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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 continuous functions).
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 …
http://www.math.ncku.edu.tw/~fjmliou/Complex/spbounded.pdf
space of all real-valued continuous functions on X:Since Xis compact, every continuous function on Xis bounded. Therefore C(X) is a subset of B(X):Moreover, since the sum of continuous functions on Xis continuous function on Xand the scalar multiplication of a continuous function by a real number is again continuous, it is easy to check that C(X)
https://www.ams.org/journals/tran/1971-156-00/S0002-9947-1971-0275367-4/S0002-9947-1971-0275367-4.pdf
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://en.wikipedia.org/wiki/Continuous_functions_on_a_compact_Hausdorff_space
In the non-compact case, however, C(X) is not in general a Banach space with respect to the uniform norm since it may contain unbounded functions. Hence it is more typical to consider the space, denoted here C B (X) of bounded continuous functions on X. This is a Banach space (in fact a commutative Banach algebra with identity) with respect to ...
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://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 .
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://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.
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