Finiteness of an integral if the function space changes
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In a previous post, I asked about the finiteness of $$int_{mathbb{R}} frac{f^2} {sqrt{a^2 + x^2}}$$ if $a > 0$ and $f: mathbb{R} rightarrow mathbb{R}$ is $mathcal{L}^1$ measurable. I was wondering if this was still true if $f in mathcal{L^2(mathbb{R}})$ or $f in mathcal{L^infty(mathbb{R}})$.
It seems to be true in the former case: by Holder's the integral should be bounded above by a finite number because $|f|^2$ is summable and the integral of the denominator is a finite number, so the upper bound (which is just the product) is also finite.
But what about the latter case? Can I use a similar line of reasoning?
functional-analysis
asked Dec 17 '18 at 14:01
TaliantTaliant
839
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add a comment |
$begingroup$
In a previous post, I asked about the finiteness of $$int_{mathbb{R}} frac{f^2} {sqrt{a^2 + x^2}}$$ if $a > 0$ and $f: mathbb{R} rightarrow mathbb{R}$ is $mathcal{L}^1$ measurable. I was wondering if this was still true if $f in mathcal{L^2(mathbb{R}})$ or $f in mathcal{L^infty(mathbb{R}})$.
It seems to be true in the former case: by Holder's the integral should be bounded above by a finite number because $|f|^2$ is summable and the integral of the denominator is a finite number, so the upper bound (which is just the product) is also finite.
But what about the latter case? Can I use a similar line of reasoning?
functional-analysis
asked Dec 17 '18 at 14:01
TaliantTaliant
839
$endgroup$
add a comment |
$begingroup$
In a previous post, I asked about the finiteness of $$int_{mathbb{R}} frac{f^2} {sqrt{a^2 + x^2}}$$ if $a > 0$ and $f: mathbb{R} rightarrow mathbb{R}$ is $mathcal{L}^1$ measurable. I was wondering if this was still true if $f in mathcal{L^2(mathbb{R}})$ or $f in mathcal{L^infty(mathbb{R}})$.
It seems to be true in the former case: by Holder's the integral should be bounded above by a finite number because $|f|^2$ is summable and the integral of the denominator is a finite number, so the upper bound (which is just the product) is also finite.
But what about the latter case? Can I use a similar line of reasoning?
functional-analysis
asked Dec 17 '18 at 14:01
TaliantTaliant
839
$endgroup$
In a previous post, I asked about the finiteness of $$int_{mathbb{R}} frac{f^2} {sqrt{a^2 + x^2}}$$ if $a > 0$ and $f: mathbb{R} rightarrow mathbb{R}$ is $mathcal{L}^1$ measurable. I was wondering if this was still true if $f in mathcal{L^2(mathbb{R}})$ or $f in mathcal{L^infty(mathbb{R}})$.
It seems to be true in the former case: by Holder's the integral should be bounded above by a finite number because $|f|^2$ is summable and the integral of the denominator is a finite number, so the upper bound (which is just the product) is also finite.
But what about the latter case? Can I use a similar line of reasoning?
functional-analysis
functional-analysis
asked Dec 17 '18 at 14:01
TaliantTaliant
839
asked Dec 17 '18 at 14:01
TaliantTaliant
839
asked Dec 17 '18 at 14:01
TaliantTaliant
839
asked Dec 17 '18 at 14:01
TaliantTaliant
839
asked Dec 17 '18 at 14:01
TaliantTaliant
839
839
add a comment |
add a comment |
1 Answer
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Consider $f equiv 1 in mathcal{L}^infty(mathbb{R})$
$frac{1} {sqrt{a^2 + x^2}} sim frac 1 x$ when $x to infty$.
What can you conclude ?
answered Dec 17 '18 at 14:32
nicomezinicomezi
4,1591920
$endgroup$
$begingroup$
Oh wow, I can't believe I didn't try that trivial example. In this case you just come out with infinity. Thanks!
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– Taliant
Dec 17 '18 at 14:39
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Consider $f equiv 1 in mathcal{L}^infty(mathbb{R})$
$frac{1} {sqrt{a^2 + x^2}} sim frac 1 x$ when $x to infty$.
What can you conclude ?
answered Dec 17 '18 at 14:32
nicomezinicomezi
4,1591920
$endgroup$
$begingroup$
Oh wow, I can't believe I didn't try that trivial example. In this case you just come out with infinity. Thanks!
$endgroup$
– Taliant
Dec 17 '18 at 14:39
add a comment |
$begingroup$
Consider $f equiv 1 in mathcal{L}^infty(mathbb{R})$
$frac{1} {sqrt{a^2 + x^2}} sim frac 1 x$ when $x to infty$.
What can you conclude ?
answered Dec 17 '18 at 14:32
nicomezinicomezi
4,1591920
$endgroup$
$begingroup$
Oh wow, I can't believe I didn't try that trivial example. In this case you just come out with infinity. Thanks!
$endgroup$
– Taliant
Dec 17 '18 at 14:39
add a comment |
$begingroup$
Consider $f equiv 1 in mathcal{L}^infty(mathbb{R})$
$frac{1} {sqrt{a^2 + x^2}} sim frac 1 x$ when $x to infty$.
What can you conclude ?
answered Dec 17 '18 at 14:32
nicomezinicomezi
4,1591920
$endgroup$
Consider $f equiv 1 in mathcal{L}^infty(mathbb{R})$
$frac{1} {sqrt{a^2 + x^2}} sim frac 1 x$ when $x to infty$.
What can you conclude ?
answered Dec 17 '18 at 14:32
nicomezinicomezi
4,1591920
answered Dec 17 '18 at 14:32
nicomezinicomezi
4,1591920
answered Dec 17 '18 at 14:32
nicomezinicomezi
4,1591920
answered Dec 17 '18 at 14:32
nicomezinicomezi
4,1591920
4,1591920
$begingroup$
Oh wow, I can't believe I didn't try that trivial example. In this case you just come out with infinity. Thanks!
$endgroup$
– Taliant
Dec 17 '18 at 14:39
add a comment |
$begingroup$
Oh wow, I can't believe I didn't try that trivial example. In this case you just come out with infinity. Thanks!
$endgroup$
– Taliant
Dec 17 '18 at 14:39
$begingroup$
Oh wow, I can't believe I didn't try that trivial example. In this case you just come out with infinity. Thanks!
$endgroup$
– Taliant
Dec 17 '18 at 14:39
$begingroup$
Oh wow, I can't believe I didn't try that trivial example. In this case you just come out with infinity. Thanks!
$endgroup$
– Taliant
Dec 17 '18 at 14:39
add a comment |
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