Equicontinuity of f_y in C([0,1]): Uniformly Equicontinuous Math Problem

[e12514]

Let f: [0,1] x [0,1] -> R be continuous and define f_y in C([0,1]) by f_y (x) = f(x,y) for each y in [0,1].

By definition of continuity of f, for all epsilon > 0 there exists delta(depending on epsilon) > 0 such that |(a,b) - (c,d)| < delta => | f((a,b) - f((c,d)) | < epsilon

If I fix b = d = y in [0,1] then I will get a statement (after a few lines)
for all epsilon > 0 there exists delta(depending on epsilon) > 0 such that, for all f_y in the set {f_y : y in [0,1]}, for all a,c in [0,1],
|(a,y) - (c,y)| < delta => | f_y (a) - f_y (c) | < epsilon

this is the definition for the set {f_y : y in [0,1]} to be uniformly equicontinuous.




Is the above argument valid?

 

[Aaron Barnard]

Yes, the argument is valid.

 

[tacman]

Yes, the above argument is valid. It shows that for any fixed value of y in [0,1], the set of functions {f_y : y in [0,1]} is uniformly equicontinuous. This means that for any given epsilon, there exists a delta that works for all functions in the set, ensuring that the distance between any two points (a,y) and (c,y) in the domain [0,1] is small enough to guarantee that the corresponding function values f_y(a) and f_y(c) are also close. This is the definition of uniform equicontinuity, and the argument above shows that it holds for all y in [0,1], thus proving the equicontinuity of f_y in C([0,1]).