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https://math.stackexchange.com/questions/1344706/are-continuous-functions-with-compact-support-bounded
While studying measure theory I came across the following fact: $\mathcal{K}(X) \subset C_b(X)$ (meaning the continuous functions with compact support are a subset of the bounded …
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 a number greater than , …
https://www.sciencedirect.com/topics/mathematics/function-with-compact-support
B = C 00 (X) (continuous functions with compact support on a locally compact Hausdorff space X). B is a Stonian lattice ring of bounded functions. Any nonnegative linear functional on C 00 (X) (it is customary to term such a functional a positive Radon measure on X) is a Bourbaki integral on C 00 (X).
https://ncatlab.org/nlab/show/compact+support
A function f: X → V f\colon X \to V on a topological space with values in a vector space V V (or really any pointed set with the basepoint called 0 0) has compact support (or is compactly supported) if the closure of its support, the set of points where it is non-zero, is a compact subset.
https://en.wikipedia.org/wiki/Bounded_function
Every continuous function f : [0, 1] → R is bounded. More generally, any continuous function from a compact space into a metric space is bounded. All complex-valued functions f : C → C which are entire are either unbounded or constant as a consequence of Liouville's theorem. In particular, the complex sin : C → C must be unbounded since ...
http://www.msc.uky.edu/ken/ma570/lectures/lecture2/html/compact.htm
Theorem 5: (Heine-Borel Theorem) With the usual topology on , a subset of is compact if and only if it both closed and bounded. Note: The Extreme Value Theorem follows: If is continuous, then is the image of a compact set and so is compact by Proposition 2. So, it is both closed and bounded by Exercise 5.
https://www.ams.org/journals/tran/1971-156-00/S0002-9947-1971-0275367-4/S0002-9947-1971-0275367-4.pdf
support. We are interested in functions with compact support. A weaker condition on a subset 5 of X is that it be pseudocompact, i.e., every function in C(S) is bounded. Still weaker is the following condition. Definition. A subset S of a space X is relatively pseudocompact in X if every function in CiX) is bounded …
https://www.math.ucdavis.edu/~hunter/m127c/hmwk6_solutions.pdf
If f, g are continuous functions with compact support, prove that kf ∗gk 1 ≤ kfk 1kgk 1. Solution. • (a) If f, g are continuous functions with compact support, then they are bounded and uniformly continuous, since the functions are zero outside a compact (i.e. closed, bounded) interval, and a continuous
https://mathoverflow.net/questions/159853/rieszs-representation-theorem-for-non-locally-compact-spaces
Riesz's representation theorem for non-locally compact spaces ... recent statement concerning the dual of the algebra of bounded continuous functions on non-locally-compact spaces? What is lost when one gives up local-compactness? (Please notice that I am not interested in the algebra of functions with compact support or vanishing at infinity.) ...
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