Mixing
showing an inequality using Holder's inequality
$begingroup$
Let $f$ be a function continuous on the real line such that $f(x) = 0$ for all $|x|geq T$ (T being some positive number). I want to show the following inequality: $int_R |f(x)|dx leq [int_R(1+|x|)^2|f(x)|^2dx]^{1/2} [int_R(1+|x|)^{-2}dx]^{1/2}$.
I know that I have to use the Holder's inequality, but I don't know how to deal with the second $(1+|x|)^{-2}$. If it was just +2 as it's power I could have taken the (1+|x|)^2 out and just used the Holder's inequality on f(x) and 1.
Any help is appreciated.
real-analysis
asked Jan 30 at 14:01
JackJack
1218
$endgroup$
add a comment |
$begingroup$
Let $f$ be a function continuous on the real line such that $f(x) = 0$ for all $|x|geq T$ (T being some positive number). I want to show the following inequality: $int_R |f(x)|dx leq [int_R(1+|x|)^2|f(x)|^2dx]^{1/2} [int_R(1+|x|)^{-2}dx]^{1/2}$.
I know that I have to use the Holder's inequality, but I don't know how to deal with the second $(1+|x|)^{-2}$. If it was just +2 as it's power I could have taken the (1+|x|)^2 out and just used the Holder's inequality on f(x) and 1.
Any help is appreciated.
real-analysis
asked Jan 30 at 14:01
JackJack
1218
$endgroup$
add a comment |
$begingroup$
Let $f$ be a function continuous on the real line such that $f(x) = 0$ for all $|x|geq T$ (T being some positive number). I want to show the following inequality: $int_R |f(x)|dx leq [int_R(1+|x|)^2|f(x)|^2dx]^{1/2} [int_R(1+|x|)^{-2}dx]^{1/2}$.
I know that I have to use the Holder's inequality, but I don't know how to deal with the second $(1+|x|)^{-2}$. If it was just +2 as it's power I could have taken the (1+|x|)^2 out and just used the Holder's inequality on f(x) and 1.
Any help is appreciated.
real-analysis
asked Jan 30 at 14:01
JackJack
1218
$endgroup$
Let $f$ be a function continuous on the real line such that $f(x) = 0$ for all $|x|geq T$ (T being some positive number). I want to show the following inequality: $int_R |f(x)|dx leq [int_R(1+|x|)^2|f(x)|^2dx]^{1/2} [int_R(1+|x|)^{-2}dx]^{1/2}$.
I know that I have to use the Holder's inequality, but I don't know how to deal with the second $(1+|x|)^{-2}$. If it was just +2 as it's power I could have taken the (1+|x|)^2 out and just used the Holder's inequality on f(x) and 1.
Any help is appreciated.
real-analysis
real-analysis
asked Jan 30 at 14:01
JackJack
1218
asked Jan 30 at 14:01
JackJack
1218
asked Jan 30 at 14:01
JackJack
1218
asked Jan 30 at 14:01
JackJack
1218
asked Jan 30 at 14:01
JackJack
1218
1218
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
Holder inequality tells you that
$$
int |g(x)|cdot |h(x)| dx le (int |g(x)|^2)^{1/2}(int |h(x)|^2)^{1/2}.
$$
Write
$$
|f(x)|=frac{1}{1+|x|}cdot (1+|x|)|f(x)|.
$$
Now use Holder to $g(x)=frac{1}{1+|x|}$ and $h(x)=(1+|x|)|f(x)|$.
$endgroup$
$begingroup$
I got it. I'm just curious what is the use for $f(x)=0$ for all $|x| geq T$.
$endgroup$
– Jack
Jan 30 at 21:40
1
$begingroup$
With that assumption, you have $fin L^p$ for all $pgeq 1$.
$endgroup$
– user587192
Jan 30 at 21:45
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Holder inequality tells you that
$$
int |g(x)|cdot |h(x)| dx le (int |g(x)|^2)^{1/2}(int |h(x)|^2)^{1/2}.
$$
Write
$$
|f(x)|=frac{1}{1+|x|}cdot (1+|x|)|f(x)|.
$$
Now use Holder to $g(x)=frac{1}{1+|x|}$ and $h(x)=(1+|x|)|f(x)|$.
$endgroup$
$begingroup$
I got it. I'm just curious what is the use for $f(x)=0$ for all $|x| geq T$.
$endgroup$
– Jack
Jan 30 at 21:40
1
$begingroup$
With that assumption, you have $fin L^p$ for all $pgeq 1$.
$endgroup$
– user587192
Jan 30 at 21:45
add a comment |
$begingroup$
Holder inequality tells you that
$$
int |g(x)|cdot |h(x)| dx le (int |g(x)|^2)^{1/2}(int |h(x)|^2)^{1/2}.
$$
Write
$$
|f(x)|=frac{1}{1+|x|}cdot (1+|x|)|f(x)|.
$$
Now use Holder to $g(x)=frac{1}{1+|x|}$ and $h(x)=(1+|x|)|f(x)|$.
$endgroup$
$begingroup$
I got it. I'm just curious what is the use for $f(x)=0$ for all $|x| geq T$.
$endgroup$
– Jack
Jan 30 at 21:40
1
$begingroup$
With that assumption, you have $fin L^p$ for all $pgeq 1$.
$endgroup$
– user587192
Jan 30 at 21:45
add a comment |
$begingroup$
Holder inequality tells you that
$$
int |g(x)|cdot |h(x)| dx le (int |g(x)|^2)^{1/2}(int |h(x)|^2)^{1/2}.
$$
Write
$$
|f(x)|=frac{1}{1+|x|}cdot (1+|x|)|f(x)|.
$$
Now use Holder to $g(x)=frac{1}{1+|x|}$ and $h(x)=(1+|x|)|f(x)|$.
$endgroup$
Holder inequality tells you that
$$
int |g(x)|cdot |h(x)| dx le (int |g(x)|^2)^{1/2}(int |h(x)|^2)^{1/2}.
$$
Write
$$
|f(x)|=frac{1}{1+|x|}cdot (1+|x|)|f(x)|.
$$
Now use Holder to $g(x)=frac{1}{1+|x|}$ and $h(x)=(1+|x|)|f(x)|$.
answered Jan 30 at 14:13
user587192
$begingroup$
I got it. I'm just curious what is the use for $f(x)=0$ for all $|x| geq T$.
$endgroup$
– Jack
Jan 30 at 21:40
1
$begingroup$
With that assumption, you have $fin L^p$ for all $pgeq 1$.
$endgroup$
– user587192
Jan 30 at 21:45
add a comment |
$begingroup$
I got it. I'm just curious what is the use for $f(x)=0$ for all $|x| geq T$.
$endgroup$
– Jack
Jan 30 at 21:40
1
$begingroup$
With that assumption, you have $fin L^p$ for all $pgeq 1$.
$endgroup$
– user587192
Jan 30 at 21:45
$begingroup$
I got it. I'm just curious what is the use for $f(x)=0$ for all $|x| geq T$.
$endgroup$
– Jack
Jan 30 at 21:40
$begingroup$
I got it. I'm just curious what is the use for $f(x)=0$ for all $|x| geq T$.
$endgroup$
– Jack
Jan 30 at 21:40
1
1
$begingroup$
With that assumption, you have $fin L^p$ for all $pgeq 1$.
$endgroup$
– user587192
Jan 30 at 21:45
$begingroup$
With that assumption, you have $fin L^p$ for all $pgeq 1$.
$endgroup$
– user587192
Jan 30 at 21:45
add a comment |
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