Inequality between normed vectors The Next CEO of Stack OverflowProve that the norm of $E$ is generate by the inner product $langle x,y rangle =frac14left(||x+y|^2-||x-y||^2right)$Linearly independent vectors in normed vector spaceClosure of a subset of normed vector spaceInequality in normed space with normalizationConvergence of vectors in a normed vector space.Showing an Isometry between normed space.Normed vector space inequality $||x|^2 - |y|^2| le |x-y||x+y|$Open ball in a normed vector spaceCoset is open in a normed linear spaceThe span of a finite number of vectors in a normed vector space is closed
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Inequality between normed vectors
The Next CEO of Stack OverflowProve that the norm of $E$ is generate by the inner product $langle x,y rangle =frac14left(||x+y|^2-||x-y||^2right)$Linearly independent vectors in normed vector spaceClosure of a subset of normed vector spaceInequality in normed space with normalizationConvergence of vectors in a normed vector space.Showing an Isometry between normed space.Normed vector space inequality $||x|^2 - |y|^2| le |x-y||x+y|$Open ball in a normed vector spaceCoset is open in a normed linear spaceThe span of a finite number of vectors in a normed vector space is closed
$begingroup$
Let $x,y$ be two vectors of an normed vector space $E$. Prove that
$$left|fracxx-fracyright|leq2fracx$$
Any hint/help?
real-analysis
edited Mar 28 at 1:24
Brian
1,208116
asked Mar 28 at 1:18
pin_rpin_r
156
$endgroup$
add a comment |
$begingroup$
Let $x,y$ be two vectors of an normed vector space $E$. Prove that
$$left|fracxx-fracyright|leq2fracx$$
Any hint/help?
real-analysis
edited Mar 28 at 1:24
Brian
1,208116
asked Mar 28 at 1:18
pin_rpin_r
156
$endgroup$
1
$begingroup$
Hint: do you know the triangle inequality?
$endgroup$
– Alex
Mar 28 at 1:20
$begingroup$
First multiply the whole thing by $|x|$ to get an equivalent inequality. Then prove this inequality by inserting $-y+y$ and the usage of both the usual triangle inequality $|u+v|le|u|+|v|$ and the inverse one: $||u|-|v||le|u-v|$.
$endgroup$
– amsmath
Mar 28 at 1:28
add a comment |
$begingroup$
Let $x,y$ be two vectors of an normed vector space $E$. Prove that
$$left|fracxx-fracyright|leq2fracx$$
Any hint/help?
real-analysis
edited Mar 28 at 1:24
Brian
1,208116
asked Mar 28 at 1:18
pin_rpin_r
156
$endgroup$
Let $x,y$ be two vectors of an normed vector space $E$. Prove that
$$left|fracxx-fracyright|leq2fracx$$
Any hint/help?
real-analysis
real-analysis
edited Mar 28 at 1:24
Brian
1,208116
asked Mar 28 at 1:18
pin_rpin_r
156
edited Mar 28 at 1:24
Brian
1,208116
asked Mar 28 at 1:18
pin_rpin_r
156
edited Mar 28 at 1:24
Brian
1,208116
edited Mar 28 at 1:24
Brian
1,208116
edited Mar 28 at 1:24
Brian
1,208116
1,208116
asked Mar 28 at 1:18
pin_rpin_r
156
asked Mar 28 at 1:18
pin_rpin_r
156
asked Mar 28 at 1:18
pin_rpin_r
156
156
1
$begingroup$
Hint: do you know the triangle inequality?
$endgroup$
– Alex
Mar 28 at 1:20
$begingroup$
First multiply the whole thing by $|x|$ to get an equivalent inequality. Then prove this inequality by inserting $-y+y$ and the usage of both the usual triangle inequality $|u+v|le|u|+|v|$ and the inverse one: $||u|-|v||le|u-v|$.
$endgroup$
– amsmath
Mar 28 at 1:28
add a comment |
1
$begingroup$
Hint: do you know the triangle inequality?
$endgroup$
– Alex
Mar 28 at 1:20
$begingroup$
First multiply the whole thing by $|x|$ to get an equivalent inequality. Then prove this inequality by inserting $-y+y$ and the usage of both the usual triangle inequality $|u+v|le|u|+|v|$ and the inverse one: $||u|-|v||le|u-v|$.
$endgroup$
– amsmath
Mar 28 at 1:28
1
1
$begingroup$
Hint: do you know the triangle inequality?
$endgroup$
– Alex
Mar 28 at 1:20
$begingroup$
Hint: do you know the triangle inequality?
$endgroup$
– Alex
Mar 28 at 1:20
$begingroup$
First multiply the whole thing by $|x|$ to get an equivalent inequality. Then prove this inequality by inserting $-y+y$ and the usage of both the usual triangle inequality $|u+v|le|u|+|v|$ and the inverse one: $||u|-|v||le|u-v|$.
$endgroup$
– amsmath
Mar 28 at 1:28
$begingroup$
First multiply the whole thing by $|x|$ to get an equivalent inequality. Then prove this inequality by inserting $-y+y$ and the usage of both the usual triangle inequality $|u+v|le|u|+|v|$ and the inverse one: $||u|-|v||le|u-v|$.
$endgroup$
– amsmath
Mar 28 at 1:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$|xover x-yover
|leq |xover x-yover x|+|yover x-yover |$
$|yover x-yover |$=
$|y||(-overx)||=|-overx|leq -overx.$
answered Mar 28 at 1:30
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
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$begingroup$
$|xover x-yover
|leq |xover x-yover x|+|yover x-yover |$
$|yover x-yover |$=
$|y||(-overx)||=|-overx|leq -overx.$
answered Mar 28 at 1:30
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
$|xover x-yover
|leq |xover x-yover x|+|yover x-yover |$
$|yover x-yover |$=
$|y||(-overx)||=|-overx|leq -overx.$
answered Mar 28 at 1:30
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
$|xover x-yover
|leq |xover x-yover x|+|yover x-yover |$
$|yover x-yover |$=
$|y||(-overx)||=|-overx|leq -overx.$
answered Mar 28 at 1:30
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
$|xover x-yover
|leq |xover x-yover x|+|yover x-yover |$
$|yover x-yover |$=
$|y||(-overx)||=|-overx|leq -overx.$
answered Mar 28 at 1:30
Tsemo AristideTsemo Aristide
60.1k11446
answered Mar 28 at 1:30
Tsemo AristideTsemo Aristide
60.1k11446
answered Mar 28 at 1:30
Tsemo AristideTsemo Aristide
60.1k11446
answered Mar 28 at 1:30
Tsemo AristideTsemo Aristide
60.1k11446
60.1k11446
add a comment |
add a comment |
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1
$begingroup$
Hint: do you know the triangle inequality?
$endgroup$
– Alex
Mar 28 at 1:20
$begingroup$
First multiply the whole thing by $|x|$ to get an equivalent inequality. Then prove this inequality by inserting $-y+y$ and the usage of both the usual triangle inequality $|u+v|le|u|+|v|$ and the inverse one: $||u|-|v||le|u-v|$.
$endgroup$
– amsmath
Mar 28 at 1:28