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Consider a point $\hat{x}$. Say it satisfies two expressions: $$g(\hat{x}) \leq 0$$ $$g(\hat{x}) + \langle \nabla g(\hat{x}), d \rangle < 0$$

where $g$ is a convex function and $d$ is a vectorial direction. Now I want to show there exists a $\lambda d$ such that:

$$g(\hat{x} + \lambda d) <0 \quad \quad \quad \lambda \in (0,1)$$

How to show this?

DuttaA
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  • @Surb this is not an easy question (Atleast to me) such that I can enumerate all approaches. This is a part of a larger question. I have no clue how to solve this. Intuitionally I can see it happening but showing it formally will most probably require a specific form of Taylor expansion which I cannot figure out. – DuttaA Feb 16 '21 at 10:27
  • @Surb I don't think one requires regularity fo show such a thing holds (if i understan your defn of regularity correctly). We need to use the asymptotic form of Taylor series for such cases. And what you said is already has been thought by me, but how to show it formally is the real question.\ – DuttaA Feb 16 '21 at 10:33
  • "but how to show it formally is the real question?" Did you read my previous comment ? – Surb Feb 16 '21 at 10:36
  • @Surb yes. What is 'small enough'? I can't really say 'small enough' in a fromal proof. – DuttaA Feb 16 '21 at 10:38
  • @Surb Maybe it is correct (atleast I can't find a mistake in the argument), but the other equation is not useless. I don't like a solution without using it. In the original question $g(\hat{x}) < 0$ is not there. I have added it as i feel it was missed, but now I think I maybe wrong (although it should be correct). – DuttaA Feb 16 '21 at 10:45
  • @Surb yep now i see the error it was supposed to be $\leq$. I made a grave error, sorry if i wasted your time – DuttaA Feb 16 '21 at 10:50
  • Sorry, I had in mind $g(\hat x)\geq 0$. This mean that my answer is completely wrong... I erased it... (and sorry to have wasted your time). – Surb Feb 16 '21 at 12:34
  • @Surb you didn't. I actually figured out the solution while discussing with you. Maybe you can formalize it. So the idea is there can be 2 cases. In 1st case $g(x)<0$ which means there exists a ball like you said and inthe second case $g(x)=0$ which means the $\langle \nabla g(x), d \rangle < 0$ and thus a small step in direction of $d$ will give me $g(x + \lambda d)<0$ (this can be shown by the method in your answer which I originally saw in Boyd Vandenberghe). If you are interested you can formalize this answer and I'll accept it. – DuttaA Feb 16 '21 at 19:01

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