Mixing
Norms and Parallelogram Identity
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I have recently started learning about Norms and Inner Products. I have came across the idea that for an inner product to introduce a norm the parallelogram Identity must be true.
The proof my books gives is just that it is apparently easy to see that the parallelogram identity fails when the norm, p is not = 2. Also that you can easily find a counterexample in R^2
I am quite confused here any help would be great.
Thanks
linear-algebra normed-spaces inner-product-space
asked Jan 8 at 21:30
jake walshjake walsh
808
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add a comment |
$begingroup$
I have recently started learning about Norms and Inner Products. I have came across the idea that for an inner product to introduce a norm the parallelogram Identity must be true.
The proof my books gives is just that it is apparently easy to see that the parallelogram identity fails when the norm, p is not = 2. Also that you can easily find a counterexample in R^2
I am quite confused here any help would be great.
Thanks
linear-algebra normed-spaces inner-product-space
asked Jan 8 at 21:30
jake walshjake walsh
808
$endgroup$
add a comment |
$begingroup$
I have recently started learning about Norms and Inner Products. I have came across the idea that for an inner product to introduce a norm the parallelogram Identity must be true.
The proof my books gives is just that it is apparently easy to see that the parallelogram identity fails when the norm, p is not = 2. Also that you can easily find a counterexample in R^2
I am quite confused here any help would be great.
Thanks
linear-algebra normed-spaces inner-product-space
asked Jan 8 at 21:30
jake walshjake walsh
808
$endgroup$
I have recently started learning about Norms and Inner Products. I have came across the idea that for an inner product to introduce a norm the parallelogram Identity must be true.
The proof my books gives is just that it is apparently easy to see that the parallelogram identity fails when the norm, p is not = 2. Also that you can easily find a counterexample in R^2
I am quite confused here any help would be great.
Thanks
linear-algebra normed-spaces inner-product-space
linear-algebra normed-spaces inner-product-space
asked Jan 8 at 21:30
jake walshjake walsh
808
asked Jan 8 at 21:30
jake walshjake walsh
808
edited Jan 8 at 21:46
jake walsh
asked Jan 8 at 21:30
jake walshjake walsh
808
asked Jan 8 at 21:30
jake walshjake walsh
808
asked Jan 8 at 21:30
jake walshjake walsh
808
808
add a comment |
add a comment |
1 Answer
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In $mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:
Let $p geq 1$, and $VertcdotVert_p$ denote the $p$-norm. Let $x,y in mathbb R$. Then
$$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = Vert x+yVert_p^2 + Vert x-yVert_p^2.$$
However, in general, the parallelogram identity will fail for $p neq 2$. Let $n > 1$. Then $mathbb R^2$ sits in $mathbb R^n$. Therefore, by finding a counter-example in $mathbb R^2$, we prove that the identity fails for $mathbb R^n$.
Let $x = (1,0), y = (0,1)$. Let $p > 1, p neq 2$. Then
$$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$
However,
$$Vert x+ yVert_p^2 + Vert x - yVert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2times4^{1/p}.$$
Clearly $2times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p neq 2$.
answered Jan 8 at 23:50
HarambeHarambe
6,06321843
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1 Answer
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1 Answer
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$begingroup$
In $mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:
Let $p geq 1$, and $VertcdotVert_p$ denote the $p$-norm. Let $x,y in mathbb R$. Then
$$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = Vert x+yVert_p^2 + Vert x-yVert_p^2.$$
However, in general, the parallelogram identity will fail for $p neq 2$. Let $n > 1$. Then $mathbb R^2$ sits in $mathbb R^n$. Therefore, by finding a counter-example in $mathbb R^2$, we prove that the identity fails for $mathbb R^n$.
Let $x = (1,0), y = (0,1)$. Let $p > 1, p neq 2$. Then
$$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$
However,
$$Vert x+ yVert_p^2 + Vert x - yVert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2times4^{1/p}.$$
Clearly $2times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p neq 2$.
answered Jan 8 at 23:50
HarambeHarambe
6,06321843
$endgroup$
add a comment |
$begingroup$
In $mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:
Let $p geq 1$, and $VertcdotVert_p$ denote the $p$-norm. Let $x,y in mathbb R$. Then
$$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = Vert x+yVert_p^2 + Vert x-yVert_p^2.$$
However, in general, the parallelogram identity will fail for $p neq 2$. Let $n > 1$. Then $mathbb R^2$ sits in $mathbb R^n$. Therefore, by finding a counter-example in $mathbb R^2$, we prove that the identity fails for $mathbb R^n$.
Let $x = (1,0), y = (0,1)$. Let $p > 1, p neq 2$. Then
$$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$
However,
$$Vert x+ yVert_p^2 + Vert x - yVert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2times4^{1/p}.$$
Clearly $2times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p neq 2$.
answered Jan 8 at 23:50
HarambeHarambe
6,06321843
$endgroup$
add a comment |
$begingroup$
In $mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:
Let $p geq 1$, and $VertcdotVert_p$ denote the $p$-norm. Let $x,y in mathbb R$. Then
$$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = Vert x+yVert_p^2 + Vert x-yVert_p^2.$$
However, in general, the parallelogram identity will fail for $p neq 2$. Let $n > 1$. Then $mathbb R^2$ sits in $mathbb R^n$. Therefore, by finding a counter-example in $mathbb R^2$, we prove that the identity fails for $mathbb R^n$.
Let $x = (1,0), y = (0,1)$. Let $p > 1, p neq 2$. Then
$$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$
However,
$$Vert x+ yVert_p^2 + Vert x - yVert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2times4^{1/p}.$$
Clearly $2times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p neq 2$.
answered Jan 8 at 23:50
HarambeHarambe
6,06321843
$endgroup$
In $mathbb R$, the parallelogram identity will always hold for any $p$-norm, as I am about to show:
Let $p geq 1$, and $VertcdotVert_p$ denote the $p$-norm. Let $x,y in mathbb R$. Then
$$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2x^2 + 2y^2 = (x+y)^2 + (x-y)^2 = Vert x+yVert_p^2 + Vert x-yVert_p^2.$$
However, in general, the parallelogram identity will fail for $p neq 2$. Let $n > 1$. Then $mathbb R^2$ sits in $mathbb R^n$. Therefore, by finding a counter-example in $mathbb R^2$, we prove that the identity fails for $mathbb R^n$.
Let $x = (1,0), y = (0,1)$. Let $p > 1, p neq 2$. Then
$$2Vert xVert_p^2 + 2Vert yVert_p^2 = 2(0^p+1^p)^{2/p} + 2(1^p + 0^p)^{2/p} = 4.$$
However,
$$Vert x+ yVert_p^2 + Vert x - yVert_p^2 = (1^p+1^p)^{2/p} + (1^p + 1^p)^{2/p} = 2times4^{1/p}.$$
Clearly $2times4^{1/p} = 4$ only if $1/p = 1/2$. But this is not the case, so we have found a counterexample to the parallelogram identity for $p neq 2$.
answered Jan 8 at 23:50
HarambeHarambe
6,06321843
answered Jan 8 at 23:50
HarambeHarambe
6,06321843
answered Jan 8 at 23:50
HarambeHarambe
6,06321843
answered Jan 8 at 23:50
HarambeHarambe
6,06321843
6,06321843
add a comment |
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