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minwaardekn

minwaardekn

Answered question

2022-06-14

if I know that a function f L 2 ( R d ) is supported in a compact ball in R d , how do I prove it is in L 1 ( R d )

From the square integrability I know that
for some constant C, then I know that f ( s u p p ( f ) ) is bounded, I need to show that it is closed as well to conclude that it is compact, but that will not be true in general if f is not continuous.

How am I supposed to start given what I have?

Answer & Explanation

Braedon Rivas

Braedon Rivas

Beginner2022-06-15Added 24 answers

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document WLOG, assume f 0, let X = f 1 ( [ 1 , ) ). Then,
R d f d x = X f d x + Supp(f) X f d x
X f 2 d x + m ( Supp ( f ) ) f 2 2 + m ( Supp ( f ) ) <
pokoljitef2

pokoljitef2

Beginner2022-06-16Added 9 answers

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document By using Holder's inequality,
R d | f h | d μ ( R d | f | 2 d μ ) 1 / 2 . ( R d | h | 2 d μ ) 1 / 2
By taking h = χ s u p p ( f )
R d | f | d μ = s u p p ( f ) | f | d μ + ( s u p p ( f ) ) c | f | d μ ( s u p p ( f ) | f | 2 d μ ) 1 / 2 . ( s u p p ( f ) 1 d μ ) 1 / 2
Since s u p p ( f ) is compact, μ ( s u p p ( f ) ) is finite, f L 2 ( R d ), then its norm is finite, then R d | f | d μ <

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