Timeline for On increasing the penalty term in convex optimization with regularization
Current License: CC BY-SA 3.0
5 events
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| Aug 16, 2014 at 22:34 | history | edited | borntotry83 | CC BY-SA 3.0 |
improved the way of asking the question
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| Aug 13, 2014 at 2:38 | comment | added | borntotry83 | Nice ways to tackle this problem. The second route looked more promising, but doing similar steps to the ones you described, now with the spectral norm involved, and finding another inequality seems challenging. | |
| Aug 12, 2014 at 16:17 | comment | added | user35593 | other ansatz: As $f$ is convex there is Matrix $A$ and $c>0$ s.t. $$f(X) \geq f(X_1)+\langle A, X-X_1 \rangle + c \|X-X_1\|_F^2$$ where $\langle A,B \rangle:=trace(A^TB)$. Note that $\|A\|_F^2=\langle A,A \rangle$ and hence $$f(X)+\|X\|_F^2=f(X_1)+\|X_1\|_F^2+\langle A+2X_1,X-X_1\rangle+(1+c) \|X-X_1\|_F^2$$ and by the minimality of $X_1$ we get $A=-2X_1$. Now to prove that $f(X_2)\geq f(X_1)$ it would suffice by the first inequ. that $$\langle 2X_1,X_1-X_2\rangle+\|X-X_1\|_F^2\geq 0.$$ For this it would suffice to prove that $$\langle 2X_1,X_1-X_2\rangle\geq 0.$$ | |
| Aug 12, 2014 at 15:39 | comment | added | user35593 | An idea: define $$X_t=\text{argmin} f(X)+t ||X||_F^2+(1-t)n||X||_2^2.$$ and then try to prove that $t \mapsto f(X_t)$ is monoton. However I dont know if this works or not. | |
| Aug 12, 2014 at 12:17 | history | asked | borntotry83 | CC BY-SA 3.0 |