Mixing
Norm in a Vector Space
$begingroup$
A vector space with norm $parallelcdotparallel$ Satisfy for two vectors the following
$parallel x+yparallel=parallel xparallel +parallel yparallel$
i need to proof the fallowing statement
$parallel alpha x+beta yparallel=alphaparallel xparallel +betaparallel yparallel$
for all $alphageq0$ and $betageq0$, and also this norm it's not necessary induced by an inner product
The fact that
$alphaparallel xparallel +betaparallel yparallelgeqparallel alpha x + beta yparallel$
it's clear because the property of the norm, what i can't proof it's the fact that
$alphaparallel xparallel +betaparallel yparallelleqparallel alpha x + beta yparallel$
i will appreciate any help.
analysis
asked Jan 19 at 4:32
PechPech
83
$endgroup$
add a comment |
$begingroup$
A vector space with norm $parallelcdotparallel$ Satisfy for two vectors the following
$parallel x+yparallel=parallel xparallel +parallel yparallel$
i need to proof the fallowing statement
$parallel alpha x+beta yparallel=alphaparallel xparallel +betaparallel yparallel$
for all $alphageq0$ and $betageq0$, and also this norm it's not necessary induced by an inner product
The fact that
$alphaparallel xparallel +betaparallel yparallelgeqparallel alpha x + beta yparallel$
it's clear because the property of the norm, what i can't proof it's the fact that
$alphaparallel xparallel +betaparallel yparallelleqparallel alpha x + beta yparallel$
i will appreciate any help.
analysis
asked Jan 19 at 4:32
PechPech
83
$endgroup$
$begingroup$
Hint: Use the definition of the norm $||x|| = sqrt {sum_{i=1}^{n}x_i^2}$ to verify your inequality.
$endgroup$
– Victoria M
Jan 19 at 4:36
$begingroup$
@VictoriaM I forgot to say that the norm it's not necessary induced by an inner product
$endgroup$
– Pech
Jan 19 at 4:39
add a comment |
$begingroup$
A vector space with norm $parallelcdotparallel$ Satisfy for two vectors the following
$parallel x+yparallel=parallel xparallel +parallel yparallel$
i need to proof the fallowing statement
$parallel alpha x+beta yparallel=alphaparallel xparallel +betaparallel yparallel$
for all $alphageq0$ and $betageq0$, and also this norm it's not necessary induced by an inner product
The fact that
$alphaparallel xparallel +betaparallel yparallelgeqparallel alpha x + beta yparallel$
it's clear because the property of the norm, what i can't proof it's the fact that
$alphaparallel xparallel +betaparallel yparallelleqparallel alpha x + beta yparallel$
i will appreciate any help.
analysis
asked Jan 19 at 4:32
PechPech
83
$endgroup$
A vector space with norm $parallelcdotparallel$ Satisfy for two vectors the following
$parallel x+yparallel=parallel xparallel +parallel yparallel$
i need to proof the fallowing statement
$parallel alpha x+beta yparallel=alphaparallel xparallel +betaparallel yparallel$
for all $alphageq0$ and $betageq0$, and also this norm it's not necessary induced by an inner product
The fact that
$alphaparallel xparallel +betaparallel yparallelgeqparallel alpha x + beta yparallel$
it's clear because the property of the norm, what i can't proof it's the fact that
$alphaparallel xparallel +betaparallel yparallelleqparallel alpha x + beta yparallel$
i will appreciate any help.
analysis
analysis
asked Jan 19 at 4:32
PechPech
83
asked Jan 19 at 4:32
PechPech
83
edited Jan 19 at 4:53
Pech
asked Jan 19 at 4:32
PechPech
83
asked Jan 19 at 4:32
PechPech
83
asked Jan 19 at 4:32
PechPech
83
83
$begingroup$
Hint: Use the definition of the norm $||x|| = sqrt {sum_{i=1}^{n}x_i^2}$ to verify your inequality.
$endgroup$
– Victoria M
Jan 19 at 4:36
$begingroup$
@VictoriaM I forgot to say that the norm it's not necessary induced by an inner product
$endgroup$
– Pech
Jan 19 at 4:39
add a comment |
$begingroup$
Hint: Use the definition of the norm $||x|| = sqrt {sum_{i=1}^{n}x_i^2}$ to verify your inequality.
$endgroup$
– Victoria M
Jan 19 at 4:36
$begingroup$
@VictoriaM I forgot to say that the norm it's not necessary induced by an inner product
$endgroup$
– Pech
Jan 19 at 4:39
$begingroup$
Hint: Use the definition of the norm $||x|| = sqrt {sum_{i=1}^{n}x_i^2}$ to verify your inequality.
$endgroup$
– Victoria M
Jan 19 at 4:36
$begingroup$
Hint: Use the definition of the norm $||x|| = sqrt {sum_{i=1}^{n}x_i^2}$ to verify your inequality.
