If $0<alpha < 1$ and $d(x,y)$ is metric, show that $((d(x,y))^alpha$ is metric. The 2019 Stack Overflow Developer Survey Results Are InA Particular Metric: $(mathbbR^2,d_2)$Show that dist(A,B) is a metricProving an algebraic inequalityProve that $|x_1-y_1|+|x_2-y_2|$ is a metricProve that square metric on $mathbbR$ is in fact a metric.Is this proof sufficient to show that a concave function of a metric is also a metric?Show that a definite integral of $f(x)$ from $-1$ to $1$ is greater than or equal to $2f(0)$Convergence of a sequence in complete metric space when distance between subsequent terms is $< alpha^n$, $0<alpha<1$.Show that $|d(m,n) -d(n,o) | leq d(m,o)$ for a metric spaceMetric on Cartesian Product
What do I do when my TA workload is more than expected?
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If $0
The 2019 Stack Overflow Developer Survey Results Are InA Particular Metric: $(mathbbR^2,d_2)$Show that dist(A,B) is a metricProving an algebraic inequalityProve that $|x_1-y_1|+|x_2-y_2|$ is a metricProve that square metric on $mathbbR$ is in fact a metric.Is this proof sufficient to show that a concave function of a metric is also a metric?Show that a definite integral of $f(x)$ from $-1$ to $1$ is greater than or equal to $2f(0)$Convergence of a sequence in complete metric space when distance between subsequent terms is $< alpha^n$, $0<alpha<1$.Show that $|d(m,n) -d(n,o) | leq d(m,o)$ for a metric spaceMetric on Cartesian Product
$begingroup$
If $0<alpha < 1$ and $d(x,y)$ is metric, show that $((d(x,y))^alpha$ is metric.
It remains for me to show that the triangle inequality, or $((d(x,y))^alphaleq ((d(x,z))^alpha+((d(z,y))^alpha$, holds.
I begin with the fact that $d(x,y)leq d(x,z)+d(z,y)$. Then I apply the function:
$(d(x,y))^alphaleq (d(x,z)+d(z,y))^alpha$
What now? I was thinking of expanding the right side but maybe it is not the best idea? I got a hint where if $ageq 0, bgeq 0$ and $a+b=1$, then $a^alpha+b^alphageq 1$ but I don't really know how to apply this hint.
real-analysis metric-spaces
asked Apr 8 at 0:59
numericalorangenumericalorange
1,939314
$endgroup$
add a comment |
$begingroup$
If $0<alpha < 1$ and $d(x,y)$ is metric, show that $((d(x,y))^alpha$ is metric.
It remains for me to show that the triangle inequality, or $((d(x,y))^alphaleq ((d(x,z))^alpha+((d(z,y))^alpha$, holds.
I begin with the fact that $d(x,y)leq d(x,z)+d(z,y)$. Then I apply the function:
$(d(x,y))^alphaleq (d(x,z)+d(z,y))^alpha$
What now? I was thinking of expanding the right side but maybe it is not the best idea? I got a hint where if $ageq 0, bgeq 0$ and $a+b=1$, then $a^alpha+b^alphageq 1$ but I don't really know how to apply this hint.
real-analysis metric-spaces
asked Apr 8 at 0:59
numericalorangenumericalorange
1,939314
$endgroup$
$begingroup$
The way to apply the hint would be to take $a=d(x,z)/big(d(x,z)+d(z,y)big)$ and $b=d(z,y)/big(d(x,z)+d(z,y)big)$. (I acknowledge that you still have to prove that the hint is true!)
$endgroup$
– Greg Martin
Apr 8 at 1:11
add a comment |
$begingroup$
If $0<alpha < 1$ and $d(x,y)$ is metric, show that $((d(x,y))^alpha$ is metric.
It remains for me to show that the triangle inequality, or $((d(x,y))^alphaleq ((d(x,z))^alpha+((d(z,y))^alpha$, holds.
I begin with the fact that $d(x,y)leq d(x,z)+d(z,y)$. Then I apply the function:
$(d(x,y))^alphaleq (d(x,z)+d(z,y))^alpha$
What now? I was thinking of expanding the right side but maybe it is not the best idea? I got a hint where if $ageq 0, bgeq 0$ and $a+b=1$, then $a^alpha+b^alphageq 1$ but I don't really know how to apply this hint.
real-analysis metric-spaces
asked Apr 8 at 0:59
numericalorangenumericalorange
1,939314
$endgroup$
If $0<alpha < 1$ and $d(x,y)$ is metric, show that $((d(x,y))^alpha$ is metric.
