Calculating angles - what's my mistake?
$begingroup$
Let $C[a,b]$ be the space of all continuous functions $[a,b]rightarrow mathbb R$. Then, functional
$$||f||=int^b_a|f(x)| dx$$
Clearly satisfies all the axioms of a norm, with its corresponding dot product given by: $left<f,gright>=frac12(||f+g||^2-||f||^2-||g||^2)$
Then, $frac{left<f,gright>}{||f||cdot||g||}$ should give the angle between $f$ ang $g$. But, if $f,g>0$, we have:
$left<f,gright>overset{def}=frac12((int^b_af(x)+g(x) dx)^2-(int_a^b f(x) dx)^2-(int_a^bg(x) dx)^2)=int_a^bf(x)dxint_a^bg(x)dx$
$||f||cdot||g||overset{def}=int_a^b f(x) dx int_a^bg(x) dx$
$frac{left<f,gright>}{||f||cdot||g||}=1$
So it would mean that all continouous functions with positive values are parallel to each other in this norm, which is nonsense.
Where's my mistake? How can I properly calculate angles in $(C[a,b],||cdot||)$? I know that the error must be something trivial, but I cannot find it.
Edit: Let $v,w in V: ||v||=||w||$. I know that element of $text{span}(v)$ closest to $w$ is in the form $theta v$, $theta in [0,1]$. If dot product is not a legit option, what is a general way of finding $theta$?
real-analysis vector-spaces
asked Jan 31 at 16:29
Michał ZapałaMichał Zapała
13812
$endgroup$
add a comment |
$begingroup$
Let $C[a,b]$ be the space of all continuous functions $[a,b]rightarrow mathbb R$. Then, functional
$$||f||=int^b_a|f(x)| dx$$
Clearly satisfies all the axioms of a norm, with its corresponding dot product given by: $left<f,gright>=frac12(||f+g||^2-||f||^2-||g||^2)$
Then, $frac{left<f,gright>}{||f||cdot||g||}$ should give the angle between $f$ ang $g$. But, if $f,g>0$, we have:
$left<f,gright>overset{def}=frac12((int^b_af(x)+g(x) dx)^2-(int_a^b f(x) dx)^2-(int_a^bg(x) dx)^2)=int_a^bf(x)dxint_a^bg(x)dx$
$||f||cdot||g||overset{def}=int_a^b f(x) dx int_a^bg(x) dx$
$frac{left<f,gright>}{||f||cdot||g||}=1$
So it would mean that all continouous functions with positive values are parallel to each other in this norm, which is nonsense.
Where's my mistake? How can I properly calculate angles in $(C[a,b],||cdot||)$? I know that the error must be something trivial, but I cannot find it.
Edit: Let $v,w in V: ||v||=||w||$. I know that element of $text{span}(v)$ closest to $w$ is in the form $theta v$, $theta in [0,1]$. If dot product is not a legit option, what is a general way of finding $theta$?
real-analysis vector-spaces
asked Jan 31 at 16:29
Michał ZapałaMichał Zapała
13812
$endgroup$
add a comment |
$begingroup$
Let $C[a,b]$ be the space of all continuous functions $[a,b]rightarrow mathbb R$. Then, functional
$$||f||=int^b_a|f(x)| dx$$
Clearly satisfies all the axioms of a norm, with its corresponding dot product given by: $left<f,gright>=frac12(||f+g||^2-||f||^2-||g||^2)$
Then, $frac{left<f,gright>}{||f||cdot||g||}$ should give the angle between $f$ ang $g$. But, if $f,g>0$, we have:
$left<f,gright>overset{def}=frac12((int^b_af(x)+g(x) dx)^2-(int_a^b f(x) dx)^2-(int_a^bg(x) dx)^2)=int_a^bf(x)dxint_a^bg(x)dx$
$||f||cdot||g||overset{def}=int_a^b f(x) dx int_a^bg(x) dx$
$frac{left<f,gright>}{||f||cdot||g||}=1$
So it would mean that all continouous functions with positive values are parallel to each other in this norm, which is nonsense.
