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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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    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. – 2017-01-21
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    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? – 2017-01-23
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    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. – 2017-01-23
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    Thanks for pointing out the direction, I will look for that. – 2017-01-23

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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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    Thanks for your answer. Is it possible to make the Hessian non-continuous everywhere? Not only on one point? – 2017-01-21
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    Sorry for misunderstanding the question (I was a bit unsure with your title) I think I'd agree with Michael Grant's comment above – 2017-01-22