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Just want to add a cautionary note to Iosif Pinelis' answer. The convexity of $n$ is sufficient but not necessary for the super-additive condition $n(r)+n(s)\le n(r+s)$. An example where the necessity breaks down is $$n(t)=t(1-e^{-t}).$$$$n(t)=-te^{-t},$$ or $n(t)=t(1-e^{-t})$ if we need $n$ to be nonnegative.

Just want to add a cautionary note to Iosif Pinelis' answer. The convexity of $n$ is sufficient but not necessary for the super-additive condition $n(r)+n(s)\le n(r+s)$. An example where the necessity breaks down is $$n(t)=t(1-e^{-t}).$$

Just want to add a cautionary note to Iosif Pinelis' answer. The convexity of $n$ is sufficient but not necessary for the super-additive condition $n(r)+n(s)\le n(r+s)$. An example where the necessity breaks down is $$n(t)=-te^{-t},$$ or $n(t)=t(1-e^{-t})$ if we need $n$ to be nonnegative.

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Just want to add a cautionary note to Iosif Pinelis' answer. The convexity of $n$ being convex is sufficient but not necessary for the super-additive condition $n(r)+n(s)\le n(r+s)$. An example where the necessity breaks down is $$n(t)=t(1-e^{-t}).$$

Just want to add a cautionary note to Iosif Pinelis' answer. The convexity of $n$ being convex is sufficient but not necessary for the super-additive condition $n(r)+n(s)\le n(r+s)$. An example where the necessity breaks down is $$n(t)=t(1-e^{-t}).$$

Just want to add a cautionary note to Iosif Pinelis' answer. The convexity of $n$ is sufficient but not necessary for the super-additive condition $n(r)+n(s)\le n(r+s)$. An example where the necessity breaks down is $$n(t)=t(1-e^{-t}).$$

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Hans
  • 2.2k
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  • 29

Just want to add a cautionary note to Iosif Pinelis' answer. The convexity of $n$ being convex is sufficient but not necessary for the super-additive condition $n(r)+n(s)\le n(r+s)$. An example where the necessity breaks down is $$n(t)=t(1-e^{-t}).$$