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Mar 3, 2021 at 3:22 comment added Willie Wong @Iosif: you are correct. I have the right asymptotics but wrong proof. Unfortunately I cannot delete an accepted answer. I've edited the post to highlight what you pointed out.
Mar 3, 2021 at 3:21 history edited Willie Wong CC BY-SA 4.0
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Mar 3, 2021 at 0:23 comment added Iosif Pinelis When you say "Since $f$ is non-decreasing, the function $\alpha\mapsto \alpha f(\alpha)$ is strictly increasing and hence invertible with strictly increasing inverse", it is not quite clear to me what you mean here by the inverse. I mean, $f$ was not assumed to be continuous, and so, the function $\alpha\mapsto \alpha f(\alpha)$ may fail to be continuous. So, the range of the latter function may be disconnected.
Mar 2, 2021 at 23:19 vote accept char
Mar 2, 2021 at 22:35 history answered Willie Wong CC BY-SA 4.0