For all $f,g |f+g|_pleq |f|_p+|g|_p$ then $pgeq 1.$
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Consider $L^p(mathbb{R}^d$) when $0<p<infty$.
Show that if for all $f,g, ||f+g||_pleq ||f||_p+||g||_p$ then $pgeq 1$.
Hint. If $p<1$ then for all $x,y>0$ then $(x+y)^p<x^p+y^p$.
I dont understand the hint.
If $p<1$, with $f(x)=(x+y)^p-x^p+y^p$, $f'(x)<0$ then $f$ decreacing. Therefore $f(0)<f(x)$ then $(x+y)^p<x^p+y^p$.
But, why $(x+y)^p<x^p+y^p$ proves the affirmation?
real-analysis functional-analysis lp-spaces
asked Nov 20 at 4:25
eraldcoil
27919
add a comment |
up vote
0
down vote
favorite
Consider $L^p(mathbb{R}^d$) when $0<p<infty$.
Show that if for all $f,g, ||f+g||_pleq ||f||_p+||g||_p$ then $pgeq 1$.
Hint. If $p<1$ then for all $x,y>0$ then $(x+y)^p<x^p+y^p$.
I dont understand the hint.
If $p<1$, with $f(x)=(x+y)^p-x^p+y^p$, $f'(x)<0$ then $f$ decreacing. Therefore $f(0)<f(x)$ then $(x+y)^p<x^p+y^p$.
But, why $(x+y)^p<x^p+y^p$ proves the affirmation?
real-analysis functional-analysis lp-spaces
asked Nov 20 at 4:25
eraldcoil
27919
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider $L^p(mathbb{R}^d$) when $0<p<infty$.
Show that if for all $f,g, ||f+g||_pleq ||f||_p+||g||_p$ then $pgeq 1$.
Hint. If $p<1$ then for all $x,y>0$ then $(x+y)^p<x^p+y^p$.
I dont understand the hint.
If $p<1$, with $f(x)=(x+y)^p-x^p+y^p$, $f'(x)<0$ then $f$ decreacing. Therefore $f(0)<f(x)$ then $(x+y)^p<x^p+y^p$.
But, why $(x+y)^p<x^p+y^p$ proves the affirmation?
real-analysis functional-analysis lp-spaces
asked Nov 20 at 4:25
eraldcoil
27919
Consider $L^p(mathbb{R}^d$) when $0<p<infty$.
Show that if for all $f,g, ||f+g||_pleq ||f||_p+||g||_p$ then $pgeq 1$.
Hint. If $p<1$ then for all $x,y>0$ then $(x+y)^p<x^p+y^p$.
I dont understand the hint.
If $p<1$, with $f(x)=(x+y)^p-x^p+y^p$, $f'(x)<0$ then $f$ decreacing. Therefore $f(0)<f(x)$ then $(x+y)^p<x^p+y^p$.
But, why $(x+y)^p<x^p+y^p$ proves the affirmation?
real-analysis functional-analysis lp-spaces
real-analysis functional-analysis lp-spaces
asked Nov 20 at 4:25
eraldcoil
27919
asked Nov 20 at 4:25
eraldcoil
27919
asked Nov 20 at 4:25
eraldcoil
27919
asked Nov 20 at 4:25
eraldcoil
27919
asked Nov 20 at 4:25
eraldcoil
27919
27919
add a comment |
add a comment |
1 Answer
1
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1
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accepted
Let $A,B$ be disjoint sets each with measure $1$ and $f=aI_A, g=bI_B$. Then $|f+g|_p=(int |f+g|^{p})^{1/p} =(|a|^{p}+|b|^{p})^{1/p}$ and $|f|_p=|a|, |g|_p|b|$, hence the given inequality gives $(|a|^{p}+|b|^{p})^{1/p} leq |a|+|b|$ or $|a|^{p}+|b|^{p} leq (|a|+|b|)^{p}$ which is a contradiction to the inequality you have proved.
answered Nov 20 at 6:05
Kavi Rama Murthy
44.4k31852
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
accepted
Let $A,B$ be disjoint sets each with measure $1$ and $f=aI_A, g=bI_B$. Then $|f+g|_p=(int |f+g|^{p})^{1/p} =(|a|^{p}+|b|^{p})^{1/p}$ and $|f|_p=|a|, |g|_p|b|$, hence the given inequality gives $(|a|^{p}+|b|^{p})^{1/p} leq |a|+|b|$ or $|a|^{p}+|b|^{p} leq (|a|+|b|)^{p}$ which is a contradiction to the inequality you have proved.
answered Nov 20 at 6:05
Kavi Rama Murthy
44.4k31852
add a comment |
up vote
1
down vote
accepted
Let $A,B$ be disjoint sets each with measure $1$ and $f=aI_A, g=bI_B$. Then $|f+g|_p=(int |f+g|^{p})^{1/p} =(|a|^{p}+|b|^{p})^{1/p}$ and $|f|_p=|a|, |g|_p|b|$, hence the given inequality gives $(|a|^{p}+|b|^{p})^{1/p} leq |a|+|b|$ or $|a|^{p}+|b|^{p} leq (|a|+|b|)^{p}$ which is a contradiction to the inequality you have proved.
answered Nov 20 at 6:05
Kavi Rama Murthy
44.4k31852
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $A,B$ be disjoint sets each with measure $1$ and $f=aI_A, g=bI_B$. Then $|f+g|_p=(int |f+g|^{p})^{1/p} =(|a|^{p}+|b|^{p})^{1/p}$ and $|f|_p=|a|, |g|_p|b|$, hence the given inequality gives $(|a|^{p}+|b|^{p})^{1/p} leq |a|+|b|$ or $|a|^{p}+|b|^{p} leq (|a|+|b|)^{p}$ which is a contradiction to the inequality you have proved.
answered Nov 20 at 6:05
Kavi Rama Murthy
44.4k31852
Let $A,B$ be disjoint sets each with measure $1$ and $f=aI_A, g=bI_B$. Then $|f+g|_p=(int |f+g|^{p})^{1/p} =(|a|^{p}+|b|^{p})^{1/p}$ and $|f|_p=|a|, |g|_p|b|$, hence the given inequality gives $(|a|^{p}+|b|^{p})^{1/p} leq |a|+|b|$ or $|a|^{p}+|b|^{p} leq (|a|+|b|)^{p}$ which is a contradiction to the inequality you have proved.
answered Nov 20 at 6:05
Kavi Rama Murthy
44.4k31852
answered Nov 20 at 6:05
Kavi Rama Murthy
44.4k31852
answered Nov 20 at 6:05
Kavi Rama Murthy
44.4k31852
answered Nov 20 at 6:05
Kavi Rama Murthy
44.4k31852
44.4k31852
add a comment |
add a comment |
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