It remains for me to show that the triangle inequality, or $((d(x,y))^alphaleq ((d(x,z))^alpha+((d(z,y))^alpha$, holds.
I begin with the fact that $d(x,y)leq d(x,z)+d(z,y)$. Then I apply the function:
$(d(x,y))^alphaleq (d(x,z)+d(z,y))^alpha$
What now? I was thinking of expanding the right side but maybe it is not the best idea? I got a hint where if $ageq 0, bgeq 0$ and $a+b=1$, then $a^alpha+b^alphageq 1$ but I don't really know how to apply this hint.
real-analysis metric-spaces
real-analysis metric-spaces
asked Apr 8 at 0:59
numericalorangenumericalorange
1,939314
asked Apr 8 at 0:59
numericalorangenumericalorange
1,939314
asked Apr 8 at 0:59
numericalorangenumericalorange
1,939314
asked Apr 8 at 0:59
numericalorangenumericalorange
1,939314
asked Apr 8 at 0:59
numericalorangenumericalorange
1,939314
1,939314
$begingroup$
The way to apply the hint would be to take $a=d(x,z)/big(d(x,z)+d(z,y)big)$ and $b=d(z,y)/big(d(x,z)+d(z,y)big)$. (I acknowledge that you still have to prove that the hint is true!)
$endgroup$
– Greg Martin
Apr 8 at 1:11
add a comment |
$begingroup$
The way to apply the hint would be to take $a=d(x,z)/big(d(x,z)+d(z,y)big)$ and $b=d(z,y)/big(d(x,z)+d(z,y)big)$. (I acknowledge that you still have to prove that the hint is true!)
$endgroup$
– Greg Martin
Apr 8 at 1:11
$begingroup$
The way to apply the hint would be to take $a=d(x,z)/big(d(x,z)+d(z,y)big)$ and $b=d(z,y)/big(d(x,z)+d(z,y)big)$. (I acknowledge that you still have to prove that the hint is true!)
$endgroup$
– Greg Martin
Apr 8 at 1:11
$begingroup$
The way to apply the hint would be to take $a=d(x,z)/big(d(x,z)+d(z,y)big)$ and $b=d(z,y)/big(d(x,z)+d(z,y)big)$. (I acknowledge that you still have to prove that the hint is true!)
$endgroup$
– Greg Martin
Apr 8 at 1:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you plot a graph, you can clearly see that when $a+b=1$, $a^alpha+b^alphageq 1$, because the curve $a^alpha+b^alpha= 1$ is "concave".
You want to prove $(d(x,y))^alphaleq (d(x,z))^alpha+(d(z,y))^alpha$. This can be done by noting that $(d(x,y))^alphaleq (d(x,z)+d(z,y))^alpha$. Now you can let $a=fracd(x,z)d(x,z)+d(z,y), b=fracd(y,z)d(x,z)+d(z,y)$, so $a+b=1$,
$$
left(fracd(x,z)d(x,z)+d(z,y)right)^alpha+left(fracd(y,z)d(x,z)+d(z,y)right)^alpha=frac(d(x,z))^alpha+(d(z,y))^alpha(d(x,z)+d(z,y))^alphageq 1.\
Rightarrow (d(x,z))^alpha+(d(z,y))^alphaleq (d(x,z)+d(z,y))^alphaleq (d(x,z)+d(z,y))^alpha leq (d(x,y))^alpha,
$$
as required.
answered Apr 8 at 1:14
Holding ArthurHolding Arthur
1,565417
$endgroup$
$begingroup$
This answer is so slick, thanks!
