Using concexity property of norm function
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Let $(X,|.|)$ be a Banach space. Let $xneq 0$ and $yneq 0$ be in $X$ such that there exists a non zero $lambda_0inmathbb{R}$, $lambda_0$ to be<0, such that $|x+lambda_0y|<|x|$. Can anyone tell me how by convexity of the norm function $|x+lambda y|<|x|$
for $lambda in [lambda_0, 0).$?
functional-analysis convex-analysis
asked Dec 18 '18 at 6:39
user534666user534666
224
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add a comment |
$begingroup$
Let $(X,|.|)$ be a Banach space. Let $xneq 0$ and $yneq 0$ be in $X$ such that there exists a non zero $lambda_0inmathbb{R}$, $lambda_0$ to be<0, such that $|x+lambda_0y|<|x|$. Can anyone tell me how by convexity of the norm function $|x+lambda y|<|x|$
for $lambda in [lambda_0, 0).$?
functional-analysis convex-analysis
asked Dec 18 '18 at 6:39
user534666user534666
224
$endgroup$
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@D.B. I don't see your point. I think you misread the question
$endgroup$
– mathworker21
Dec 18 '18 at 7:03
1
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@user535666 If $f(t) := ||x+ty||$, then $f$ is convex and so $f(lambda) le alpha f(lambda_0)+(1-alpha)f(0) < ||x||$ where $alpha in (0,1]$ is s.t. $lambda = alpha lambda_0+(1-alpha)_0$.
$endgroup$
– mathworker21
Dec 18 '18 at 7:05
add a comment |
$begingroup$
Let $(X,|.|)$ be a Banach space. Let $xneq 0$ and $yneq 0$ be in $X$ such that there exists a non zero $lambda_0inmathbb{R}$, $lambda_0$ to be<0, such that $|x+lambda_0y|<|x|$. Can anyone tell me how by convexity of the norm function $|x+lambda y|<|x|$
for $lambda in [lambda_0, 0).$?
functional-analysis convex-analysis
asked Dec 18 '18 at 6:39
user534666user534666
224
$endgroup$
Let $(X,|.|)$ be a Banach space. Let $xneq 0$ and $yneq 0$ be in $X$ such that there exists a non zero $lambda_0inmathbb{R}$, $lambda_0$ to be<0, such that $|x+lambda_0y|<|x|$. Can anyone tell me how by convexity of the norm function $|x+lambda y|<|x|$
for $lambda in [lambda_0, 0).$?
functional-analysis convex-analysis
functional-analysis convex-analysis
asked Dec 18 '18 at 6:39
user534666user534666
224
asked Dec 18 '18 at 6:39
user534666user534666
224
asked Dec 18 '18 at 6:39
user534666user534666
224
asked Dec 18 '18 at 6:39
user534666user534666
224
asked Dec 18 '18 at 6:39
user534666user534666
224
224
$begingroup$
@D.B. I don't see your point. I think you misread the question
$endgroup$
– mathworker21
Dec 18 '18 at 7:03
1
$begingroup$
@user535666 If $f(t) := ||x+ty||$, then $f$ is convex and so $f(lambda) le alpha f(lambda_0)+(1-alpha)f(0) < ||x||$ where $alpha in (0,1]$ is s.t. $lambda = alpha lambda_0+(1-alpha)_0$.
$endgroup$
– mathworker21
Dec 18 '18 at 7:05
add a comment |
$begingroup$
@D.B. I don't see your point. I think you misread the question
$endgroup$
– mathworker21
Dec 18 '18 at 7:03
1
$begingroup$
@user535666 If $f(t) := ||x+ty||$, then $f$ is convex and so $f(lambda) le alpha f(lambda_0)+(1-alpha)f(0) < ||x||$ where $alpha in (0,1]$ is s.t. $lambda = alpha lambda_0+(1-alpha)_0$.
$endgroup$
– mathworker21
Dec 18 '18 at 7:05
$begingroup$
@D.B. I don't see your point. I think you misread the question
$endgroup$
– mathworker21
Dec 18 '18 at 7:03
$begingroup$
@D.B. I don't see your point. I think you misread the question
$endgroup$
– mathworker21
Dec 18 '18 at 7:03
1
1
$begingroup$
@user535666 If $f(t) := ||x+ty||$, then $f$ is convex and so $f(lambda) le alpha f(lambda_0)+(1-alpha)f(0) < ||x||$ where $alpha in (0,1]$ is s.t. $lambda = alpha lambda_0+(1-alpha)_0$.
