- #1
Oxymoron
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- 0
If I have a function [itex]f \in L^2([1,\infty))[/itex] then is it true that all I know about that function is that
[tex]\sqrt{\int_{[1,\infty)}|f|^2\mbox{d}\mu} < \infty[/tex]
I ask this because I am asked to show that if I have a function as above, then the function [itex]g(x):= f(x)/x[/itex] is in [itex]L^1([1,\infty))[/itex]. Now, is it true that g is simply a composition of functions from [itex]L^2([1,\infty))[/itex]?
[tex]g(x):=x^{-1}\circ f(x)[/tex]
If I want to prove that a function is in [itex]L^1([1,\infty))[/itex] then do I have to show that
[tex]\int_{[1,\infty)} |g|\mbox{d}\mu[/tex]
is finite? Is this all I have to do?
[tex]\sqrt{\int_{[1,\infty)}|f|^2\mbox{d}\mu} < \infty[/tex]
I ask this because I am asked to show that if I have a function as above, then the function [itex]g(x):= f(x)/x[/itex] is in [itex]L^1([1,\infty))[/itex]. Now, is it true that g is simply a composition of functions from [itex]L^2([1,\infty))[/itex]?
[tex]g(x):=x^{-1}\circ f(x)[/tex]
If I want to prove that a function is in [itex]L^1([1,\infty))[/itex] then do I have to show that
[tex]\int_{[1,\infty)} |g|\mbox{d}\mu[/tex]
is finite? Is this all I have to do?
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