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Can someone give some examples of convex functions with positive semi-definite Hessian, where the Hessian is non-continuous everywhere?

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    $\begingroup$ I don't have a sufficiently strong command of integral theory here. But I'm pretty sure the second fundamental theorem of calculus would say no; that a function discontinuous everywhere cannot be the derivative of a continuous function---and, therefore, it can't be the second derivative of one, either. Hence it can't be the Hessian of any function, convex or otherwise. $\endgroup$ – Michael Grant Jan 21 '17 at 5:25
  • $\begingroup$ Thanks for your answer. However, my understanding of the second fundamental theorem of calculus is that if f is continuous then f can be F's derivative. It does not imply directly "a function discontinuous everywhere cannot be the derivative of a continuous function". Do you have a reference for this statement? $\endgroup$ – Sisi Jan 23 '17 at 5:56
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    $\begingroup$ I do not, which is why it was a comment, not an answer ;-) However I'm reasonably sure that a Riemannian integrable function can only have a countable number of discontinuities. $\endgroup$ – Michael Grant Jan 23 '17 at 12:02
  • $\begingroup$ Thanks for pointing out the direction, I will look for that. $\endgroup$ – Sisi Jan 23 '17 at 18:26
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How about for example:

$$ f(x) = \mathbf 1_{x\le 0} {x^2\over 2} + \mathbf 1_{x\ge 0} x^2 $$

blue line is the function, green is gradient, red is hessian. I think it meets your requirements.

enter image description here

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  • $\begingroup$ Thanks for your answer. Is it possible to make the Hessian non-continuous everywhere? Not only on one point? $\endgroup$ – Sisi Jan 21 '17 at 2:46
  • $\begingroup$ Sorry for misunderstanding the question (I was a bit unsure with your title) I think I'd agree with Michael Grant's comment above $\endgroup$ – tibL Jan 22 '17 at 8:30

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