Where's my mistake? How can I properly calculate angles in $(C[a,b],||cdot||)$? I know that the error must be something trivial, but I cannot find it.
Edit: Let $v,w in V: ||v||=||w||$. I know that element of $text{span}(v)$ closest to $w$ is in the form $theta v$, $theta in [0,1]$. If dot product is not a legit option, what is a general way of finding $theta$?
real-analysis vector-spaces
asked Jan 31 at 16:29
Michał ZapałaMichał Zapała
13812
$endgroup$
Let $C[a,b]$ be the space of all continuous functions $[a,b]rightarrow mathbb R$. Then, functional
$$||f||=int^b_a|f(x)| dx$$
Clearly satisfies all the axioms of a norm, with its corresponding dot product given by: $left<f,gright>=frac12(||f+g||^2-||f||^2-||g||^2)$
Then, $frac{left<f,gright>}{||f||cdot||g||}$ should give the angle between $f$ ang $g$. But, if $f,g>0$, we have:
$left<f,gright>overset{def}=frac12((int^b_af(x)+g(x) dx)^2-(int_a^b f(x) dx)^2-(int_a^bg(x) dx)^2)=int_a^bf(x)dxint_a^bg(x)dx$
$||f||cdot||g||overset{def}=int_a^b f(x) dx int_a^bg(x) dx$
$frac{left<f,gright>}{||f||cdot||g||}=1$
So it would mean that all continouous functions with positive values are parallel to each other in this norm, which is nonsense.
Where's my mistake? How can I properly calculate angles in $(C[a,b],||cdot||)$? I know that the error must be something trivial, but I cannot find it.
Edit: Let $v,w in V: ||v||=||w||$. I know that element of $text{span}(v)$ closest to $w$ is in the form $theta v$, $theta in [0,1]$. If dot product is not a legit option, what is a general way of finding $theta$?
real-analysis vector-spaces
real-analysis vector-spaces
asked Jan 31 at 16:29
Michał ZapałaMichał Zapała
13812
asked Jan 31 at 16:29
Michał ZapałaMichał Zapała
13812
edited Jan 31 at 17:27
Michał Zapała
asked Jan 31 at 16:29
Michał ZapałaMichał Zapała
13812
asked Jan 31 at 16:29
Michał ZapałaMichał Zapała
13812
asked Jan 31 at 16:29
Michał ZapałaMichał Zapała
13812
13812
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Not every norm is compatible with an inner product, and you have discovered an example.
Here's a simpler but very closely related example. Define the sup norm on $mathbb R^2$:
$$|(x,y)| = |x| + |y|
$$
Suppose one adopts the same formula for the inner product of two vectors as given in your question. If you then take two positive vectors $vec v_1 = (x_1,y_1)$ and $vec v_2 = (x_2,y_2)$, you will similarly discover that the quantity $frac{langle vec v_1,vec v_2 rangle}{|v_1| |v_2|}$ misbehaves.
You can see what is happening on a geometric level by examining the unit ball of the norm $|(x,y) | = |x| + |y|$: it is a diamond, not an ellipse as it would be if it were defined by a positive definite inner product. On an algebraic level, the explanation is that your inner product is not bilinear, as is said by @csprun.
answered Jan 31 at 16:51
Lee MosherLee Mosher
49.2k33685
$endgroup$
2
$begingroup$
To add to this, what makes a norm compatible with an inner product is the parallelogram law. If there is an inner product, you can easily prove this identity. However if you look at your norm, it is not hard to find two points for which it does not hold, thus there cannot be an inner product.
