I suppose we could do that if we stated the problem a bit differently:
I have succeeded in proving from first principles thatfor such a function, which may, or may not, be an intermediate result that is required.
By the way, last night, while thinking in my sleep (yes, thinking in my sleep ) about uniform continuity of real-valued functions, I remembered this thread, and then I conjectured if a function whose first and second derivatives are positive cannot be uniformly continuous. I might come up with a proof soon.
Since f'(x) is always +ve, f is a monotone increasing function.
Similarly f"(x) is a monotone increasing function.
Now, it is easy to prove that f(x+c)-f(x)=c f'(x+t), where 0<t<c, for any c and x.
We have f'(a)> (f"(a)-ε)*k+f'(a-k)=m for a point a in R.
Then, when x>a, f(x+nc)>ncm+f(x), which is unbounded above. Hence proved.
Jane Fairfax, consider the function f(x)=x^3 which is differentiable at every point of R. f'(x) is positive and differentiable at every point and so is f"(x), but f'(x) does not tend to 0 as x tends to infinity.