$endgroup$
– mathworker21
Dec 18 '18 at 7:05
$begingroup$
@user535666 If $f(t) := ||x+ty||$, then $f$ is convex and so $f(lambda) le alpha f(lambda_0)+(1-alpha)f(0) < ||x||$ where $alpha in (0,1]$ is s.t. $lambda = alpha lambda_0+(1-alpha)_0$.
$endgroup$
– mathworker21
Dec 18 '18 at 7:05
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Set $y_0=lambda_0y$, then $|x+y_0|<|x|$. By triangle inequality (or the convexity property as you mention) we have, for $gammain(0,1]$,
$$
|x+gamma y_0|=|gamma(x+y_0)+(1-gamma) x|leqgamma|x+y_0|+(1-gamma)|x|<|x|,
$$
that is to say, for $lambdainlambda_0(0,1]$, we have
$$
|x+lambda y|<|x|.
$$
answered Dec 18 '18 at 7:41
JRenJRen
1845
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Set $y_0=lambda_0y$, then $|x+y_0|<|x|$. By triangle inequality (or the convexity property as you mention) we have, for $gammain(0,1]$,
$$
|x+gamma y_0|=|gamma(x+y_0)+(1-gamma) x|leqgamma|x+y_0|+(1-gamma)|x|<|x|,
$$
that is to say, for $lambdainlambda_0(0,1]$, we have
$$
|x+lambda y|<|x|.
$$
answered Dec 18 '18 at 7:41
JRenJRen
1845
$endgroup$
add a comment |
$begingroup$
Set $y_0=lambda_0y$, then $|x+y_0|<|x|$. By triangle inequality (or the convexity property as you mention) we have, for $gammain(0,1]$,
$$
|x+gamma y_0|=|gamma(x+y_0)+(1-gamma) x|leqgamma|x+y_0|+(1-gamma)|x|<|x|,
$$
that is to say, for $lambdainlambda_0(0,1]$, we have
$$
|x+lambda y|<|x|.
$$
answered Dec 18 '18 at 7:41
JRenJRen
1845
$endgroup$
add a comment |
$begingroup$
Set $y_0=lambda_0y$, then $|x+y_0|<|x|$. By triangle inequality (or the convexity property as you mention) we have, for $gammain(0,1]$,
$$
|x+gamma y_0|=|gamma(x+y_0)+(1-gamma) x|leqgamma|x+y_0|+(1-gamma)|x|<|x|,
$$
that is to say, for $lambdainlambda_0(0,1]$, we have
$$
|x+lambda y|<|x|.
$$
answered Dec 18 '18 at 7:41
JRenJRen
1845
$endgroup$
Set $y_0=lambda_0y$, then $|x+y_0|<|x|$. By triangle inequality (or the convexity property as you mention) we have, for $gammain(0,1]$,
$$
|x+gamma y_0|=|gamma(x+y_0)+(1-gamma) x|leqgamma|x+y_0|+(1-gamma)|x|<|x|,
$$
that is to say, for $lambdainlambda_0(0,1]$, we have
$$
|x+lambda y|<|x|.
$$
answered Dec 18 '18 at 7:41
JRenJRen
1845
answered Dec 18 '18 at 7:41
JRenJRen
1845
answered Dec 18 '18 at 7:41
JRenJRen
1845
answered Dec 18 '18 at 7:41
JRenJRen
1845
1845
add a comment |
add a comment |
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$begingroup$
@D.B. I don't see your point. I think you misread the question
$endgroup$
– mathworker21
Dec 18 '18 at 7:03
1
$begingroup$
@user535666 If $f(t) := ||x+ty||$, then $f$ is convex and so $f(lambda) le alpha f(lambda_0)+(1-alpha)f(0) < ||x||$ where $alpha in (0,1]$ is s.t. $lambda = alpha lambda_0+(1-alpha)_0$.
$endgroup$
– mathworker21
Dec 18 '18 at 7:05