Consider function $f:\mathbb{R}_{\geq 0} \times [0,1] \rightarrow \mathbb{R}$ defined with: $$ f(z,\alpha) = (1+\alpha) \left [ z- \frac{z^{\alpha}}{2} - z^{\alpha+1} \right].$$ Prove that $f(z,\alpha) <\frac{1}{2}$ on its domain ($z \geq 0, \alpha \in [0,1]$). Numerically, the claim seems to be correct. I'm unable to prove it analytically.
1 Answer
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1 Here's an analytic solution that I hope is not too convoluted:
First, observe that when $z > 1$, we have:
\begin{align} f(z,a) &= (1 + \alpha) (z - \frac{z^{\alpha}}{2} - z^{1+\alpha}) \\ &<(1 + \alpha) \Big(- \frac{z^{\alpha}}{2}\Big) \\ &< 0 \end{align}
So we only need to be concerned when $z \leq 1$. When $z \leq 1$, we have:
\begin{align} f(z,\alpha) &= (1 + \alpha) (z - \frac{z^{\alpha}}{2} - z^{1+\alpha}) \\ &\leq (1 + \alpha) (z - z^{1+\alpha}) \\ &:= g(z, \alpha) \end{align} Now, differentiating $g$ with respect to $z$ and setting to zero (essentially finding the maximum) gives us the following value of $z$: $$z = (\alpha+1)^{-\frac{1}{\alpha}} $$
So, substituting this back into $g(z,\alpha)$, we obtain:
\begin{align} g(z, \alpha) &\leq (1+\alpha) ((\alpha+1)^{-\frac{1}{\alpha}} - (\alpha+1)^{-\frac{1}{\alpha}(\alpha + 1)} ) \\ &= \frac{\alpha}{ (1 + \alpha)^{\frac{1}{\alpha} } }\\ &\leq \frac{\alpha}{1 + \alpha} \\ &\leq 0.5 \end{align} Where the last inequality follows from the fact that $\frac{\alpha}{1+\alpha}$ is an increasing function and is equal to $0.5$ when $\alpha = 1$.
- 1$\begingroup$ Thanks @SeanLee. Very clear solution! $\endgroup$Nidjsi– Nidjsi2019-04-11 10:36:06 +00:00Commented Apr 11, 2019 at 10:36