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.
