2楼: Originally posted by hank612 at 2013-10-10 03:46:59
你能说说函数空间的度量是如何定义的吗? 还有, C 空间的定义是什么, 你想要比较的两个空间谁大谁小, 我怎么觉得 C交 L^{Infinity}更小呢?

3楼: Originally posted by 荞桥 at 2013-10-10 08:13:10
\omega是有界开集，C_0(\omega)是连续有紧支集的，C(\omega)是连续，L无穷\omega是有界，C_0当然是连续有界的...

4楼: Originally posted by hank612 at 2013-10-10 12:38:46
how about the definition of metric?

If use the Sup_x |f(x)-g(x)| for the distance, then the constant x=1 cannot be approximated by any finite-supported continuous function.

If use Integral( | ...

5楼: Originally posted by 荞桥 at 2013-10-11 17:35:29
范数你当然不能随便取了，C_0在L^p(1<=p<\infinite)里稠密，但是在L^{\infinite}里面不稠密，这是基本性质...

6楼: Originally posted by hank612 at 2013-10-11 22:51:25
When omega is a bounded open domain, continuous function set with compact support is dense. Otherwise, can you show a counter example ?...

7楼: Originally posted by hank612 at 2013-10-11 23:24:01
不好意思啊，又想当然了。L^Infinity范数和积分没什么关系，本质上就是连续函数空间那个点点收敛的范数，因此常数函数就没法被紧支的逼近了。
既然你早就知道这个结论了，那不就是求助的问题么？ 不明白。...

8楼: Originally posted by 荞桥 at 2013-10-12 09:33:49
不是这样的，C_0在L^p(1<=p<\infinite)里稠密，但是在L^{\infinite}里面不稠密，但是我的问题是C_0在（C交L无穷）里是否稠密怎么证明，我觉得常函数C可以用有紧支集的函数逼近，取一列有紧支集的函数f_k，在 ...