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Linear space

  1. Nov 18, 2011 #1
    why arent non continuous functions in an interval a linear space?
     
  2. jcsd
  3. Nov 18, 2011 #2

    CompuChip

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    Why don't you try to find a counterexample?

    Hint: try looking at closure under addition: can you find two discontinuous functions f and g such that f + g is continuous?
     
  4. Nov 18, 2011 #3
    why does f+g have to be continuous? I can see that even for discontinuous functions they are closed under addition and multiplication
     
  5. Nov 18, 2011 #4

    Deveno

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    suppose our interval is [0,1]. define f:[0,1]→R by:

    f(x) = -1, for 0 ≤ x < 1/2
    f(x) = 1, for 1/2 ≤ x ≤ 1.

    clearly, f is discontinuous (at 1/2).

    now define g:[0,1]→R by:

    g(x) = 1 for 0 ≤ x < 1/2
    gx) = -1, for 1/2 ≤ x ≤ 1.

    again, g(x) is discontinuous (at 1/2).

    but (f+g)(x) = 0, for all x in [0,1], and constant functions are continuous.
     
  6. Nov 18, 2011 #5

    HallsofIvy

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    In other words, "discontinuous functions" are NOT closed under addition. For a counter example to closure under multiplication, let f(x)= 1 if x is rational, -1 if x is irrational and let g(x)= -1 if x is rational, 1 if x is irrational.
     
  7. Nov 18, 2011 #6

    CompuChip

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    Because you wanted to show that the space of discontinuous is not linear. If it were linear, it would mean that f + g would be discontinuous if f and g are.
     
  8. Nov 18, 2011 #7
    If it were linear, it would mean that f + g would be discontinuous if f and g are. i dont understand this part
     
  9. Nov 18, 2011 #8

    micromass

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    By definition, a linear space V must satisfy: if x and y are in V, so is x+y.

    If V would be the set of discontinuous functions, then the above becomes: if x and y are discontinuous functions, so is x+y.

    The above example shows that this is false, hence the discontinuous functions do not form a linear space.
     
  10. Nov 18, 2011 #9
  11. Nov 19, 2011 #10

    HallsofIvy

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    Or, perhaps much more simply, every linear space must contain an additive identity. Since the addition here is ordinary addition of functions, the "additive identity" is the 0 function (f(x)= 0 for all x). That is a continuous function and so is NOT in the set of discontinuous functions.
     
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