Constant first-moment weight is outside the proved square-root regret schedule

The interval $[-1,1]$ is nonempty, closed, and bounded by [F9], hence compact by [F5]. For $x,y\in [-1,1]$ and $\theta \in [0,1]$, [F8] gives $-1\le (1-\theta )x+\theta y\le 1$, so the domain is convex. Also ${\mathrm{\ell }}_t((1-\theta )x+\theta y)=(1-\theta ){\mathrm{\ell }}_t(x)+\theta {\mathrm{\ell }}_t(y)$, so each loss is convex by [F3]. Since $\mathrm{\mid }{\mathrm{\ell }}_t(x)-{\mathrm{\ell }}_t(y)\mathrm{\mid }=\mathrm{\mid }x-y\mathrm{\mid }$, choosing $\delta =\eta$ in [F4] proves continuity for every tolerance $\eta >0$. The gradient $g_t=1$ is a subgradient because ${\mathrm{\ell }}_t(y)={\mathrm{\ell }}_t(x)+(y-x)$ by [F6], and $\gamma ={\beta }_1\mathrm{/}\sqrt{{\beta }_2}=\sqrt{2}\mathrm{/}4<1$. Thus the hypotheses for [F2] hold. Iterating [F1] gives $v_t=1-2^{-t}$, which is nondecreasing, so ${\widehat{v}}_t=v_t\ge 1\mathrm{/}2$ for every $t\ge 1$.

Here ${\alpha }_t=1\mathrm{/}\sqrt{t}$ and ${\beta }_{1,t}=1\mathrm{/}4$. The second term in [F2] is therefore at least $\frac{4}{(3\mathrm{/}4{)}^2}{\sum }_{t=1}^T\frac{1}{4\sqrt{2}}\sqrt{t}=\frac{16}{9\sqrt{2}}{\sum }_{t=1}^T\sqrt{t}\mathrm{.}$ For $T\ge 2$, at least $T\mathrm{/}2$ indices satisfy $t\ge T\mathrm{/}2$, so ${\sum }_{t=1}^T\sqrt{t}\ge T^{3\mathrm{/}2}\mathrm{/}(2\sqrt{2})$. The correction term is at least $4T^{3\mathrm{/}2}\mathrm{/}9$.

The constant schedule is neither ${\beta }_{1,t}={\beta }_1{\lambda }^{t-1}$ with $\lambda <1$ nor ${\beta }_{1,t}={\beta }_1\mathrm{/}t$. Since the theorem's upper-bound expression contains this positive term, that expression cannot be $O(\sqrt{T})$ for this schedule: dividing its lower bound by $\sqrt{T}$ gives $4T\mathrm{/}9$, which is unbounded by [F7]. The computation concerns the stated bound only and gives no lower bound on actual regret.