# 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 Statement

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

## 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!

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

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)|?

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