In case of non-self mapping it is possible and the required point is such that d(x,fx)=d(A,B) if f:A\to B.
If the mapping f has a fixed point then the minimum distance is zero*. Therefore you assumed f to be a fixed point free. We are looking for the nearest point of the curve of f to the line f(x)=x. Here the distance between the curve and the line f(x)=x gives the minimum distance (say b). Now, by solving the equation f(x)-x=b we can get the required points.
*Observe that the infimum could be zero while the mapping f has no fixed points.
The first derivative will give you the rate of change (slope) and the second derivative will give you the curvature (maximums and minimums) on your metric space. If the second derivative f” is positive , then the function f is concave up (looks like a U shape) and your point x is a minimum. In a GIS system, curvature can calculate the second derivative of the whole surface (the slope of the slope), that is, whether a given part of a surface is convex or concave. This will give you local minimums from which you can select the domain minimum.
If the mapping f has a fixed point then the minimum distance is zero*. Therefore you assumed f to be a fixed point free. We are looking for the nearest point of the curve of f to the line f(x)=x. Here the distance between the curve and the line f(x)=x gives the minimum distance (say b). Now, by solving the equation f(x)-x=b we can get the required points.
*Observe that the infimum could be zero while the mapping f has no fixed points.
Yes, I also agree with Professor Rqeeb Gubran that "the minimum could be zero while the mapping f has no fixed points" and it is very closely related to
the topic of AFPP(approximate fixed point property). For instant, consider f(x)=x+\fract{1}{2x+1}, for all x\in [0, \infty).
In the answer given by Prof Rqeeb Gubran, if b=0 at some x, then we have f(x)=x. That means x is a fixed point for f.
d(x,fx) is minimum if its second derivative is positive at a point at which first derivative is zero.
The answers of Prof. Pavon and Prof. Tomar are correct in case the map is having second derivative which is very strong condition...
No smoothness condition is assumed in the question statement. In this case the answer is no. Take a contraction T on a complete metric space (X, d). Then T has a unique fixed point x. Now, take another self function of X coinciding with T on the whole X except on x in which it takes an arbitrary y different from x. Then S has no fixed point and the minimum of d(S(t), t) is 0. However 0 is not attained in any point.
What is t? How the distance becomes 0 at any point if you exclude the unique point which make this distance 0 (namely: the fixed point).
The distance d(t, S(t)) is never 0, but the minimum of all distances d(t, S(t)), t in X, is 0.
Perhaps I have misunderstood your question. In the example I have given, there is no x such that d(x,f(x))
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