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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 …
https://math.stackexchange.com/questions/1259433/prove-that-a-continuous-function-of-compact-support-defined-on-rn-is-bounded
$\begingroup$ @JoeJohnson126 Is the image just the set of all values that the function can take? If so, then all that is required for this exercise is to state what you stated and then say that the image is compact and therefore bounded (and closed)?
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 , where is sufficiently small.
https://en.wikipedia.org/wiki/Bounded_function
A family of bounded functions may be uniformly bounded. A bounded operator T : X → Y is not a bounded function in the sense of this page's definition (unless T = 0), but has the weaker property of preserving boundedness: Bounded sets M ⊆ X are mapped to bounded sets T(M) ⊆ Y.
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.math.ucdavis.edu/~hunter/m127c/hmwk6_solutions.pdf
• (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 function on a compact interval is bounded and uniformly continuous.
http://www-users.math.umn.edu/~polacik/Publications/examples-1d.pdf
bounded nonnegative solutions whose initial value u 0 has compact support [10, 13, 14, 41] and for the solutions, not necessarily nonnegative, which are localized in the sense that they decay to zero at x= 1 uniformly in t>0 (see [16]; in this result, it is also assumed that f0(0) <0). Recently, a qua-
https://ncatlab.org/nlab/show/compact+support
sequentially compact metric spaces are totally bounded. continuous metric space valued function on compact metric space is uniformly continuous. ... 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://www.math.ucdavis.edu/~hunter/pdes/ch1.pdf
be bounded even if Ω is bounded; for example, ... almost everywhere to a function with compact support. Next we summarize some fundamental inequalities for integrals, in addition to Minkowski’s inequality which is implicit in the statement that k·kLp is a norm for p ≥ …
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