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Defining of function in equation

  1. Jan 4, 2012 #1

    [tex]K(x,y)→[a,b]\times [a,b]→ℝ[/tex]

    Is [tex][a,b]\times [a,b][/tex] Deckart product? Is that the way to construct [tex]ℝ^2[/tex] space?

    If I say [tex]f\in C([a,b])[/tex], [tex]K\in C([a,b]\times [a,b])[/tex] that means that [tex]f[/tex] and [tex]K[/tex] are differentiable on this intervals. Right?
  2. jcsd
  3. Jan 4, 2012 #2


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    Yes, [itex][a, b]\times [a, b][/itex] is the "Cartesian product" (named for DesCartes so what you mean by "Dekart product"), the set of all ordered pairs of numbers from the interval [a, b].

    However, C([a, b]) is NOT the set of differentiable functions. It means simply functions that are continuous on [a, b], not necessarily differentiable. C1([a, b]) is the set of functions that are at least once differentiable on [a, b].
  4. Jan 4, 2012 #3
    Tnx for the answer.
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