Positive diagonal maps preserve closed convex sets

Choose x_0 in C. Then Dx_0 in DC, so DC is nonempty. If x,y in C and t in [0,1], then by linearity D((1-t)x+ty)=(1-t)Dx+tDy; [F1] puts the left side in DC, so DC is convex.

Fix $z\mathrm{\notin }DC$. Since each $d_j>0$, the definition of negative integer powers gives $d_j^{-2}=(d_j^2{)}^{-1}$. The square $d_j^2$ is positive by [F8], so $d_j^{-2}>0$ by [F9]. Thus ${\sum }_j{d}_j^{-2}>0$ by [F11], and $M=\sqrt{{\sum }_j{d}_j^{-2}}$ exists by [F6] and is positive. The inverse image $x=D^{-1}z$ is not in $C$. Since $C$ is closed, [F2] supplies $r>0$ with $B(x,r)$ disjoint from $C$. For every $q\in {\mathbb{R}}^d$, $\mathrm{\parallel }{D}^{-1}q{\mathrm{\parallel }}_2^2={\sum }_j{d}_j^{-2}{q}_j^2\le \left({\sum }_j{d}_j^{-2}\right)\left({\sum }_j{q}_j^2\right)=M^2\mathrm{\parallel }q{\mathrm{\parallel }}_2^2\mathrm{.}$ Indeed each $q_j^2$ is nonnegative, is at most its finite sum by [F7], and multiplication by $d_j^{-2}>0$ preserves that inequality; summing uses [F7]. Both norms and $M\mathrm{\parallel }q{\mathrm{\parallel }}_2$ are nonnegative, so [F10] yields $\mathrm{\parallel }{D}^{-1}q{\mathrm{\parallel }}_2\le M\mathrm{\parallel }q{\mathrm{\parallel }}_2$.

If ||w-z||_2<r/M, then step 1.2 gives ||D^{-1}w-x||_2<r, hence D^{-1}w notin C and w notin DC. Therefore B(z,r/M) lies in the complement of DC. Every point outside DC has such a ball, so [F2] makes DC closed; step 1.1 already proved DC nonempty and convex.