$endgroup$
– mlk
Jan 31 at 21:39
add a comment |
$begingroup$
You dropped absolute values in your computation of $langle f,grangle$ and $||f||cdot||g||$, but when you put them back in, the dot product is not bilinear, which is a problem. And if you do leave them out, then it is not positive-definite:
$$||f|| = sqrt{langle f,f rangle} = sqrt{left(int_a^b f(x) , dxright)^2} = 0$$
does not imply $f=0$.
You might want
$$||f|| := int_a^b f(x)^2 , dx$$
and
$$langle f,g rangle := int_a^b f(x)g(x) , dx.$$
answered Jan 31 at 16:44
cspruncsprun
1,04318
$endgroup$
add a comment |
$begingroup$
The scalar product you introduce is not a scalar product.
answered Jan 31 at 16:46
AlessioDVAlessioDV
2387
$endgroup$
$begingroup$
While true, this doesn't really provide any additional information beyond what is stated in the other two answers. @AlessioDV
$endgroup$
– Tyberius
Jan 31 at 20:13
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not every norm is compatible with an inner product, and you have discovered an example.
Here's a simpler but very closely related example. Define the sup norm on $mathbb R^2$:
$$|(x,y)| = |x| + |y|
$$
Suppose one adopts the same formula for the inner product of two vectors as given in your question. If you then take two positive vectors $vec v_1 = (x_1,y_1)$ and $vec v_2 = (x_2,y_2)$, you will similarly discover that the quantity $frac{langle vec v_1,vec v_2 rangle}{|v_1| |v_2|}$ misbehaves.
You can see what is happening on a geometric level by examining the unit ball of the norm $|(x,y) | = |x| + |y|$: it is a diamond, not an ellipse as it would be if it were defined by a positive definite inner product. On an algebraic level, the explanation is that your inner product is not bilinear, as is said by @csprun.
answered Jan 31 at 16:51
Lee MosherLee Mosher
49.2k33685
$endgroup$
2
$begingroup$
To add to this, what makes a norm compatible with an inner product is the parallelogram law. If there is an inner product, you can easily prove this identity. However if you look at your norm, it is not hard to find two points for which it does not hold, thus there cannot be an inner product.
$endgroup$
– mlk
Jan 31 at 21:39
add a comment |
$begingroup$
Not every norm is compatible with an inner product, and you have discovered an example.
Here's a simpler but very closely related example. Define the sup norm on $mathbb R^2$:
$$|(x,y)| = |x| + |y|
$$
Suppose one adopts the same formula for the inner product of two vectors as given in your question. If you then take two positive vectors $vec v_1 = (x_1,y_1)$ and $vec v_2 = (x_2,y_2)$, you will similarly discover that the quantity $frac{langle vec v_1,vec v_2 rangle}{|v_1| |v_2|}$ misbehaves.
You can see what is happening on a geometric level by examining the unit ball of the norm $|(x,y) | = |x| + |y|$: it is a diamond, not an ellipse as it would be if it were defined by a positive definite inner product. On an algebraic level, the explanation is that your inner product is not bilinear, as is said by @csprun.
answered Jan 31 at 16:51
Lee MosherLee Mosher
49.2k33685
$endgroup$
2
$begingroup$
To add to this, what makes a norm compatible with an inner product is the parallelogram law. If there is an inner product, you can easily prove this identity. However if you look at your norm, it is not hard to find two points for which it does not hold, thus there cannot be an inner product.
$endgroup$
– mlk
Jan 31 at 21:39
add a comment |
$begingroup$
Not every norm is compatible with an inner product, and you have discovered an example.
Here's a simpler but very closely related example. Define the sup norm on $mathbb R^2$:
$$|(x,y)| = |x| + |y|
$$
Suppose one adopts the same formula for the inner product of two vectors as given in your question. If you then take two positive vectors $vec v_1 = (x_1,y_1)$ and $vec v_2 = (x_2,y_2)$, you will similarly discover that the quantity $frac{langle vec v_1,vec v_2 rangle}{|v_1| |v_2|}$ misbehaves.
