I was thinking about this and it makes sense to me for it to be true but I was wondering if their is anyway to prove this?
Posted Jan 21, 2015, 2:06 AM UTC
[deleted]
It's possible to prove that for continuous functions over the reals. One way to do it is to invoke the intermediate value theorem. (http://en.wikipedia.org/wiki/Intermediate_value_theorem)
Let's say we have x,y,z and a function f with the properties that:
x<y<z and f(x)<f(y) and f(z)<f(y)
Now let m be the maximum of f(x) and f(z). Since both f(x) and f(z) are strictly less than f(y), we know that m < f(y). Let n=(f(y)+m)/2.
Now we have that f(x)<n<f(y) and f(z)<n<f(y).
By the intermediate value theorem f takes the value n on the interval (x,y) and on the interval (y,z) since those intervals are disjoint, there are at least two points where f takes the value n, so f^-1 (n) can't be well-defined.
Assuming you mean a continuous function: For any function f with the property you described, it must be the case that there is some a and some b (with a != b) and some K such that f(a) = f(b) = K.
Then that would mean f^(-1)(K) = a, and f^(-1)(K) = b, hence f^(-1) is not a function.
If you allow for discontinuities, then you can have a function that is naturally split into an increasing interval and a decreasing interval which do not overlap in values for f(x), and would then have an inverse.