ into $E_0$ such that $x-u(x)$ belongs to U for every $x\in C'$. Put $v(x) =\frac{1}{2}(u(x)-iu(ix))$ for every $x$ in E. Thus one defines a continuous $\mathbf{C}$-linear mapping of finite rank from E into E. For every $x\in C$, we have $x\in C',ix\in C'$ and $x-v(x) =\frac{1}{2}(x-u(x))-\frac{i}{2}(ix-u(ix))$, so that $x-v(x)$ belongs to U. This proves that E is an approximation space.