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7Â¥: Originally posted by hank612 at 2013-10-11 23:24:01
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8Â¥2013-10-12 09:33:49
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hank612

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ÄãÄÜ˵˵º¯Êý¿Õ¼äµÄ¶ÈÁ¿ÊÇÈçºÎ¶¨ÒåµÄÂð? »¹ÓÐ, C ¿Õ¼äµÄ¶¨ÒåÊÇʲô, ÄãÏëÒª±È½ÏµÄÁ½¸ö¿Õ¼äË­´ó˭С, ÎÒÔõô¾õµÃ C½» L^{Infinity}¸üÐ¡ÄØ?
We_must_know. We_will_know.
2Â¥2013-10-10 03:46:59
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2Â¥: Originally posted by hank612 at 2013-10-10 03:46:59
ÄãÄÜ˵˵º¯Êý¿Õ¼äµÄ¶ÈÁ¿ÊÇÈçºÎ¶¨ÒåµÄÂð? »¹ÓÐ, C ¿Õ¼äµÄ¶¨ÒåÊÇʲô, ÄãÏëÒª±È½ÏµÄÁ½¸ö¿Õ¼äË­´ó˭С, ÎÒÔõô¾õµÃ C½» L^{Infinity}¸üÐ¡ÄØ?

\omegaÊÇÓн翪¼¯£¬C_0(\omega)ÊÇÁ¬ÐøÓнôÖ§¼¯µÄ£¬C(\omega)ÊÇÁ¬Ðø£¬LÎÞÇî\omegaÊÇÓн磬C_0µ±È»ÊÇÁ¬ÐøÓнçµÄ
3Â¥2013-10-10 08:13:10
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hank612

ÖÁ×ðľ³æ (ÖøÃûдÊÖ)

ÒýÓûØÌû:
3Â¥: Originally posted by ÜñÇÅ at 2013-10-10 08:13:10
\omegaÊÇÓн翪¼¯£¬C_0(\omega)ÊÇÁ¬ÐøÓнôÖ§¼¯µÄ£¬C(\omega)ÊÇÁ¬Ðø£¬LÎÞÇî\omegaÊÇÓн磬C_0µ±È»ÊÇÁ¬ÐøÓнçµÄ...

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( |f(x)-g(x)|, dx), then of course C_0 closure is the full space.
We_must_know. We_will_know.
4Â¥2013-10-10 12:38:46
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