Inequality in $L^2$
up vote
1
down vote
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Let $u,vin L^2(mathbb{R}^d)$,
I want to prove the following ineqality
$$|u|^2_{L^2(mathbb{R}^d)}ge a |v|^2_{L^2(mathbb{R}^d)}-|u-v|^2_{L^2(mathbb{R}^d)}$$
for all $u,vin L^2(mathbb{R}^d)$ where $a>0$ is an optimal number to be determined.
Thanks.
real-analysis functional-analysis inequality norm
edited Nov 14 at 12:56
Davide Giraudo
123k16149253
asked Nov 13 at 16:24
aymen
63
add a comment |
up vote
1
down vote
favorite
Let $u,vin L^2(mathbb{R}^d)$,
I want to prove the following ineqality
$$|u|^2_{L^2(mathbb{R}^d)}ge a |v|^2_{L^2(mathbb{R}^d)}-|u-v|^2_{L^2(mathbb{R}^d)}$$
for all $u,vin L^2(mathbb{R}^d)$ where $a>0$ is an optimal number to be determined.
Thanks.
real-analysis functional-analysis inequality norm
edited Nov 14 at 12:56
Davide Giraudo
123k16149253
asked Nov 13 at 16:24
aymen
63
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $u,vin L^2(mathbb{R}^d)$,
I want to prove the following ineqality
$$|u|^2_{L^2(mathbb{R}^d)}ge a |v|^2_{L^2(mathbb{R}^d)}-|u-v|^2_{L^2(mathbb{R}^d)}$$
for all $u,vin L^2(mathbb{R}^d)$ where $a>0$ is an optimal number to be determined.
Thanks.
real-analysis functional-analysis inequality norm
edited Nov 14 at 12:56
Davide Giraudo
123k16149253
asked Nov 13 at 16:24
aymen
63
Let $u,vin L^2(mathbb{R}^d)$,
I want to prove the following ineqality
$$|u|^2_{L^2(mathbb{R}^d)}ge a |v|^2_{L^2(mathbb{R}^d)}-|u-v|^2_{L^2(mathbb{R}^d)}$$
for all $u,vin L^2(mathbb{R}^d)$ where $a>0$ is an optimal number to be determined.
Thanks.
real-analysis functional-analysis inequality norm
real-analysis functional-analysis inequality norm
edited Nov 14 at 12:56
Davide Giraudo
123k16149253
asked Nov 13 at 16:24
aymen
63
edited Nov 14 at 12:56
Davide Giraudo
123k16149253
asked Nov 13 at 16:24
aymen
63
edited Nov 14 at 12:56
Davide Giraudo
123k16149253
edited Nov 14 at 12:56
Davide Giraudo
123k16149253
edited Nov 14 at 12:56
Davide Giraudo
123k16149253
123k16149253
asked Nov 13 at 16:24
aymen
63
asked Nov 13 at 16:24
aymen
63
asked Nov 13 at 16:24
aymen
63
63
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
It is easy to see that $(a+b)^2 le 2(a^2 + b^2)$ for any two real numbers $a$ and $b$.
Thus with $a = u$ and $b=v-u$ you get
$$frac 12 v^2 le u^2 + (u-v)^2.$$ Integrate over $mathbb R^d$ to discover $$frac 12 |v|_{L^2(mathbb R^d)}^2 le |u|_{L^2(mathbb R^d)}^2 + |u-v|_{L^2(mathbb R^d)}^2$$
answered Nov 13 at 18:24
Umberto P.
37.9k13063
P, do you know what is the name of the following equality?: $$2(|x|^2+|y|^2)-|x+y|^2=|x-y|^2$$ Thanks
– aymen
Nov 13 at 18:52
Parallelogram law.
– Tomath
Nov 13 at 18:58
thank you Tomath!
– aymen
Nov 13 at 19:12
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It is easy to see that $(a+b)^2 le 2(a^2 + b^2)$ for any two real numbers $a$ and $b$.
Thus with $a = u$ and $b=v-u$ you get
$$frac 12 v^2 le u^2 + (u-v)^2.$$ Integrate over $mathbb R^d$ to discover $$frac 12 |v|_{L^2(mathbb R^d)}^2 le |u|_{L^2(mathbb R^d)}^2 + |u-v|_{L^2(mathbb R^d)}^2$$
answered Nov 13 at 18:24
Umberto P.
