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Continuous Functions, Closed Sets

  1. Apr 8, 2008 #1
    1. The problem statement, all variables and given/known data

    A mapping f from a metric space X to another metric space Y is continuous if and only if [itex]f^{-1}(V)[/itex] is closed (open) for every closed (open) V in Y.

    Use this and the metric space (X,d), where X=C[0,1] (continuous functions on the interval [0,1]) with the metric [itex]d(f,g)=\sup _{x\in[0,1]}|f(x)-g(x)|[/itex] to show that the set
    [tex]A=\left\{f\in C[0,1]:\int_0^1 tf^2(t)dt\ge 3\right\}[/tex]

    is closed.

    2. Relevant equations
    None other than the theorem given.

    3. The attempt at a solution

    My first step is to define a function from C[0,1] to R by choosing the easiest function, [itex]F(f)=\int_0^1 tf^2(t)dt[/itex]. If I can show that this is continuous, then it follows that [itex]\left\{f\in C[0,1]:\int_0^1 t f^2(t)dt\ge 3\right\}[/itex] is closed since [itex][3,\infty)[/itex] is closed. That is, A is closed.

    So, starting from the definition of a continuous function, I have:
    [tex]|F(f)-F(g)|=\left|\int_0^1 t\left(f^2(t)-g^2(t)\right)\right|dt \le \int_0^1\left|t\left(f^2(t)-g^2(t)\right)\right|dt \le \int_0^1 |f^2(t)-g^2(t)|dt=\int_0^1 |f(t)-g(t)||f(t)+g(t)|dt[/tex]
    [tex]\le \int_0^1 d(f,g)|f(t)+g(t)|dt=d(f,g)\int_0^1 |f(t)+g(t)|dt<\epsilon[/tex].

    My trouble is, I can't get rid of the last part of the integral that has |f+g| without restricting the domain of F. Any suggestions? Also, is my reasoning here fine? Thanks in advance!
  2. jcsd
  3. Apr 8, 2008 #2


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    This may be easier to see using sequences. A useful thing to keep in mind is that convergence w.r.t. d is uniform convergence. Now, if f_n -> f uniformly, can we say that |f_n| -> |f| uniformly, too? How can this help us here? (Note: We also have that f_n+f -> 2f uniformly, since d(f_n+f,2f)=d(f_n,f)->0.)
  4. Apr 9, 2008 #3
    Okay. Sorry if I seem slow here, but I'm trying to restate what you're saying so that I understand it.

    Reinterpreting what I've written with a sequence instead of g, I have

    |F(f)-F(f_n)|=\left|\int_0^1 t\left(f^2(t)-f_n^2(t)\right)\right|dt \le \int_0^1\left|t\left(f^2(t)-f_n^2(t)\right)\right|dt \le \int_0^1 |f^2(t)-f_n^2(t)|dt=\int_0^1 |f(t)-f_n(t)||f(t)+f_n(t)|dt
    \le \int_0^1 d(f,f_n)|f(t)+f_n(t)|dt=d(f,f_n)\int_0^1 |f(t)+f_n(t)|dt<\epsilon

    Now, [itex]\int_0^1 |f(t)+f_n(t)| dt[/itex] goes to a finite number, so I can pick [itex]\delta=\frac{\epsilon}{\int_0^1 |2(f)| dt}[/itex]. Thus F is uniformly continuous (and hence continuous), and so I can apply the theorem. So A is closed.

    This doesn't totally make sense to me, so if what I've done doesn't make sense then that's why.

    Thanks again!
  5. Apr 9, 2008 #4


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    Since the sequence of real numbers [itex]\{\int_0^1 |f(t)+f_n(t)| dt\}_n[/itex] converges, it's bounded by some M>0.

    Hence, given [itex]\epsilon > 0[/itex], we can find an N such that if n>N then [itex]d(f,f_n) < \epsilon/M[/itex], and consequently, [itex]|F(f) - F(f_n)| \leq d(f,f_n) \cdot M < \epsilon[/itex]. In other words, F(f) -> F(f_n).
  6. Apr 9, 2008 #5
    Okay, that makes more sense now. I see where you're pulling things from. Thank you!
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