Timeline for How can we show that if $f$ is convex, then $\liminf_{|x|\to\infty}\frac{x\cdot\nabla f(x)}{|x|}>0$?
Current License: CC BY-SA 4.0
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| Jan 27, 2019 at 16:09 | comment | added | Iosif Pinelis | @0xbadf00d : "However, what is ∇f(x) in nguyen0610's answer if it's not the "subgradient"?" -- I think this question should be addressed to nguyen0610. Is there anything in my answer that is still not quite clear to you? | |
| Jan 27, 2019 at 16:04 | comment | added | 0xbadf00d | @IosifPinelis Yes, actually I've intended to comment your answer. However, what is $\nabla f(x)$ in nguyen0610's answer if it's not the "subgradient"? | |
| Jan 27, 2019 at 15:26 | comment | added | Iosif Pinelis | @0xbadf00d : Perhaps you meant this comment rather as a comment to my answer, since the above answer does not contain the term "subgradient". Anyhow, I have now added notes to my answer in response to your comment. Please let me know if anything is still not clear enough. | |
| Jan 27, 2019 at 11:33 | comment | added | 0xbadf00d | I'm not familiar with the concept of a subgradient. From Wikipedia, $\nabla f(x)$ is actually a set. So, how do we need to read $\liminf_{|x|\to \infty} \frac{\langle \nabla f(x),x\rangle}{|x|} >0$? Does it mean, for any choice $g(x)\in\nabla f(x)$, it holds $\liminf_{|x|\to \infty} \frac{\langle g(x),x\rangle}{|x|} >0$? And we clearly need that each $\nabla f(x)$ is nonempty. | |
| Jan 27, 2019 at 11:27 | vote | accept | 0xbadf00d | ||
| Jan 27, 2019 at 7:21 | history | answered | nguyen0610 | CC BY-SA 4.0 |