Why always $x^{\alpha}-x\geq 1-\lambda$ is satisfied?

Question

Prove for any $0<\lambda<1$, there exists $0<\alpha<1$ and $0<x<1$ such that $x^{\alpha}-x\geq 1-\lambda$.

My solution: The maximum of the function $x^{\alpha}-x- 1$, over $x>0$ tends to zero as $\alpha \to 0$. Thus, there exist such $\alpha$.

Even if my solution is correct, it is so ugly. Is there a better way of solving this problem? Any inequality I can use? I would appreciate any helpful comments. Feel free to vote up!

Answer 1

Since $(0,1)$ is open, for any $\lambda$ there exists epsilon such that $(\lambda-\epsilon,\lambda+\epsilon) \subset (0,1)$. Pick $y \in (0,\lambda - \epsilon)$. Again, since $(0,1)$ is open, there exists $r$ with $(y-r,y+r)\subset (0,1)$. Let $\delta = \min(r,\epsilon/4)$, meaning $(y-\delta, y+\delta) \subseteq (y-r,y+r) \subset (0,1)$ and pick $x \in (y-\delta, y+\delta)$. Since $x^a \to 1$ as $a \to 0$, we can find $a$ such that $|x^a-1|\le \epsilon/2$, then $$\begin{align}|x^a-x-1+y|\le |x^a-1|+|x-y|\le \epsilon &\implies -\epsilon \le x^a-x-1+y \\ &\implies x^a-x \ge 1-y-\epsilon \ge 1-\lambda \end{align}$$

Answer 2

Intuitively, as $\alpha$ gets small for $x \lt 1$, $x^\alpha \to 1$. We have $x^\alpha-x=x(x^{\alpha-1}-1)$ Since $\alpha \lt 1, \alpha-1 \lt 0$ and $x^{\alpha-1} \gt 1$ and having $x$ small is a good thing. We will certainly get there.

For a proper proof, given $\lambda$ we can find $x, \alpha$. Let $x=\lambda/10$. Then we want $x^\alpha-x \gt 1-10x, x^{\alpha-1}\gt 1-9x, \alpha\gt 1+\frac{\log(1-9x)}{\log x}$