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Scharze space asked in 科學數學 · 1 decade ago

分析(連續函數性質)

Let D be the dense subset of C[0,1] Prove that for all f€C[0,1] there exists an increasing sequence of element of D

that converges uniformly to f

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    1 decade ago
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    I suppose that the topology of C[0,1] is given by the sup norm, since you have not stated it.

    Since D is dense in C[0,1], and f is in C[0,1], there is a sequence {g_n} in D that converges to f in C[0,1]. Since the norm on C[0,1] is the sup norm, this means that gn converges to f uniformly.

    Let hn = inf(k => n) gk. Note that hn is increasing.

    We want to prove that hn is continuous and hn converges to f uniformly.

    We first prove uniform continuity. For ε>0, there is N such that

    |f(x)-gn(x)|<ε for all n>N and x in [0,1]. ie

    f(x)-ε < gn(x) < f(x)+ε for all n>N and x in [0,1].

    Then for n>N and x in [0,1], we have

    hn(x) =< gn(x) < f(x)+ε, and

    hn(x)=inf(k => n) gk =>inf(k > N) gk => f(x)-ε.

    Thus, |f(x)-hn(x)| =< εfor n>N and x in [0,1]. This proved that hn converges to f uniformly.

    Next we prove hn is continuous. From above, we have for ε>0, there is N such that

    |f(x)-hn(x)|<ε for all n>N and x in [0,1].

    Let x in [0,1]. Since f is continuous, there is δ>0 such that

    |f(y)-f(x)|<ε for |y-x|<δ.

    Then for n>N and for all y with |y-x|<δ, we have

    |hn(y)-hn(x)|

    =< |hn(y)-f(y)| + |f(y)-f(x)| + |f(x)-hn(x)|

    < 3ε.

    This shows that hn is continuous for all n>N.

    Indeed this is enough because we can then consider the sequence

    {hN+1, hN+2, ....} which is increasing and converges uniformly to f.

    But in fact hn is also continuous for n=<N, because then we have

    hn = inf{gn, gn+1, .... , gN, hN+1},

    and use the fact that the inf of finitely many continuous functions is also continuous.

    2009-04-29 16:16:52 補充:

    I made a mistake in the last part when proving hn is continuous. The correction is as follows.

    Fix n and fix x in [0,1].

    From above, we have for ε>0, there is N such that

    |f(x)-hm(x)|<ε/3 for all m>N and x in [0,1].

    2009-04-29 16:16:58 補充:

    For the fixed x, since f is continuous, there is δ'>0 such that

    |f(y)-f(x)|<ε/3 for |y-x|<δ'.

    Then for m>N and for all y with |y-x|<δ', we have

    |hm(y)-hm(x)| =< |hm(y)-f(y)| + |f(y)-f(x)| + |f(x)-hm(x)|< ε.

    In particular, |hN+1(y)-hN+1(x)| <εfor |y-x|<δ'.

    2009-04-29 16:17:19 補充:

    Now consider gn, gn+1, ..., gN. Note that we have

    hn = min{gn, gn+1, ... , gN, hN+1}

    For each k between n and N, since gk is continuous, there is δk such that

    |gk(y)-gk(x)|<ε for |y-x|<δk.

    2009-04-29 16:17:30 補充:

    Let δ= min{δ', δn, δn+1, ... , δN}, then when |y-x|<δ,

    hn(y) = min{gn(y), gn+1(y), ... , gN(y), hN+1(y)}

    < min{gn(x)+ε, gn+1(x)+ε, ... , gN(x)+ε, hN+1(x)+ε}

    = min{gn(x), gn+1(x), ... , gN(x), hN+1(x)}+ε

    =hn(x)+ε

    2009-04-29 16:17:40 補充:

    Also

    hn(y) = min{gn(y), gn+1(y), ... , gN(y), hN+1(y)}

    > min{gn(x)-ε, gn+1(x)-ε, ... , gN(x)-ε, hN+1(x)-ε}

    = min{gn(x), gn+1(x), ... , gN(x), hN+1(x)}-ε

    =hn(x)-ε

    Thus |hn(y)-hn(x)| < ε for |y-x|<δ. This shows that hn is continuous.

    2009-04-30 11:42:06 補充:

    One more correction. In the fifth paragraph, I typed "uniform continuity", but what I mean is "uniform convergence".

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