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# 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

### 1 Answer

- ?Lv 51 decade agoFavorite Answer
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".