- #1

Oxymoron

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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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