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Sep
30
awarded  Nice Answer
Sep
29
comment Question about normed vector spaces and quotients
Nice answer, thank you!
Sep
29
accepted Question about normed vector spaces and quotients
Sep
29
asked Question about normed vector spaces and quotients
Sep
29
comment $C_c(X)$ dense in $L_1(X)$
thanks so much for your help!
Sep
25
comment $C_c(X)$ dense in $L_1(X)$
Actually, I wonder why the Tietze extension is needed. Once one has $s_n|_C$ continuous and $C$ closed (and therefore compact) I'm finished I think.
Sep
25
comment $C_c(X)$ dense in $L_1(X)$
Thank you! Of course, the support is not where the function is non-zero but the closure of that...!
Sep
25
comment $C_c(X)$ dense in $L_1(X)$
Do you not need to show that $h\bar{s_n}$ has compact support?
Sep
25
accepted $C_c(X)$ dense in $L_1(X)$
Sep
25
comment $C_c(X)$ dense in $L_1(X)$
Yay!! I get it! Finally!! Thank you!
Sep
25
comment $C_c(X)$ dense in $L_1(X)$
Oh. That follows from $s_n$ being in $L^1$, right?
Sep
25
comment $C_c(X)$ dense in $L_1(X)$
What I still don't understand is: if I want to assume I have simple functions $s_n$ that converge to $f$ pointwise, without spelling them out, how do I get $\tilde{s_n}$ with finite support?
Sep
25
comment $C_c(X)$ dense in $L_1(X)$
OK, maybe not : ( The sets in the support of $s_n$ are not compact : (
Sep
25
comment $C_c(X)$ dense in $L_1(X)$
But if "finite support" is used to mean "support of finite measure" question a) becomes trivial: a simple function already has support of finite measure in $\Omega$.
Sep
25
comment Smooth functions with compact support are dense in $L^1$
Thanks for pointing it out, that was a typo, I corrected it!
Sep
25
revised Smooth functions with compact support are dense in $L^1$
added 2 characters in body
Sep
25
accepted Homology relative to a point
Sep
25
asked Smooth functions with compact support are dense in $L^1$
Sep
25
comment $C_c(X)$ dense in $L_1(X)$
What does "I should something to you" mean?
Sep
25
comment $C_c(X)$ dense in $L_1(X)$
Thank you! About the measurable function with compact support: the step function $s_n$ has finite support and therefore compact support. This is what I meant.