Real Analysis Proof: Suppose f:[a,b] - R and define g:[a,b] - R as follows:...

[maxwell]

From: Introduction To Analysis, fifth edition, by Edward Gaughan (1998), Exercise 2.24.

Suppose f:[a,b] - R and define g:[a,b] - R as follows: g(x)= sup { f(t): a<= t <= x }.
Prove that g has a limit at x_o if f has a limit at x_o and lim_t-xof(t) = f(x_o).

In the course of trying to work out the proof of this problem I considered the following scenario:

Let a = 0, b=1. Let x₀=3\4 and let f(t) = (2t-1)^-1 for all t in [0,1/2) U (1/2,1], while f(t)=0 if t = 1/2.

Then lim _{t - xo} f(t) = 2 but g(x) is unbounded on [a,b] = [0,1] (because of the singularity at t = 1/2). Then f(t) has a limit at t= x_o but g is not defined. Therefore the statement to be proved appears to be false. Where is my logical error? Thanks for your help.

Max.

 

[aj111000]

From: Introduction To Analysis, fifth edition, by Edward Gaughan (1998), Exercise 2.24.

Suppose f:[a,b] - R and define g:[a,b] - R as follows: g(x)= sup { f(t): a<= t <= x }.
Prove that g has a limit at x_o if f has a limit at x_o and lim_t-xof(t) = f(x_o).

In the course of trying to work out the proof of this problem I considered the following scenario:

Let a = 0, b=1. Let x₀=3\4 and let f(t) = (2t-1)^-1 for all t in [0,1/2) U (1/2,1], while f(t)=0 if t = 1/2.

Then lim _{t - xo} f(t) = 2 but g(x) is unbounded on [a,b] = [0,1] (because of the singularity at t = 1/2). Then f(t) has a limit at t= x_o but g is not defined. Therefore the statement to be proved appears to be false. Where is my logical error? Thanks for your help.

Max.

I may be wrong, but I think it’s because your function doesn’t follow the rules of the proof. The proof involves two functions that mapped from the same continuous domain, specifically [a,b]. However, your function has a discontinuity in the middle, meaning it’s domain doesn’t have the same form as the proof describes. In fact, any function with a vertical asumptote would not meet the criteria of the problem. However, we could redefine the problem on only the domain of [1/2 + a bit, 1], where sup{f(t)}=f(1/2+a bit) on the whole domain and, therefore, would both follow the premise of the proof and it’s expected result

 

[maxwell]

Continity

Can a domain be continuous? A function can be continuous at a point in the domain (or it may be continuouson an interval such as [a,b]), but a closed interval of real numbers cannot, to my knowledge, be said to be continuous. I believe what you are referring to is the fact that any closed interval of real numbers has the cardinality of the continuum and so there are no "gaps" in the interval.
However, I agree that the question of continuity is at the heart of the problem here. The exercise, however, does not specify that f(t) must be continuous on [a,b]. The requirement is that f(t) be defined on each point of [a,b] and that it maps to some subset of the real numbers. As far as I can tell, the function f(t) that I described meets these requirements.
The function g(x) follows from f(t). For example g(1\2) = 0 since f(t) is bounded above by 0 for all 0 <= x <= 1/2, and g(1/2) = sup {f(t): 0 <= t <= 1\2} = 0 = f(1\2). The fact that f(t) is unbounded above once t passes 1/2 is where the problem lies because that is where g is undefined.

Thanks for your response!

 

[Dr.Peterson]

Can a domain be continuous? A function can be continuous at a point in the domain (or it may be continuouson an interval such as [a,b]), but a closed interval of real numbers cannot, to my knowledge, be said to be continuous. I believe what you are referring to is the fact that any closed interval of real numbers has the cardinality of the continuum and so there are no "gaps" in the interval.
However, I agree that the question of continuity is at the heart of the problem here. The exercise, however, does not specify that f(t) must be continuous on [a,b]. The requirement is that f(t) be defined on each point of [a,b] and that it maps to some subset of the real numbers. As far as I can tell, the function f(t) that I described meets these requirements.
The function g(x) follows from f(t). For example g(1\2) = 0 since f(t) is bounded above by 0 for all 0 <= x <= 1/2, and g(1/2) = sup {f(t): 0 <= t <= 1\2} = 0 = f(1\2). The fact that f(t) is unbounded above once t passes 1/2 is where the problem lies because that is where g is undefined.

Thanks for your response!

I think the problem with your example is just that g(x) is undefined for x>1/2, so it doesn't fit the requirement that its domain must be [0,1], since the theorem requires g:[a,b]->R.