Can you find f,g st sup{|f-g|}=0 but f is not equal to g

[happybear]

Homework Statement

For all real number x, can you find a function f and g such that
sup|f(x)-g(x)|=0

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

Homework Statement

For all real number x, can you find a function f and g such that
sup|f(x)-g(x)|=0

Homework Equations

The Attempt at a Solution

What have you done on this problem yourself? In particular, what does sup |f(x)- g(x)| mean? And if you are going to put "f is not equal to g" please put it in the statement of the problem!

 

[happybear]

What have you done on this problem yourself? In particular, what does sup |f(x)- g(x)| mean? And if you are going to put "f is not equal to g" please put it in the statement of the problem!

sup{|f-g|} means that the maximum of the value. I try to find a function f approaching g, but this eem not to be the case

 

[HallsofIvy]

sup{|f-g|} means that the maximum of the value. I try to find a function f approaching g, but this eem not to be the case

Strictly speaking, sup does NOT mean 'maximum', it means "least upper bound" which may or may not be a maximum. What do you mean "f approaching g"? As x approaches what value? This has to be true over all x.

Suppose f were NOT equal to g. Then there must exist some x such that $f(x)\ne g(x)$. Let M= |f(x)- g(x)| for that x. Now, what can you say about sup|f(x)- g(x)|?

 

[happybear]

so does that mean that no matter whether X is compact or not, there is no such a function?