We want to show that for any point ${\alpha }_0\ne {\alpha }_m$ the interval $J_2$ coming from any small interval $I_2$ centred on ${\alpha }_0$ is larger than the interval $J_1$ coming from the interval $I_1$ of the same width as $I_2$, but centred on ${\alpha }_m$ (see the figure). Since ${\alpha }_m$ is the only point with $D_f^{\prime }(\alpha )=0$ we know that $D_f({\alpha }_0)=s\ne 0.$ To match with our figure we'll assume that $s0.$ An analogous argument works when $s>0.$ Now choose a positive number $ϵ$ with $ϵ-2s/3$ (you'll see later why we make this choice). From the definition of a derivative we know that there exists ${\delta }_1>0$ such that for all $\alpha$ with $|\alpha -{\alpha }_m|{\delta }_1$ we have $$-ϵ\frac{D_f(\alpha )-D_f({\alpha }_m)}{\alpha -{\alpha }_m}ϵ.$$ Similarly, there exists ${\delta }_2>0$ such that for all $\alpha$ with $|\alpha -{\alpha }_0|{\delta }_2$ we have \begin{equation} s-\epsilon \frac{D_f(\alpha)-D_f(\alpha_0)}{\alpha-\alpha_0} D_f(\alpha^\prime_1)-D_f(\alpha_0)$$and$$(s+\epsilon)(\alpha_0-\alpha_2^\prime) D_f(\alpha_0)-D_f(\alpha^\prime_2).$$\text{Erroneous nesting of equation structures}$$D_f(\alpha^\prime_1)-D_f(\alpha^\prime_2)>-2\delta(s+\epsilon)>\epsilon\delta>D_f(\alpha^\prime)-D_f(\alpha_m).$$ Thus, interval $J_2$ is wider than interval $J_1$. QED.

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