Timeline for How can we calculate the generalized gradient of $L^2\ni x\mapsto a\min(x(s),by(t))$?
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29 events
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| Oct 27, 2019 at 19:53 | comment | added | 0xbadf00d | I was able to reduce my problem to a smooth optimization problem. It would be great if you could took a look mathoverflow.net/q/344621/91890. | |
| Sep 24, 2019 at 15:28 | comment | added | 0xbadf00d | Give me one last shot: $x\mapsto x^+:=\max(0,x)$ can be smoothly approximated by $x\mapsto p_\alpha(x):=x+\frac1\alpha\ln\left(1+e^{-\alpha x}\right)$, $\alpha>0$. We could use this to approximate $x\mapsto|x|=x^++(-x)^+$ and hence to approximate $(a,b)\mapsto\min(a,b)=\frac12(a+b-|a-b|)$. We could solve the optimization problem with an accordingly modified objective function. If $w^{(\alpha)}$ is the resulting maximizer, how would it be related to the maximizer $w$ of the actual objective function? (If you know it, this might be an answer to one of the other questions.) | |
| Sep 24, 2019 at 6:24 | comment | added | 0xbadf00d | interchange of minimization and integration given in Theorem 14.60 of the book Variational Analysis by Rockafellar and Wets (see also Theorem 2.1 in this paper) might be useful. Do you think it is applicable here? | |
| Sep 22, 2019 at 18:55 | comment | added | 0xbadf00d | Just worked out the left details (mathoverflow.net/q/342173/91890) and noticed that $w$ disappears when we differentiate the integral function. So, it seems like this approach is of no help to determine the maximizing $w$. Maybe you can take a look if I missed something crucial. | |
| Sep 21, 2019 at 14:53 | comment | added | 0xbadf00d | I've tried to figure out as much as possible in the special case $I=\{1,2\}$ and asked a separate question for how we can proceed: mathoverflow.net/q/342173/91890. Would be great if you could take a look. | |
| Sep 19, 2019 at 18:23 | comment | added | 0xbadf00d | Thanks for your comment. I don't necessarily need an answer with all the details, but an outline or sketch how one could proceed. (I could imagine that a consideration in the spirit of Example 10.49 of Clarke's book might work.) | |
| Sep 19, 2019 at 16:55 | comment | added | Iosif Pinelis | @0xbadf00d : I agree with your latter comment. Concerning the previous one, I think that may go beyond the usual framework for MO answers and possibly require a full-blown paper. Even for the much simpler question posted on this page, the answer with complete details is quite long. | |
| Sep 19, 2019 at 14:34 | comment | added | 0xbadf00d | (BTW, we can conclude that in the case $x(s)=bx(t)$, $$\partial_Cf(x)=\{A\delta_s+(1-A)b\delta_t:A\in[0,1]\},$$ no matter whether $s=t$ or not, right? And it's worth to mention that your argument should still hold if we replace $L^2(\tau)$ by $L^p(\tau)$, $p\in[1,\infty]$.) | |
| Sep 19, 2019 at 13:02 | comment | added | 0xbadf00d | In light of the equation $(\ast)$ in the screenshot I've provided in the question math.stackexchange.com/q/3360750/47771, shouldn't we be able to use the result of this question? | |
| Sep 19, 2019 at 12:11 | comment | added | Iosif Pinelis | @0xbadf00d : I have briefly looked at those questions, but at the moment I don't have good ideas about them yet. | |
| Sep 19, 2019 at 4:16 | comment | added | 0xbadf00d | Thanks, once again. Did you had a chance to take a look at the other two questions? | |
| Sep 18, 2019 at 21:04 | comment | added | Iosif Pinelis | @0xbadf00d : I have provided the details you requested, and also a correction: the previous formulas were good only for the more difficult case $x(s)=bx(t)$; the considerations are quite a bit simpler if $x(s)<bx(t)$ or $x(s)>bx(t)$. | |
| Sep 18, 2019 at 20:57 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 18, 2019 at 20:52 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 18, 2019 at 18:21 | comment | added | 0xbadf00d | One last query for this question: (a) How exactly did you obtain the result for the directional derivative? (b) It's clear to me that if $\{t\}\in\mathcal T$ and $\tau(\{t\})>0$, then $$|x(t)|=\frac1{\tau(\{t\})}\int_{\{t\}}|x|\:{\rm d}\tau$$ for all $x\in L^1(\tau)$. Thus, assuming that $\tau$ is finite, $L^p(\tau)\ni x\mapsto x(\{t\})$ is a bounded linear functional by Hölder's inequality for all $p\ge1$. In particular, it is (globally) Lipschitz continuous. But what do we do, when $\tau$ is not finite? I guess we need at least something like local compactness or am I wrong? | |
| Sep 18, 2019 at 16:51 | comment | added | 0xbadf00d | Thanks, once again. BTW: My motivation for the question was this question: mathoverflow.net/q/339511/91890. Can we utilize the result here to answer the other question? See this question: math.stackexchange.com/q/3360750/47771 for a situation which is closer to the situation here as well. | |
| Sep 18, 2019 at 16:50 | comment | added | Iosif Pinelis | @0xbadf00d : I'll try to look at that other question as well. | |
| Sep 18, 2019 at 16:50 | vote | accept | 0xbadf00d | ||
| Sep 18, 2019 at 16:49 | comment | added | Iosif Pinelis | @0xbadf00d : I have added the case $s=t$ as well. | |
| Sep 18, 2019 at 16:49 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 18, 2019 at 16:40 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 18, 2019 at 16:26 | comment | added | 0xbadf00d | That's great, thank you. What can we do in the diagonal case $s=t$? | |
| Sep 18, 2019 at 16:04 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 18, 2019 at 16:00 | comment | added | Iosif Pinelis | @0xbadf00d : I have added the calculation of the generalized gradient in the case when $\tau$ does not have atoms at $s$ or $t$. | |
| Sep 18, 2019 at 15:58 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 18, 2019 at 15:36 | comment | added | 0xbadf00d | I see, thanks. It's my fault that I didn't check whether $f$ at least satisfies the necessary conditions. However, I'm still interested in the question under the additional assumption you provided. | |
| Sep 18, 2019 at 15:33 | comment | added | Iosif Pinelis | @0xbadf00d : Yes, the necessary and sufficient condition for the evaluation map at $s$ to be Lipshcitz is that $\tau(\{s\})>0$. | |
| Sep 18, 2019 at 15:29 | comment | added | 0xbadf00d | Thank you for your answer. Are there useful sufficient conditions on $\tau$ ensuring that the evaluation maps are locally Lipschitz (or even continuously differentiable)? | |
| Sep 18, 2019 at 15:15 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |