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Linear Algebra - set of piecewise continuous functions is a vector space

  1. Jan 25, 2013 #1
    1. The problem statement, all variables and given/known data
    A function f:[a,b] [itex]\rightarrow[/itex] ℝ is called piecewise continuous if there exists a finite number of points a = x0 < x1 < x2 < ... < xk-1 < xk = b such that
    (a) f is continuous on (xi-1, xi) for i = 0, 1, 2, ..., k
    (b) the one sided limits exist as finite numbers

    Let V be the set of all piecewise continuous functions on [a, b]. Prove that V is a vector space over ℝ, with addition and scalar multiplication defined as usual for functions.


    2. Relevant equations
    The axioms for fields and vector spaces.


    3. The attempt at a solution
    If function addition and scalar multiplication work in the same way as usual, don't I just have to make an argument that the limits still exist as finite numbers (since our scalars are finite)? And a similar argument that if I add two functions that those limits will just be the sum of the two limits of each function?
     
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  3. Jan 25, 2013 #2

    jbunniii

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    Suppose [itex]f, g\in V[/itex], and let [itex]h = f + g[/itex]. If [itex]f[/itex] and [itex]g[/itex] are both continuous at a point [itex]x[/itex], what can you say about [itex]h[/itex] at that point? Is it continuous at [itex]x[/itex]?
     
  4. Jan 25, 2013 #3
    It will also be continuous because the sum of two continuous functions will also be continuous. But do I need extra arguments besides that because V is the set of piecewise continuous functions?
     
  5. Jan 25, 2013 #4

    jbunniii

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    Yes, but we have learned something already. [itex]h[/itex] will be continuous at any point where both [itex]f[/itex] and [itex]g[/itex] are continuous, so [itex]h[/itex] can only be discontinuous at points where [itex]f[/itex] or [itex]g[/itex] are discontinuous, and there are only finitely many such points. Thus [itex]h[/itex] can have only finitely many discontinuities. So [itex]h[/itex] satisfies condition (a).

    Now consider condition (b). If [itex]h[/itex] is discontinuous at [itex]x[/itex], what are [itex]\lim_{y \rightarrow x^+} h(x)[/itex] and [itex]\lim_{y \rightarrow x^-} h(x)[/itex]? Can you write these in terms of the corresponding one-sided limits of [itex]f[/itex] and [itex]g[/itex]?
     
  6. Jan 25, 2013 #5
    Would that not be just the sum of the two limits of f and g at x, or if one or both of those two functions is discontinuous at x, then I would just use the whatever sided-limit that I need?
     
  7. Jan 25, 2013 #6

    jbunniii

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    Right, you have [itex]\lim_{y \rightarrow x^+} h(y) = \lim_{y \rightarrow x^+} f(y) + \lim_{y \rightarrow x^+} g(y)[/itex], and similarly with the left hand limit. Since these limits exist for [itex]f[/itex] and [itex]g[/itex], they also exist for [itex]h[/itex]. This shows that [itex]h[/itex] satisfies (b).
     
  8. Jan 25, 2013 #7

    jbunniii

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    So that shows that [itex]V[/itex] is closed under addition. What else do you need to show in order to prove that [itex]V[/itex] is a vector space?
     
  9. Jan 25, 2013 #8
    Closure under scalar multiplication.
    Commutative/Associative for addition/multiplication.
    Additive identity/inverse.
    Multiplicative identity.
    Distributive property.
    Associative property for scalar multiplication.
     
  10. Jan 25, 2013 #9

    jbunniii

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    OK, these three are true for any functions so they remain true for piecewise continuous functions.
    Which functions do these correspond to? Are they piecewise continuous?
    If [itex]f \in V[/itex] and [itex]c[/itex] is a scalar, is [itex]cf[/itex] piecewise continuous? Should be pretty easy to show that it is.
     
  11. Jan 25, 2013 #10
    For the 2nd item you listed, would the additive inverse of f be -f and the additive identity of f be the 0 function that is discontinuous where ever f is discontinuous?

    As for the 3rd item, I'm not entirely sure how I would go about showing that. It makes sense intuitively as to why it would be true, but I'm not sure how to put that on paper.
     
  12. Jan 25, 2013 #11

    jbunniii

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    Yes. Note that [itex]-f[/itex] is discontinuous at exactly the same points as [itex]f[/itex], and the left and right limits exist because they exist for [itex]f[/itex]. (Insert easy proofs as needed.)
    The additive identity of [itex]V[/itex] would be just the standard 0 function, which is continuous everywhere. Note that all continuous functions are in [itex]V[/itex] because the number of discontinuities (zero) is certainly finite, and the left and right limits exist due to continuity.
    Claim: [itex]cf[/itex] is continuous at any point [itex]x[/itex] where [itex]f[/itex] is continuous. (Proof?) Therefore [itex]cf[/itex] has at most finitely many discontinuities. What about [itex]\lim_{y \rightarrow x^+} cf(y)[/itex]? What does this equal?
     
  13. Jan 25, 2013 #12
    Since that is just a constant (scalar in our case), it would equal c times whatever the limit of just f(x) was.
     
  14. Jan 25, 2013 #13

    jbunniii

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    Right. Expressing it formally:
    $$\lim_{y \rightarrow x^+} cf(y) = c \lim_{y \rightarrow x^+}f(y)$$
    by the linearity property of limits, and the right hand side exists because [itex]f \in V[/itex].
     
  15. Jan 25, 2013 #14
    You have been more than helpful! I feel like I have a much greater understanding of this proof now!
     
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