$endgroup$
– numericalorange
Apr 8 at 1:55
add a comment |
$begingroup$
Basically, you have to prove that
$$
x^alpha+y^alphage (x+y)^alpha
$$
for $x,yge 0$. If $y=0$ this is trivial. Assume it is not and divide by $y^alpha$. Calling $x/y=z$ you get the equivalent inequality
$$
z^alpha+1ge (1+z)^alpha
$$
which must be true for $zge 0$. Consider the function
$$
f(z)= z^alpha+1-(1+z)^alpha
$$
You should prove $f(z)ge 0$. But $f(0)=0$ and $f(infty)=0$. You can easily check that $f$ has only one critical point on $(0,infty)$ and it is a local maximum. Therefore, $f$ is never negative.
answered Apr 8 at 1:17
GReyesGReyes
2,43815
$endgroup$
add a comment |
$begingroup$
Let $S = d(x,z) + d(z,y)$. By dividing both sides of the inequality you had by $S^alpha$, you get
$$
frac(d(x,y))^alphaS^alpha
leq left( fracd(x,z)S + fracd(z,y)S right)^alpha
leq left( fracd(x,z)S right)^alpha + left( fracd(z,y)S right)^alpha
,
$$
where we use the fact that $ fracd(x,z)S + fracd(z,y)S = 1$.
Multiplying again by $S^alpha$ gives you the result you need.
answered Apr 8 at 1:16
Hotdog2000Hotdog2000
678
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you plot a graph, you can clearly see that when $a+b=1$, $a^alpha+b^alphageq 1$, because the curve $a^alpha+b^alpha= 1$ is "concave".
You want to prove $(d(x,y))^alphaleq (d(x,z))^alpha+(d(z,y))^alpha$. This can be done by noting that $(d(x,y))^alphaleq (d(x,z)+d(z,y))^alpha$. Now you can let $a=fracd(x,z)d(x,z)+d(z,y), b=fracd(y,z)d(x,z)+d(z,y)$, so $a+b=1$,
$$
left(fracd(x,z)d(x,z)+d(z,y)right)^alpha+left(fracd(y,z)d(x,z)+d(z,y)right)^alpha=frac(d(x,z))^alpha+(d(z,y))^alpha(d(x,z)+d(z,y))^alphageq 1.\
Rightarrow (d(x,z))^alpha+(d(z,y))^alphaleq (d(x,z)+d(z,y))^alphaleq (d(x,z)+d(z,y))^alpha leq (d(x,y))^alpha,
$$
as required.
answered Apr 8 at 1:14
Holding ArthurHolding Arthur
1,565417
$endgroup$
$begingroup$
This answer is so slick, thanks!
$endgroup$
– numericalorange
Apr 8 at 1:55
add a comment |
$begingroup$
If you plot a graph, you can clearly see that when $a+b=1$, $a^alpha+b^alphageq 1$, because the curve $a^alpha+b^alpha= 1$ is "concave".
You want to prove $(d(x,y))^alphaleq (d(x,z))^alpha+(d(z,y))^alpha$. This can be done by noting that $(d(x,y))^alphaleq (d(x,z)+d(z,y))^alpha$. Now you can let $a=fracd(x,z)d(x,z)+d(z,y), b=fracd(y,z)d(x,z)+d(z,y)$, so $a+b=1$,
$$
left(fracd(x,z)d(x,z)+d(z,y)right)^alpha+left(fracd(y,z)d(x,z)+d(z,y)right)^alpha=frac(d(x,z))^alpha+(d(z,y))^alpha(d(x,z)+d(z,y))^alphageq 1.\
Rightarrow (d(x,z))^alpha+(d(z,y))^alphaleq (d(x,z)+d(z,y))^alphaleq (d(x,z)+d(z,y))^alpha leq (d(x,y))^alpha,
$$
as required.
answered Apr 8 at 1:14
Holding ArthurHolding Arthur
1,565417
$endgroup$
$begingroup$
This answer is so slick, thanks!
$endgroup$
– numericalorange
Apr 8 at 1:55
add a comment |
$begingroup$
If you plot a graph, you can clearly see that when $a+b=1$, $a^alpha+b^alphageq 1$, because the curve $a^alpha+b^alpha= 1$ is "concave".
You want to prove $(d(x,y))^alphaleq (d(x,z))^alpha+(d(z,y))^alpha$. This can be done by noting that $(d(x,y))^alphaleq (d(x,z)+d(z,y))^alpha$. Now you can let $a=fracd(x,z)d(x,z)+d(z,y), b=fracd(y,z)d(x,z)+d(z,y)$, so $a+b=1$,
$$
left(fracd(x,z)d(x,z)+d(z,y)right)^alpha+left(fracd(y,z)d(x,z)+d(z,y)right)^alpha=frac(d(x,z))^alpha+(d(z,y))^alpha(d(x,z)+d(z,y))^alphageq 1.\
Rightarrow (d(x,z))^alpha+(d(z,y))^alphaleq (d(x,z)+d(z,y))^alphaleq (d(x,z)+d(z,y))^alpha leq (d(x,y))^alpha,
$$
as required.
answered Apr 8 at 1:14
Holding ArthurHolding Arthur
1,565417
$endgroup$
If you plot a graph, you can clearly see that when $a+b=1$, $a^alpha+b^alphageq 1$, because the curve $a^alpha+b^alpha= 1$ is "concave".
