# Functional analysis

Hello
Would anyone out there be able to help me with a problem I'm having? I have to prove that a function is open and that another is closed. The question is:

Consider C [0,1] with the sup metric. Let f:[0,1]→R be the function given by f(x)=x²+2
Let A={g Є C[0,1]: d(g,f) > 3}. Prove that A is an open set
Let B={g Є C[0,1]: 1 ≤ d(g,f) ≤ 3}. Prove that B is a closed set

I'm new to all of this and just don't know what to do even with the f(x)=x²+2 part so if anyone out there can shed some light, I'd be really grateful!

Thanks

EnumaElish
Homework Helper
A is the all functions that belong to C[0,1] and whose distance to f is at least 3. That's where f(x) should come in.

Ok, to prove that A is open you need to show that given a arbitrary g in A, you can find a ball around g contained in A.

First note that we are considering continous functions from a closed interval, so the supremum always exist.

Now let g be some function in A, that is

3 < d(g,f) = sup|g-f| = b

b is simply the real number we already have noted exist, that is different from infinity.

now define a = b - 3, note this is greater than zero. Then define the open ball around g

B_a(g) = {h in C[0,1] | d(g,h) < a}

we now show that this is contained in A. So for h in B_a(g), because d is a metric it satisfy the triangle inequality, we have

d(h,f) => |d(g,h) - d(g,f)| > |a - b| = |b-3-b| = 3

so B_a(g) is contained in A, this proves that A is open.

To show that B is closed, first in the same way for A show that

C = {g in C[0,1] | d(g,f) < 1}

is open.

then A intersect C is open, because the intersection of two open sets are open. We also know that the complement of an open set is closed, so by noting that

B = Compliment of (A intersect C)

you are done. It have been some time before i did this kind of thing so someone might need to check that I done it right. As you see you don't need to know an explicit formula for f(x), I guees the only reason you got that was to get a fealing of what the sup norm looks like.

a little correction to the line

d(h,f) => |d(g,h) - d(g,f)| > |a - b| = |b-3-b| = 3

write

d(h,f) => |d(g,h) - d(g,f)| > |a - b| = |-3| = 3

a little correction to the line

d(h,f) => |d(g,h) - d(g,f)| > |a - b| = |b-3-b| = 3

write

d(h,f) => |d(g,h) - d(g,f)| > |a - b| = |-3| = 3