Relationship between exponents inside Lebesgue integral and evaluation
I am trying to understand the properties of Lebesgue integration, and in particular how exponents play between the integrand and the value of the integral.
For instance, suppose that it is known $$int_{[0,1]} f_n(x)^2 dlambda leq frac{1}{n^4}$$ for all $n in mathbb{N}$, where $lambda$ is the Lebesgue measure on $[0,1]$.
It seems to me, since $+sqrt{f_n(x)^2} = |f_n(x)|$ for every $x$ (where we consider only the positive values of $sqrt{f_n(x)^2}$) then we should have that $$int_{[0,1]} |f_n(x)| d lambda leq frac{1}{n^2}.$$ I am attempting to prove this, but I am not sure how.
I realize that in the $L^2$ norm, the statement $||f_n||_2 leq frac{1}{n^2}$ holds (since $f_n^2 = |f_n|^2$), but it is not in general true that $$int|f_n(x)|^2 d lambda = left( int |f_n(x)| d lambda right )^2.$$ Would anyone be able to provide a hint on how to bound the statement above, or hint at a counterexample?
Thank you!
real-analysis lebesgue-integral lebesgue-measure
asked Nov 25 at 14:47
tekay-squared
625
add a comment |
I am trying to understand the properties of Lebesgue integration, and in particular how exponents play between the integrand and the value of the integral.
For instance, suppose that it is known $$int_{[0,1]} f_n(x)^2 dlambda leq frac{1}{n^4}$$ for all $n in mathbb{N}$, where $lambda$ is the Lebesgue measure on $[0,1]$.
It seems to me, since $+sqrt{f_n(x)^2} = |f_n(x)|$ for every $x$ (where we consider only the positive values of $sqrt{f_n(x)^2}$) then we should have that $$int_{[0,1]} |f_n(x)| d lambda leq frac{1}{n^2}.$$ I am attempting to prove this, but I am not sure how.
I realize that in the $L^2$ norm, the statement $||f_n||_2 leq frac{1}{n^2}$ holds (since $f_n^2 = |f_n|^2$), but it is not in general true that $$int|f_n(x)|^2 d lambda = left( int |f_n(x)| d lambda right )^2.$$ Would anyone be able to provide a hint on how to bound the statement above, or hint at a counterexample?
Thank you!
real-analysis lebesgue-integral lebesgue-measure
asked Nov 25 at 14:47
tekay-squared
625
add a comment |
I am trying to understand the properties of Lebesgue integration, and in particular how exponents play between the integrand and the value of the integral.
For instance, suppose that it is known $$int_{[0,1]} f_n(x)^2 dlambda leq frac{1}{n^4}$$ for all $n in mathbb{N}$, where $lambda$ is the Lebesgue measure on $[0,1]$.
It seems to me, since $+sqrt{f_n(x)^2} = |f_n(x)|$ for every $x$ (where we consider only the positive values of $sqrt{f_n(x)^2}$) then we should have that $$int_{[0,1]} |f_n(x)| d lambda leq frac{1}{n^2}.$$ I am attempting to prove this, but I am not sure how.
I realize that in the $L^2$ norm, the statement $||f_n||_2 leq frac{1}{n^2}$ holds (since $f_n^2 = |f_n|^2$), but it is not in general true that $$int|f_n(x)|^2 d lambda = left( int |f_n(x)| d lambda right )^2.$$ Would anyone be able to provide a hint on how to bound the statement above, or hint at a counterexample?
Thank you!
real-analysis lebesgue-integral lebesgue-measure
asked Nov 25 at 14:47
tekay-squared
625
I am trying to understand the properties of Lebesgue integration, and in particular how exponents play between the integrand and the value of the integral.
For instance, suppose that it is known $$int_{[0,1]} f_n(x)^2 dlambda leq frac{1}{n^4}$$ for all $n in mathbb{N}$, where $lambda$ is the Lebesgue measure on $[0,1]$.
It seems to me, since $+sqrt{f_n(x)^2} = |f_n(x)|$ for every $x$ (where we consider only the positive values of $sqrt{f_n(x)^2}$) then we should have that $$int_{[0,1]} |f_n(x)| d lambda leq frac{1}{n^2}.$$ I am attempting to prove this, but I am not sure how.
I realize that in the $L^2$ norm, the statement $||f_n||_2 leq frac{1}{n^2}$ holds (since $f_n^2 = |f_n|^2$), but it is not in general true that $$int|f_n(x)|^2 d lambda = left( int |f_n(x)| d lambda right )^2.$$ Would anyone be able to provide a hint on how to bound the statement above, or hint at a counterexample?
Thank you!
real-analysis lebesgue-integral lebesgue-measure
real-analysis lebesgue-integral lebesgue-measure
asked Nov 25 at 14:47
tekay-squared
625
asked Nov 25 at 14:47
tekay-squared
625
asked Nov 25 at 14:47
tekay-squared
625
asked Nov 25 at 14:47
tekay-squared
625
asked Nov 25 at 14:47
tekay-squared
625
625
add a comment |
add a comment |
1 Answer
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Use Schwarz's inequality:
$$
int_{[0,1]} |f_n(x)|, dlambdaleBigl(int_{[0,1]} |f_n(x)|^2, dlambdaBigr)^{1/2}Bigl(int_{[0,1]} dlambdaBigr)^{1/2}.
$$
answered Nov 25 at 16:48
Julián Aguirre
67.4k24094
add a comment |
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1 Answer
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1 Answer
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active
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Use Schwarz's inequality:
$$
int_{[0,1]} |f_n(x)|, dlambdaleBigl(int_{[0,1]} |f_n(x)|^2, dlambdaBigr)^{1/2}Bigl(int_{[0,1]} dlambdaBigr)^{1/2}.
$$
answered Nov 25 at 16:48
Julián Aguirre
67.4k24094
add a comment |
Use Schwarz's inequality:
$$
int_{[0,1]} |f_n(x)|, dlambdaleBigl(int_{[0,1]} |f_n(x)|^2, dlambdaBigr)^{1/2}Bigl(int_{[0,1]} dlambdaBigr)^{1/2}.
$$
answered Nov 25 at 16:48
Julián Aguirre
67.4k24094
add a comment |
Use Schwarz's inequality:
$$
int_{[0,1]} |f_n(x)|, dlambdaleBigl(int_{[0,1]} |f_n(x)|^2, dlambdaBigr)^{1/2}Bigl(int_{[0,1]} dlambdaBigr)^{1/2}.
$$
answered Nov 25 at 16:48
Julián Aguirre
67.4k24094
Use Schwarz's inequality:
$$
int_{[0,1]} |f_n(x)|, dlambdaleBigl(int_{[0,1]} |f_n(x)|^2, dlambdaBigr)^{1/2}Bigl(int_{[0,1]} dlambdaBigr)^{1/2}.
$$
answered Nov 25 at 16:48
Julián Aguirre
67.4k24094
answered Nov 25 at 16:48
Julián Aguirre
67.4k24094
answered Nov 25 at 16:48
Julián Aguirre
67.4k24094
answered Nov 25 at 16:48
Julián Aguirre
67.4k24094
67.4k24094
add a comment |
add a comment |
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