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Functional analysis

  1. Jan 27, 2008 #1
    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
     
  2. jcsd
  3. Jan 27, 2008 #2

    EnumaElish

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    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.
     
  4. Jan 27, 2008 #3
    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.
     
  5. Jan 27, 2008 #4
    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

    instead
     
  6. Jan 28, 2008 #5
    Thank you so so much for your prompt reply. You've really helped me out and I really appreciate your time.

    Thanks again
     
  7. Jan 28, 2008 #6
    no problem, as long as you learn something from an not just copy it.
     
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