> To be clear though, it does not imply that NP is subexponential

Is there a subset of NP problems for which a subexponential algorithm would witness the entire NP class being subexponential?

~~Yeah, that's NP-complete problems.~~

EDIT: DISREGARD THAT I WAS MISTAKEN

	▲	csheehan10 4 hours ago \| parent \| next [-]
		It's NP-complete problems if "subexponential" means O(2^{n^{\epsilon}}) for every \epsilon>0. If subexponential just means 2^{o(n)}, (as it is here, O(2^{n/\sqrt{\log n}})), then no. E.g. if the cost of the reduction took you from n to \sqrt{\log n}n, then you'd be back to exponential time. The Subexponential Reduction Family is the class of problems that preserve subexponential time in the second sense.
	▲	LPisGood 4 hours ago \| parent \| prev [-]
		That is unfortunately not the case