$endgroup$
– Victoria M
Jan 19 at 4:36
$begingroup$
@VictoriaM I forgot to say that the norm it's not necessary induced by an inner product
$endgroup$
– Pech
Jan 19 at 4:39
$begingroup$
@VictoriaM I forgot to say that the norm it's not necessary induced by an inner product
$endgroup$
– Pech
Jan 19 at 4:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I have changed $alpha, beta$ to $a,b$. Assume without loss of generality that $bleq a$. Then $leftVert ax+byrightVert =leftVert
a(x+y)+(b-a)yrightVert geq aleftVert x+yrightVert -(a-b)leftVert
yrightVert =a(leftVert xrightVert +leftVert yrightVert
)-(a-b)leftVert yrightVert $
$=aleftVert xrightVert +bleftVert yrightVert $ and $leftVert
ax+byrightVert leq aleftVert xrightVert +bleftVert yrightVert $.
answered Jan 19 at 4:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
$begingroup$
I really appreciate the help, now i see it clear thank's :)
$endgroup$
– Pech
Jan 19 at 4:59
add a comment |
$begingroup$
To show that the norm is induced by an inner product you can show that the norm satisfies the parallelogram law and then use the polarization identity to recover the norm.
answered Jan 19 at 5:53
Mustafa SaidMustafa Said
3,0461913
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
I have changed $alpha, beta$ to $a,b$. Assume without loss of generality that $bleq a$. Then $leftVert ax+byrightVert =leftVert
a(x+y)+(b-a)yrightVert geq aleftVert x+yrightVert -(a-b)leftVert
yrightVert =a(leftVert xrightVert +leftVert yrightVert
)-(a-b)leftVert yrightVert $
$=aleftVert xrightVert +bleftVert yrightVert $ and $leftVert
ax+byrightVert leq aleftVert xrightVert +bleftVert yrightVert $.
answered Jan 19 at 4:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
$begingroup$
I really appreciate the help, now i see it clear thank's :)
$endgroup$
– Pech
Jan 19 at 4:59
add a comment |
$begingroup$
I have changed $alpha, beta$ to $a,b$. Assume without loss of generality that $bleq a$. Then $leftVert ax+byrightVert =leftVert
a(x+y)+(b-a)yrightVert geq aleftVert x+yrightVert -(a-b)leftVert
yrightVert =a(leftVert xrightVert +leftVert yrightVert
)-(a-b)leftVert yrightVert $
$=aleftVert xrightVert +bleftVert yrightVert $ and $leftVert
ax+byrightVert leq aleftVert xrightVert +bleftVert yrightVert $.
answered Jan 19 at 4:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
$begingroup$
I really appreciate the help, now i see it clear thank's :)
$endgroup$
– Pech
Jan 19 at 4:59
add a comment |
$begingroup$
I have changed $alpha, beta$ to $a,b$. Assume without loss of generality that $bleq a$. Then $leftVert ax+byrightVert =leftVert
a(x+y)+(b-a)yrightVert geq aleftVert x+yrightVert -(a-b)leftVert
yrightVert =a(leftVert xrightVert +leftVert yrightVert
)-(a-b)leftVert yrightVert $
$=aleftVert xrightVert +bleftVert yrightVert $ and $leftVert
ax+byrightVert leq aleftVert xrightVert +bleftVert yrightVert $.
answered Jan 19 at 4:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
I have changed $alpha, beta$ to $a,b$. Assume without loss of generality that $bleq a$. Then $leftVert ax+byrightVert =leftVert
a(x+y)+(b-a)yrightVert geq aleftVert x+yrightVert -(a-b)leftVert
yrightVert =a(leftVert xrightVert +leftVert yrightVert
)-(a-b)leftVert yrightVert $
$=aleftVert xrightVert +bleftVert yrightVert $ and $leftVert
ax+byrightVert leq aleftVert xrightVert +bleftVert yrightVert $.
answered Jan 19 at 4:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 19 at 4:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 19 at 4:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 19 at 4:51
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
63.7k42463
$begingroup$
I really appreciate the help, now i see it clear thank's :)
$endgroup$
– Pech
Jan 19 at 4:59
add a comment |
$begingroup$
I really appreciate the help, now i see it clear thank's :)
$endgroup$
– Pech
Jan 19 at 4:59
$begingroup$
I really appreciate the help, now i see it clear thank's :)
$endgroup$
– Pech
Jan 19 at 4:59
$begingroup$
I really appreciate the help, now i see it clear thank's :)
$endgroup$
– Pech
Jan 19 at 4:59
add a comment |
$begingroup$
To show that the norm is induced by an inner product you can show that the norm satisfies the parallelogram law and then use the polarization identity to recover the norm.
answered Jan 19 at 5:53
Mustafa SaidMustafa Said
3,0461913
$endgroup$
add a comment |
$begingroup$
To show that the norm is induced by an inner product you can show that the norm satisfies the parallelogram law and then use the polarization identity to recover the norm.
answered Jan 19 at 5:53
Mustafa SaidMustafa Said
3,0461913
$endgroup$
add a comment |
$begingroup$
To show that the norm is induced by an inner product you can show that the norm satisfies the parallelogram law and then use the polarization identity to recover the norm.
answered Jan 19 at 5:53
Mustafa SaidMustafa Said
3,0461913
$endgroup$
To show that the norm is induced by an inner product you can show that the norm satisfies the parallelogram law and then use the polarization identity to recover the norm.
answered Jan 19 at 5:53
Mustafa SaidMustafa Said
3,0461913
answered Jan 19 at 5:53
Mustafa SaidMustafa Said
3,0461913
answered Jan 19 at 5:53
Mustafa SaidMustafa Said
3,0461913
answered Jan 19 at 5:53
Mustafa SaidMustafa Said
3,0461913
3,0461913
add a comment |
add a comment |
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$begingroup$
Hint: Use the definition of the norm $||x|| = sqrt {sum_{i=1}^{n}x_i^2}$ to verify your inequality.
$endgroup$
– Victoria M
Jan 19 at 4:36
$begingroup$
@VictoriaM I forgot to say that the norm it's not necessary induced by an inner product
$endgroup$
– Pech
Jan 19 at 4:39