# Upper bound of a function 2

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.

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 $$\alpha = 1$$.

• Thanks @SeanLee. Very clear solution! Apr 11, 2019 at 10:36