prove inequality $| f(x) - f(y) | < 1/16$
up vote
-4
down vote
favorite
$f: [ 4, + infty) to mathbb R$
$f(x)= 1/Ax$
Prove: $| f(x) - f(y) | le 1/16 $
I don't know what to do ...
I have no idea... Please help. I have very important test...
analysis inequality
edited Nov 25 at 17:22
Andrei
10.6k21025
asked Nov 25 at 17:16
Estera Gmiterek
306
add a comment |
up vote
-4
down vote
favorite
$f: [ 4, + infty) to mathbb R$
$f(x)= 1/Ax$
Prove: $| f(x) - f(y) | le 1/16 $
I don't know what to do ...
I have no idea... Please help. I have very important test...
analysis inequality
edited Nov 25 at 17:22
Andrei
10.6k21025
asked Nov 25 at 17:16
Estera Gmiterek
306
What is the condition on $A$ ?
– DeepSea
Nov 25 at 17:18
If it is a lipschitz condition ... A = 4
– Estera Gmiterek
Nov 25 at 17:21
1
Is it $1/(Ax)$ or $(1/A)x$?
– Andrei
Nov 25 at 17:21
add a comment |
up vote
-4
down vote
favorite
up vote
-4
down vote
favorite
$f: [ 4, + infty) to mathbb R$
$f(x)= 1/Ax$
Prove: $| f(x) - f(y) | le 1/16 $
I don't know what to do ...
I have no idea... Please help. I have very important test...
analysis inequality
edited Nov 25 at 17:22
Andrei
10.6k21025
asked Nov 25 at 17:16
Estera Gmiterek
306
$f: [ 4, + infty) to mathbb R$
$f(x)= 1/Ax$
Prove: $| f(x) - f(y) | le 1/16 $
I don't know what to do ...
I have no idea... Please help. I have very important test...
analysis inequality
analysis inequality
edited Nov 25 at 17:22
Andrei
10.6k21025
asked Nov 25 at 17:16
Estera Gmiterek
306
edited Nov 25 at 17:22
Andrei
10.6k21025
asked Nov 25 at 17:16
Estera Gmiterek
306
edited Nov 25 at 17:22
Andrei
10.6k21025
edited Nov 25 at 17:22
Andrei
10.6k21025
edited Nov 25 at 17:22
Andrei
10.6k21025
10.6k21025
asked Nov 25 at 17:16
Estera Gmiterek
306
asked Nov 25 at 17:16
Estera Gmiterek
306
asked Nov 25 at 17:16
Estera Gmiterek
306
306
What is the condition on $A$ ?
– DeepSea
Nov 25 at 17:18
If it is a lipschitz condition ... A = 4
– Estera Gmiterek
Nov 25 at 17:21
1
Is it $1/(Ax)$ or $(1/A)x$?
– Andrei
Nov 25 at 17:21
add a comment |
What is the condition on $A$ ?
– DeepSea
Nov 25 at 17:18
If it is a lipschitz condition ... A = 4
– Estera Gmiterek
Nov 25 at 17:21
1
Is it $1/(Ax)$ or $(1/A)x$?
– Andrei
Nov 25 at 17:21
What is the condition on $A$ ?
– DeepSea
Nov 25 at 17:18
What is the condition on $A$ ?
– DeepSea
Nov 25 at 17:18
If it is a lipschitz condition ... A = 4
– Estera Gmiterek
Nov 25 at 17:21
If it is a lipschitz condition ... A = 4
– Estera Gmiterek
Nov 25 at 17:21
1
1
Is it $1/(Ax)$ or $(1/A)x$?
– Andrei
Nov 25 at 17:21
Is it $1/(Ax)$ or $(1/A)x$?
– Andrei
Nov 25 at 17:21
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
Suppose $A ge 8$, then $|f(x)-f(y)| = left|dfrac{1}{Ax}-dfrac{1}{Ay}right| = dfrac{|x-y|}{Axy} le dfrac{|x|+|y|}{xyA} = dfrac{1}{A}left(dfrac{1}{x}+dfrac{1}{y}right) le dfrac{1}{A}left(dfrac{1}{4}+dfrac{1}{4}right) =dfrac{1}{2A} le dfrac{1}{16}$, as claimed.
answered Nov 25 at 17:21
DeepSea
70.7k54487
Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
– Ingix
Nov 25 at 20:59
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Suppose $A ge 8$, then $|f(x)-f(y)| = left|dfrac{1}{Ax}-dfrac{1}{Ay}right| = dfrac{|x-y|}{Axy} le dfrac{|x|+|y|}{xyA} = dfrac{1}{A}left(dfrac{1}{x}+dfrac{1}{y}right) le dfrac{1}{A}left(dfrac{1}{4}+dfrac{1}{4}right) =dfrac{1}{2A} le dfrac{1}{16}$, as claimed.
answered Nov 25 at 17:21
DeepSea
70.7k54487
Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
– Ingix
Nov 25 at 20:59
add a comment |
up vote
4
down vote
Suppose $A ge 8$, then $|f(x)-f(y)| = left|dfrac{1}{Ax}-dfrac{1}{Ay}right| = dfrac{|x-y|}{Axy} le dfrac{|x|+|y|}{xyA} = dfrac{1}{A}left(dfrac{1}{x}+dfrac{1}{y}right) le dfrac{1}{A}left(dfrac{1}{4}+dfrac{1}{4}right) =dfrac{1}{2A} le dfrac{1}{16}$, as claimed.
answered Nov 25 at 17:21
DeepSea
70.7k54487
Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
– Ingix
Nov 25 at 20:59
add a comment |
up vote
4
down vote
up vote
4
down vote
Suppose $A ge 8$, then $|f(x)-f(y)| = left|dfrac{1}{Ax}-dfrac{1}{Ay}right| = dfrac{|x-y|}{Axy} le dfrac{|x|+|y|}{xyA} = dfrac{1}{A}left(dfrac{1}{x}+dfrac{1}{y}right) le dfrac{1}{A}left(dfrac{1}{4}+dfrac{1}{4}right) =dfrac{1}{2A} le dfrac{1}{16}$, as claimed.
answered Nov 25 at 17:21
DeepSea
70.7k54487
Suppose $A ge 8$, then $|f(x)-f(y)| = left|dfrac{1}{Ax}-dfrac{1}{Ay}right| = dfrac{|x-y|}{Axy} le dfrac{|x|+|y|}{xyA} = dfrac{1}{A}left(dfrac{1}{x}+dfrac{1}{y}right) le dfrac{1}{A}left(dfrac{1}{4}+dfrac{1}{4}right) =dfrac{1}{2A} le dfrac{1}{16}$, as claimed.
answered Nov 25 at 17:21
DeepSea
70.7k54487
edited Nov 25 at 17:29
answered Nov 25 at 17:21
DeepSea
70.7k54487
answered Nov 25 at 17:21
DeepSea
70.7k54487
answered Nov 25 at 17:21
DeepSea
70.7k54487
70.7k54487
Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
– Ingix
Nov 25 at 20:59
add a comment |
Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
– Ingix
Nov 25 at 20:59
Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
– Ingix
Nov 25 at 20:59
Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
– Ingix
Nov 25 at 20:59
add a comment |
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What is the condition on $A$ ?
– DeepSea
Nov 25 at 17:18
If it is a lipschitz condition ... A = 4
– Estera Gmiterek
Nov 25 at 17:21
1
Is it $1/(Ax)$ or $(1/A)x$?
– Andrei
Nov 25 at 17:21