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Metric Spaces

  1. Dec 19, 2011 #1
    ih87l0.jpg

    The sequential characterisation of continuity says that [itex]f[/itex] is continuous at [itex]x_0[/itex] if and only if for every sequence [itex](x_n)_{n\in\mathbb{N}}[/itex] in [itex]X[/itex], [itex]f(x_n)\to f(x_0)[/itex] as [itex]x_n \to x_0[/itex]. [itex]f[/itex] is continuous on [itex]X[/itex] if this is the case for all [itex]x_0 \in X[/itex].

    I think I've done all the parts of this question up to the last 2 parts.

    For part (b) is this right:

    Suppose [itex](x_n)_{n\in\mathbb{N}}[/itex] is a sequence in [itex]X[/itex] with [itex]x_n \to x\in X[/itex]. Then for all [itex]x\in X[/itex]: [tex]f(x_n) = (f_1(x_n) , f_2(x_n) , ... , f_N (x_n)) \to (f_1(x) , f_2(x) , ... , f_N (x) ) = f(x)[/tex] since all the [itex]f_i[/itex] are continuous.

    (This is also using a theorem which says that if [itex](x^{(n)})_{n\in\mathbb{N}}[/itex] is a sequence of vectors in [itex]\mathbb{R}^N[/itex] then [itex]x^{(n)} \to x\in\mathbb{R}^N[/itex] in the Euclidean metric [itex]\iff x_j^{(n)} \to x_j[/itex] for each [itex]1\leqslant j \leqslant N[/itex] in the standard metric on [itex]\mathbb{R}[/itex].)

    How would you show in the last 2 parts that [itex]F[/itex] and [itex]H[/itex] are continuous?
     
    Last edited: Dec 19, 2011
  2. jcsd
  3. Dec 20, 2011 #2
    EDIT: I see that [itex]F = \phi \circ f[/itex] and we've already shown that the composition of 2 continuous functions is continuous.

    What function can I compose [itex]F[/itex] with to turn the product in [itex]H[/itex] into a sum?
     
  4. Dec 21, 2011 #3
    Well, to turn a sum into a product we could use ##e^x## since ##e^{x+y} = e^x e^y ##. So what might you use to do the opposite?
     
  5. Dec 22, 2011 #4
    Log!
     
    Last edited: Dec 22, 2011
  6. Dec 22, 2011 #5
    :smile: Right! And now this explains why the functions must be strictly positive for the last part.
     
  7. Dec 22, 2011 #6
    If [itex]h : (X,d_X) \to \mathbb{R}^N[/itex] is defined by [itex]h(x) = (h_1(x) , h_2(x) , ... , h_N(x) )[/itex] then [itex]h[/itex] is continuous by part (b).

    [itex]\displaystyle \log \circ H (x) = \log(H(x)) = \log \left( \prod_{j=1}^{N} h_j(x)^{a_j} \right)= \sum_{j=1}^N \log ( h_j(x)^{a_j} )= \sum_{j=1}^N a_j \log(h_j (x)) = \phi \circ (\log \circ h (x))[/itex]

    [itex]\phi , \log , h[/itex] are all continuous so their composition is continuous, so [itex]\log \circ H[/itex] is continuous.

    We know if [itex]f[/itex] and [itex]g[/itex] are 2 continuous functions then [itex]g \circ f[/itex] is continuous but [itex]g \circ f[/itex] continuous [itex]\not\Rightarrow f, g[/itex] are continuous so how do I frame the argument to show that H is continuous?
     
  8. Dec 22, 2011 #7

    Deveno

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    what continuous function can you compose with [itex]log \circ H[/itex] to recover H?
     
  9. Dec 22, 2011 #8
    [itex]\exp \circ ( \log \circ (H(x) ) = H(x) = \exp \circ (\phi \circ (\log \circ h(x)) )[/itex]

    [itex]\exp , \phi , \log , h[/itex] are all continuous so their compositions are all continuous, so H is continuous.
     
  10. Dec 22, 2011 #9

    Deveno

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    i'll buy that.
     
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