For a function to strictly decrease forever, at least one of two statements is true:

1. It decreases by vanishingly smaller amount.

2. It decreases to negative infinity.

So...which is it? Infinitesimal time units, or negative time?

Strictly speaking it either tends to negative infinity, or tends to so real number X, which could be any number positive or negative.

In that latter case, statement 1 is true.

3. Nobody said "strictly".

They said decreases, not stays constant

Plenty of O(n), O(nlogn), O(n^2) algorithms don't strictly increase.

Rounding the decrease to the nearest time unit some of the time is hardly a disqualifier for "decreasing". (Though specifically they said "faster".)

Yes, but...

The mathematically rigorous form of the posed question is whether there is an algorithm with run time f(N) for which there exists strictly decreasing g(N) such that f(N) = O(g(N)). That is, the run time is bounded by a forever decreasing function.

f(N) itself doesn't need to be strictly decreasing, but there needs to be a g(N) that is.

Bottom line, no algorithm can have a forever decreasing run time as N increases. An algorithm can exhibit that behavior over a limited range of N, but that's not how complexity works.

> The mathematically rigorous form of the posed question is whether there is an algorithm with run time f(N) for which there exists strictly decreasing g(N) such that f(N) = O(g(N)). That is, the run time is bounded by a forever decreasing function.

Let's say my minimum runtime is 7, and the strictly decreasing bound on my runtime is g(N) = 1e15/N + 7

Doesn't that fit your description fine?