Let $0<α<1$. Show that if x and y are positive real numbers, then $|x^α-y^α|≤|x-y|^α$.
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I'm having a hard time proving this statement. I do know that d(x,y)=$|x-y|^α$ defines a metric on R. Any help would be greatly appreciated.
real-analysis
asked Nov 21 at 17:10
Amanda Varvak
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up vote
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I'm having a hard time proving this statement. I do know that d(x,y)=$|x-y|^α$ defines a metric on R. Any help would be greatly appreciated.
real-analysis
asked Nov 21 at 17:10
Amanda Varvak
13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm having a hard time proving this statement. I do know that d(x,y)=$|x-y|^α$ defines a metric on R. Any help would be greatly appreciated.
real-analysis
asked Nov 21 at 17:10
Amanda Varvak
13
I'm having a hard time proving this statement. I do know that d(x,y)=$|x-y|^α$ defines a metric on R. Any help would be greatly appreciated.
real-analysis
real-analysis
asked Nov 21 at 17:10
Amanda Varvak
13
asked Nov 21 at 17:10
Amanda Varvak
13
edited Nov 21 at 17:30
asked Nov 21 at 17:10
Amanda Varvak
13
asked Nov 21 at 17:10
Amanda Varvak
13
asked Nov 21 at 17:10
Amanda Varvak
13
13
add a comment |
add a comment |
1 Answer
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Observe the inequality is symmetric in $x,y$ and satisfied with equality for $x=y$. Thus we may restrict to $0<x< y$. Define
$$t=x/yin (0,1)$$
then the inequality is
$$ 1-t^alpha le (1-t)^alpha iff 1 le (1-t)^alpha + t^alpha$$
this is true because for $tin (0,1)$ and $alpha in (0,1)$, $t^alpha ge t$ and $(1-t)^alpha ge (1-t)$.
Alternatively $|d(x,0) - d(y,0)| le d(x,y) $ purely from triangle inequality (follow the proof for the usual euclidean norm).
answered Nov 21 at 17:40
Calvin Khor
10.7k21436
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Observe the inequality is symmetric in $x,y$ and satisfied with equality for $x=y$. Thus we may restrict to $0<x< y$. Define
$$t=x/yin (0,1)$$
then the inequality is
$$ 1-t^alpha le (1-t)^alpha iff 1 le (1-t)^alpha + t^alpha$$
this is true because for $tin (0,1)$ and $alpha in (0,1)$, $t^alpha ge t$ and $(1-t)^alpha ge (1-t)$.
Alternatively $|d(x,0) - d(y,0)| le d(x,y) $ purely from triangle inequality (follow the proof for the usual euclidean norm).
answered Nov 21 at 17:40
Calvin Khor
10.7k21436
add a comment |
up vote
2
down vote
Observe the inequality is symmetric in $x,y$ and satisfied with equality for $x=y$. Thus we may restrict to $0<x< y$. Define
$$t=x/yin (0,1)$$
then the inequality is
$$ 1-t^alpha le (1-t)^alpha iff 1 le (1-t)^alpha + t^alpha$$
this is true because for $tin (0,1)$ and $alpha in (0,1)$, $t^alpha ge t$ and $(1-t)^alpha ge (1-t)$.
Alternatively $|d(x,0) - d(y,0)| le d(x,y) $ purely from triangle inequality (follow the proof for the usual euclidean norm).
answered Nov 21 at 17:40
Calvin Khor
10.7k21436
add a comment |
up vote
2
down vote
up vote
2
down vote
Observe the inequality is symmetric in $x,y$ and satisfied with equality for $x=y$. Thus we may restrict to $0<x< y$. Define
$$t=x/yin (0,1)$$
then the inequality is
$$ 1-t^alpha le (1-t)^alpha iff 1 le (1-t)^alpha + t^alpha$$
this is true because for $tin (0,1)$ and $alpha in (0,1)$, $t^alpha ge t$ and $(1-t)^alpha ge (1-t)$.
Alternatively $|d(x,0) - d(y,0)| le d(x,y) $ purely from triangle inequality (follow the proof for the usual euclidean norm).
answered Nov 21 at 17:40
Calvin Khor
10.7k21436
Observe the inequality is symmetric in $x,y$ and satisfied with equality for $x=y$. Thus we may restrict to $0<x< y$. Define
$$t=x/yin (0,1)$$
then the inequality is
$$ 1-t^alpha le (1-t)^alpha iff 1 le (1-t)^alpha + t^alpha$$
this is true because for $tin (0,1)$ and $alpha in (0,1)$, $t^alpha ge t$ and $(1-t)^alpha ge (1-t)$.
Alternatively $|d(x,0) - d(y,0)| le d(x,y) $ purely from triangle inequality (follow the proof for the usual euclidean norm).
answered Nov 21 at 17:40
Calvin Khor
10.7k21436
answered Nov 21 at 17:40
Calvin Khor
10.7k21436
answered Nov 21 at 17:40
Calvin Khor
10.7k21436
answered Nov 21 at 17:40
Calvin Khor
10.7k21436
10.7k21436
add a comment |
add a comment |
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