37.9k13063
P, do you know what is the name of the following equality?: $$2(|x|^2+|y|^2)-|x+y|^2=|x-y|^2$$ Thanks
– aymen
Nov 13 at 18:52
Parallelogram law.
– Tomath
Nov 13 at 18:58
thank you Tomath!
– aymen
Nov 13 at 19:12
add a comment |
up vote
1
down vote
It is easy to see that $(a+b)^2 le 2(a^2 + b^2)$ for any two real numbers $a$ and $b$.
Thus with $a = u$ and $b=v-u$ you get
$$frac 12 v^2 le u^2 + (u-v)^2.$$ Integrate over $mathbb R^d$ to discover $$frac 12 |v|_{L^2(mathbb R^d)}^2 le |u|_{L^2(mathbb R^d)}^2 + |u-v|_{L^2(mathbb R^d)}^2$$
answered Nov 13 at 18:24
Umberto P.
37.9k13063
P, do you know what is the name of the following equality?: $$2(|x|^2+|y|^2)-|x+y|^2=|x-y|^2$$ Thanks
– aymen
Nov 13 at 18:52
Parallelogram law.
– Tomath
Nov 13 at 18:58
thank you Tomath!
– aymen
Nov 13 at 19:12
add a comment |
up vote
1
down vote
up vote
1
down vote
It is easy to see that $(a+b)^2 le 2(a^2 + b^2)$ for any two real numbers $a$ and $b$.
Thus with $a = u$ and $b=v-u$ you get
$$frac 12 v^2 le u^2 + (u-v)^2.$$ Integrate over $mathbb R^d$ to discover $$frac 12 |v|_{L^2(mathbb R^d)}^2 le |u|_{L^2(mathbb R^d)}^2 + |u-v|_{L^2(mathbb R^d)}^2$$
answered Nov 13 at 18:24
Umberto P.
37.9k13063
It is easy to see that $(a+b)^2 le 2(a^2 + b^2)$ for any two real numbers $a$ and $b$.
Thus with $a = u$ and $b=v-u$ you get
$$frac 12 v^2 le u^2 + (u-v)^2.$$ Integrate over $mathbb R^d$ to discover $$frac 12 |v|_{L^2(mathbb R^d)}^2 le |u|_{L^2(mathbb R^d)}^2 + |u-v|_{L^2(mathbb R^d)}^2$$
answered Nov 13 at 18:24
Umberto P.
37.9k13063
answered Nov 13 at 18:24
Umberto P.
37.9k13063
answered Nov 13 at 18:24
Umberto P.
37.9k13063
answered Nov 13 at 18:24
Umberto P.
37.9k13063
37.9k13063
P, do you know what is the name of the following equality?: $$2(|x|^2+|y|^2)-|x+y|^2=|x-y|^2$$ Thanks
– aymen
Nov 13 at 18:52
Parallelogram law.
– Tomath
Nov 13 at 18:58
thank you Tomath!
– aymen
Nov 13 at 19:12
add a comment |
P, do you know what is the name of the following equality?: $$2(|x|^2+|y|^2)-|x+y|^2=|x-y|^2$$ Thanks
– aymen
Nov 13 at 18:52
Parallelogram law.
– Tomath
Nov 13 at 18:58
thank you Tomath!
– aymen
Nov 13 at 19:12
P, do you know what is the name of the following equality?: $$2(|x|^2+|y|^2)-|x+y|^2=|x-y|^2$$ Thanks
– aymen
Nov 13 at 18:52
P, do you know what is the name of the following equality?: $$2(|x|^2+|y|^2)-|x+y|^2=|x-y|^2$$ Thanks
– aymen
Nov 13 at 18:52
Parallelogram law.
– Tomath
Nov 13 at 18:58
Parallelogram law.
– Tomath
Nov 13 at 18:58
thank you Tomath!
– aymen
Nov 13 at 19:12
thank you Tomath!
– aymen
Nov 13 at 19:12
add a comment |
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