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Homework Help: Sequences and continuous functions

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data
    a) Let {s[itex]_{n}[/itex]} and {t[itex]_{n}[/itex]} be two sequences converging to s and t. Suppose that s[itex]_{n}[/itex] < (1+[itex]\frac{1}{n}[/itex])t[itex]_{n}[/itex]

    Show that s [itex]\leq[/itex]t.

    b) Let f, g be continuous functions in the interval [a, b]. If f(x)>g(x) for all x[itex]\in[/itex][a, b], then show that there exists a positive real z>1 such that f(x)[itex]\geq[/itex]zg(x) for all x[itex]\in[/itex][a, b].

    2. Relevant equations

    3. The attempt at a solution

    Ok, so I've already done part a. I'm trying to figure out part b. I think my ideas are on the right track, but I'm looking for some help fleshing them out a bit more.

    Argue by contradiction. Suppose there does not exist z such that f(x)[itex]\geq[/itex]zg(x). I'm not sure exactly what this implies in terms of deriving a contradiction; some general guidance would be appreciated it. Also, I'm using the definition that the notion that f(x) is continuous at c is equivalent to the following:

    For every sequence {s[itex]_{n}[/itex]} in the domain of f converging to c, one has

    lim, n-->[itex]\infty[/itex] f(s[itex]_{n}[/itex])=f(c). I somehow want to construct an analogous argument to that in part a, but I'm not sure how the z is similar to the (1+[itex]\frac{1}){n}) term. Can anyone help with this?
  2. jcsd
  3. Feb 2, 2012 #2
    I think this can go easier, define h:

    [tex]f(x)-g(x)=h(x) [/tex]

    because f,g are continuous in [a,b], h is continuous in [a,b] but then h(x) has a minimum >0 in [a,b] call it c thus:

    [tex] f \geq g+c = g \left( 1+\frac{c}{g(x)} \right) [/tex]

    This holds for all x in [a,b] thus:

    [tex] f \geq g+c = g \left( 1+\frac{c}{ ||g(x)||_{\infty} } \right) = zg(x)[/tex]

    with [tex] ||g(x)||_{\infty} =sup_{x \in [a,b]} g(x) [/tex] and [tex] z= 1+\frac{c}{ ||g(x)||_{\infty} }[/tex]

    Something like this should work ( I think the part with the inf term should be replaced by something else).
    Last edited: Feb 2, 2012
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