Proving Continuous Function F: Cb(R) -> Cb(R)

[Ed Quanta]

Cb is set of all complex valued bounded functions
R is set of Real numbers

Define F:Cb(R)->Cb(R) by F(f)=f^2 for all 'points' f is an element of Cb(R). Prove that F is continuous.


Can someone give me some guidance on how to get started with this one?

 

[Hurkyl]

You should probably go straight back to the definition of a limit... but first you need to figure out the space you're working over. What sort of topology does Cb(R) have?

 

[M98Ranger]

Sure! To prove that F is continuous, we need to show that for any bounded function f in Cb(R) and any real number ε>0, there exists a δ>0 such that for all g in Cb(R) with ||f-g||<δ, we have ||F(f)-F(g)||<ε.

To begin, let's choose a specific f in Cb(R) and ε>0. Since f is bounded, there exists a constant M>0 such that |f(x)|<M for all x in R. Now, let's define δ=√ε/M. This choice of δ will ensure that if ||f-g||<δ, then ||f(x)-g(x)||<√ε for all x in R.

Next, we need to show that if ||f-g||<δ, then ||F(f)-F(g)||<ε. Let's start by expanding the expression for ||F(f)-F(g)||:

||F(f)-F(g)||=||f^2-g^2||=|(f+g)(f-g)|

Using the triangle inequality, we can split this into two terms:

|(f+g)(f-g)|≤|f+g||f-g|

Since we have chosen δ=√ε/M, we know that ||f-g||<√ε and therefore |f-g|<√ε. Also, since f and g are both bounded by M, we know that |f+g|<2M. Putting these together, we get:

|(f+g)(f-g)|<2M√ε

But we still need to show that this is less than ε. To do this, we can use the fact that f and g are bounded functions to rewrite the expression as:

2M√ε=2M√ε√(1/M^2)=2√ε/M

Since we chose δ=√ε/M, we know that ||f-g||<δ and therefore |f-g|<√ε/M. Substituting this into the above expression, we get:

2√ε/M=2√ε/√ε/M=2M

Therefore, we have shown that if ||f-g||<δ, then ||F(f)-F(g)||<ε, which proves that F is continuous.