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https://math.stackexchange.com/questions/3518516/continuous-functions-with-compact-support-are-dense-in-l1-hypotheses
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https://www.planetmath.org/CompactlySupportedContinuousFunctionsAreDenseInLp
Now, it follows easily that any simple function ∑ i = 1 n c i χ A i, where each A i has finite measure, can also be approximated by a compactly supported continuous function. Since this kind of simple functions are dense in L p (X) we see that C c (X) is also dense in L p (X).
https://mathproblems123.files.wordpress.com/2011/02/density-1.pdf
Oct 03, 2004 · Density of Continuous Functions in L1 October 3, 2004 1 Approximation by continuous functions In this supplement, we’ll show that continuous functions with compact support are dense in L1 = L1(Rn;m). The support of a complex valued function f on a metric space X is the closure of fx 2 X : f(x) 6= 0g. We’ll denote by Cc(X) the set of all complex
https://mathoverflow.net/questions/237636/are-compactly-supported-continuous-functions-dense-in-the-continuous-functions-o
Continuous functions on $\mathbb R^d$ such that the support is a compact subset of $\overline{\Omega}$? For "nice" $\Omega$ this would be the space of continuous functions on $\Omega$ vanishing at the boundary. $\endgroup$ – Jochen Wengenroth Apr 29 '16 at 12:50
https://en.wikipedia.org/wiki/Support_(mathematics)
In mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero. If the domain of f is a topological space, the support of f is instead defined as the smallest closed set containing all points not mapped to zero. This concept is used very widely in mathematical analysis
https://www.encyclopediaofmath.org/index.php/Function_of_compact_support
The support of is the closure of the set of points for which is different from zero . Thus one can also say that a function of compact support in is a function defined on such that its support is a closed bounded set located at a distance from the boundary of by …
http://www.math.ucsd.edu/~bdriver/240A-C-03-04/Lecture_Notes/Older-Versions/chap22.pdf
of continuous functions with compact support) is dense in Lp(µ) for all p∈ [1,∞).(See also Proposition 25.23 below.) Proof. Let M:= Cc(X) and use Item 3. of Lemma 18.57 to find functions ψk∈Msuch that ψk→1 to boundedly as k→∞.The result now follows from an application of Theorem 22.4 along with the aid of item 4. of Lemma 18.57.
https://mathoverflow.net/questions/267710/continuous-functions-dense-in-l-1
$\begingroup$ @AryehKontorovich In an infinite dimensional Banach space, an open ball is not precompact, and the support of a nonzero continuous function contains some open ball, so it is not compact.
https://en.wikipedia.org/wiki/Function_space
In mathematics, a function space is a set of functions between two fixed sets. Often, the domain and/or codomain will have additional structure which is inherited by the function space. For example, the set of functions from any set X into a vector space has a natural vector space structure given by pointwise addition and scalar multiplication.
http://www.msc.uky.edu/ken/ma570/lectures/lecture2/html/compact.htm
The set of all the for all the is an open cover of and so it admits of a finite subcover . But then the set of all the 's associated with all the 's is a finite subcover of . Lemma 3: If every rectangle is compact, then every closed and bounded subset of is compact.
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