You want to prove $(d(x,y))^alphaleq (d(x,z))^alpha+(d(z,y))^alpha$. This can be done by noting that $(d(x,y))^alphaleq (d(x,z)+d(z,y))^alpha$. Now you can let $a=fracd(x,z)d(x,z)+d(z,y), b=fracd(y,z)d(x,z)+d(z,y)$, so $a+b=1$,
$$
left(fracd(x,z)d(x,z)+d(z,y)right)^alpha+left(fracd(y,z)d(x,z)+d(z,y)right)^alpha=frac(d(x,z))^alpha+(d(z,y))^alpha(d(x,z)+d(z,y))^alphageq 1.\
Rightarrow (d(x,z))^alpha+(d(z,y))^alphaleq (d(x,z)+d(z,y))^alphaleq (d(x,z)+d(z,y))^alpha leq (d(x,y))^alpha,
$$
as required.
answered Apr 8 at 1:14
Holding ArthurHolding Arthur
1,565417
answered Apr 8 at 1:14
Holding ArthurHolding Arthur
1,565417
answered Apr 8 at 1:14
Holding ArthurHolding Arthur
1,565417
answered Apr 8 at 1:14
Holding ArthurHolding Arthur
1,565417
1,565417
$begingroup$
This answer is so slick, thanks!
$endgroup$
– numericalorange
Apr 8 at 1:55
add a comment |
$begingroup$
This answer is so slick, thanks!
$endgroup$
– numericalorange
Apr 8 at 1:55
$begingroup$
This answer is so slick, thanks!
$endgroup$
– numericalorange
Apr 8 at 1:55
$begingroup$
This answer is so slick, thanks!
$endgroup$
– numericalorange
Apr 8 at 1:55
add a comment |
$begingroup$
Basically, you have to prove that
$$
x^alpha+y^alphage (x+y)^alpha
$$
for $x,yge 0$. If $y=0$ this is trivial. Assume it is not and divide by $y^alpha$. Calling $x/y=z$ you get the equivalent inequality
$$
z^alpha+1ge (1+z)^alpha
$$
which must be true for $zge 0$. Consider the function
$$
f(z)= z^alpha+1-(1+z)^alpha
$$
You should prove $f(z)ge 0$. But $f(0)=0$ and $f(infty)=0$. You can easily check that $f$ has only one critical point on $(0,infty)$ and it is a local maximum. Therefore, $f$ is never negative.
answered Apr 8 at 1:17
GReyesGReyes
2,43815
$endgroup$
add a comment |
$begingroup$
Basically, you have to prove that
$$
x^alpha+y^alphage (x+y)^alpha
$$
for $x,yge 0$. If $y=0$ this is trivial. Assume it is not and divide by $y^alpha$. Calling $x/y=z$ you get the equivalent inequality
$$
z^alpha+1ge (1+z)^alpha
$$
which must be true for $zge 0$. Consider the function
$$
f(z)= z^alpha+1-(1+z)^alpha
$$
You should prove $f(z)ge 0$. But $f(0)=0$ and $f(infty)=0$. You can easily check that $f$ has only one critical point on $(0,infty)$ and it is a local maximum. Therefore, $f$ is never negative.
answered Apr 8 at 1:17
GReyesGReyes
2,43815
$endgroup$
add a comment |
$begingroup$
Basically, you have to prove that
$$
x^alpha+y^alphage (x+y)^alpha
$$
for $x,yge 0$. If $y=0$ this is trivial. Assume it is not and divide by $y^alpha$. Calling $x/y=z$ you get the equivalent inequality
$$
z^alpha+1ge (1+z)^alpha
$$
which must be true for $zge 0$. Consider the function
$$
f(z)= z^alpha+1-(1+z)^alpha
$$
You should prove $f(z)ge 0$. But $f(0)=0$ and $f(infty)=0$. You can easily check that $f$ has only one critical point on $(0,infty)$ and it is a local maximum. Therefore, $f$ is never negative.
