Exercise 10, handnotes, open function example

Hi,

On the third page of hand notes of exercise 10, there is an example of an open convex function \(f(x)\) and its conjugate of conjugate. I'm wondering how the conjugate of \(f\) is computed for \(y>0\) in this case.

I see that \(x^Ty\) is always below \(f(x)\) for positive \(y\), and never intersects with \(f(x)\), so the gap is always negative. But for values of \(x\) slightly smaller than zero, \(x^Ty - f(x)\) is very close but to equal to zero. So it's not clear to me how \(f^*(y)\) is computed here.

open_funct.jpg

Thank you in advance.

Top comment

Hello,

Contrary to the lecture notes, the handwritten notes use 'sup' instead of 'max'. This is exactly to take care of the issue you raise. Indeed, for \(f(x)=0 \text{ if } x < a\), if \(u > 0\), \(\max_x u x - f(x)\) is something like \( u(a - \epsilon)\) for any small positive \(\epsilon\). 'sup' resolves this limit, so the conjugate\(f(x)\) value would be \( ua \).

Does that help?

Hi,

Yes, that answers my question. Thank you for your help.

Hello,

Where do these handwritten notes come from ? I never saw them and cannot find them on github.

Thank you for your help

Thank you very much, I missed them since they were not uploaded in the first weeks :)

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