Diagonal AdaGrad subproblem has a unique minimizer

If $d=0$, ${\mathbb{R}}^0$ contains only the empty vector by [F9], so nonempty $C$ is that singleton and its sole point is the unique minimizer, with objective value zero. For the rest of the proof suppose $d\ge 1$. Let T=diag(sqrt(a_1),...,sqrt(a_d)). By [F8] each entry is the nonnegative square root of a_j; it is nonzero because a_j>0, so T is invertible and T^2=A. Set C'=TC. Since T is positive diagonal, [F2] makes C' nonempty, closed, and convex; [F10] supplies the convex-set convention used by [F2]. Therefore [F3]–[F5] make its indicator ι_{C'} a proper closed convex function.

Put $v=Tx-\eta T^{-1}g$. Since $T$ is diagonal, $T(z-x)$ has squared norm $\mathrm{\parallel }z-x{\mathrm{\parallel }}_A^2$ and $⟨T(z-x),T^{-1}g⟩=⟨z-x,g⟩$. Coordinate expansion gives $\eta ⟨g,z⟩+\frac{1}{2}\mathrm{\parallel }z-x{\mathrm{\parallel }}_A^2=\frac{1}{2}\mathrm{\parallel }Tz-v{\mathrm{\parallel }}_2^2+\eta ⟨g,x⟩-\frac{1}{2}{\eta }^2\mathrm{\parallel }{T}^{-1}g{\mathrm{\parallel }}_2^2$. The last two terms are independent of $z$, and $z\mapsto Tz$ bijects $C$ with $C^{\mathrm{\prime }}$, so the minimizers correspond to ${\mathrm{prox}}_{{\iota }_{C^{\mathrm{\prime }}}}(v)$ by [F3] and [F6].

By [F7] the latter proximal set is a singleton. The inverse map T^{-1} is one-to-one, so its corresponding minimizer z in C is also unique. This proves the Statement.