answered Apr 8 at 1:17
GReyesGReyes
2,43815
$endgroup$
Basically, you have to prove that
$$
x^alpha+y^alphage (x+y)^alpha
$$
for $x,yge 0$. If $y=0$ this is trivial. Assume it is not and divide by $y^alpha$. Calling $x/y=z$ you get the equivalent inequality
$$
z^alpha+1ge (1+z)^alpha
$$
which must be true for $zge 0$. Consider the function
$$
f(z)= z^alpha+1-(1+z)^alpha
$$
You should prove $f(z)ge 0$. But $f(0)=0$ and $f(infty)=0$. You can easily check that $f$ has only one critical point on $(0,infty)$ and it is a local maximum. Therefore, $f$ is never negative.
answered Apr 8 at 1:17
GReyesGReyes
2,43815
edited Apr 8 at 1:26
answered Apr 8 at 1:17
GReyesGReyes
2,43815
answered Apr 8 at 1:17
GReyesGReyes
2,43815
answered Apr 8 at 1:17
GReyesGReyes
2,43815
2,43815
add a comment |
add a comment |
$begingroup$
Let $S = d(x,z) + d(z,y)$. By dividing both sides of the inequality you had by $S^alpha$, you get
$$
frac(d(x,y))^alphaS^alpha
leq left( fracd(x,z)S + fracd(z,y)S right)^alpha
leq left( fracd(x,z)S right)^alpha + left( fracd(z,y)S right)^alpha
,
$$
where we use the fact that $ fracd(x,z)S + fracd(z,y)S = 1$.
Multiplying again by $S^alpha$ gives you the result you need.
answered Apr 8 at 1:16
Hotdog2000Hotdog2000
678
$endgroup$
add a comment |
$begingroup$
Let $S = d(x,z) + d(z,y)$. By dividing both sides of the inequality you had by $S^alpha$, you get
$$
frac(d(x,y))^alphaS^alpha
leq left( fracd(x,z)S + fracd(z,y)S right)^alpha
leq left( fracd(x,z)S right)^alpha + left( fracd(z,y)S right)^alpha
,
$$
where we use the fact that $ fracd(x,z)S + fracd(z,y)S = 1$.
Multiplying again by $S^alpha$ gives you the result you need.
answered Apr 8 at 1:16
Hotdog2000Hotdog2000
678
$endgroup$
add a comment |
$begingroup$
Let $S = d(x,z) + d(z,y)$. By dividing both sides of the inequality you had by $S^alpha$, you get
$$
frac(d(x,y))^alphaS^alpha
leq left( fracd(x,z)S + fracd(z,y)S right)^alpha
leq left( fracd(x,z)S right)^alpha + left( fracd(z,y)S right)^alpha
,
$$
where we use the fact that $ fracd(x,z)S + fracd(z,y)S = 1$.
Multiplying again by $S^alpha$ gives you the result you need.
answered Apr 8 at 1:16
Hotdog2000Hotdog2000
678
$endgroup$
Let $S = d(x,z) + d(z,y)$. By dividing both sides of the inequality you had by $S^alpha$, you get
$$
frac(d(x,y))^alphaS^alpha
leq left( fracd(x,z)S + fracd(z,y)S right)^alpha
leq left( fracd(x,z)S right)^alpha + left( fracd(z,y)S right)^alpha
,
$$
where we use the fact that $ fracd(x,z)S + fracd(z,y)S = 1$.
Multiplying again by $S^alpha$ gives you the result you need.
answered Apr 8 at 1:16
Hotdog2000Hotdog2000
678
answered Apr 8 at 1:16
Hotdog2000Hotdog2000
678
answered Apr 8 at 1:16
Hotdog2000Hotdog2000
678
answered Apr 8 at 1:16
Hotdog2000Hotdog2000
678
678
add a comment |
add a comment |
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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
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Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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$begingroup$
The way to apply the hint would be to take $a=d(x,z)/big(d(x,z)+d(z,y)big)$ and $b=d(z,y)/big(d(x,z)+d(z,y)big)$. (I acknowledge that you still have to prove that the hint is true!)
$endgroup$
– Greg Martin
Apr 8 at 1:11