You can see what is happening on a geometric level by examining the unit ball of the norm $|(x,y) | = |x| + |y|$: it is a diamond, not an ellipse as it would be if it were defined by a positive definite inner product. On an algebraic level, the explanation is that your inner product is not bilinear, as is said by @csprun.
answered Jan 31 at 16:51
Lee MosherLee Mosher
49.2k33685
$endgroup$
Not every norm is compatible with an inner product, and you have discovered an example.
Here's a simpler but very closely related example. Define the sup norm on $mathbb R^2$:
$$|(x,y)| = |x| + |y|
$$
Suppose one adopts the same formula for the inner product of two vectors as given in your question. If you then take two positive vectors $vec v_1 = (x_1,y_1)$ and $vec v_2 = (x_2,y_2)$, you will similarly discover that the quantity $frac{langle vec v_1,vec v_2 rangle}{|v_1| |v_2|}$ misbehaves.
You can see what is happening on a geometric level by examining the unit ball of the norm $|(x,y) | = |x| + |y|$: it is a diamond, not an ellipse as it would be if it were defined by a positive definite inner product. On an algebraic level, the explanation is that your inner product is not bilinear, as is said by @csprun.
answered Jan 31 at 16:51
Lee MosherLee Mosher
49.2k33685
edited Jan 31 at 17:01
answered Jan 31 at 16:51
Lee MosherLee Mosher
49.2k33685
answered Jan 31 at 16:51
Lee MosherLee Mosher
49.2k33685
answered Jan 31 at 16:51
Lee MosherLee Mosher
49.2k33685
49.2k33685
2
$begingroup$
To add to this, what makes a norm compatible with an inner product is the parallelogram law. If there is an inner product, you can easily prove this identity. However if you look at your norm, it is not hard to find two points for which it does not hold, thus there cannot be an inner product.
$endgroup$
– mlk
Jan 31 at 21:39
add a comment |
2
$begingroup$
To add to this, what makes a norm compatible with an inner product is the parallelogram law. If there is an inner product, you can easily prove this identity. However if you look at your norm, it is not hard to find two points for which it does not hold, thus there cannot be an inner product.
$endgroup$
– mlk
Jan 31 at 21:39
2
2
$begingroup$
To add to this, what makes a norm compatible with an inner product is the parallelogram law. If there is an inner product, you can easily prove this identity. However if you look at your norm, it is not hard to find two points for which it does not hold, thus there cannot be an inner product.
$endgroup$
– mlk
Jan 31 at 21:39
$begingroup$
To add to this, what makes a norm compatible with an inner product is the parallelogram law. If there is an inner product, you can easily prove this identity. However if you look at your norm, it is not hard to find two points for which it does not hold, thus there cannot be an inner product.
$endgroup$
– mlk
Jan 31 at 21:39
add a comment |
$begingroup$
You dropped absolute values in your computation of $langle f,grangle$ and $||f||cdot||g||$, but when you put them back in, the dot product is not bilinear, which is a problem. And if you do leave them out, then it is not positive-definite:
$$||f|| = sqrt{langle f,f rangle} = sqrt{left(int_a^b f(x) , dxright)^2} = 0$$
does not imply $f=0$.
You might want
$$||f|| := int_a^b f(x)^2 , dx$$
and
$$langle f,g rangle := int_a^b f(x)g(x) , dx.$$
answered Jan 31 at 16:44
cspruncsprun
1,04318
$endgroup$
add a comment |
$begingroup$
You dropped absolute values in your computation of $langle f,grangle$ and $||f||cdot||g||$, but when you put them back in, the dot product is not bilinear, which is a problem. And if you do leave them out, then it is not positive-definite:
$$||f|| = sqrt{langle f,f rangle} = sqrt{left(int_a^b f(x) , dxright)^2} = 0$$
does not imply $f=0$.
You might want
$$||f|| := int_a^b f(x)^2 , dx$$
and
$$langle f,g rangle := int_a^b f(x)g(x) , dx.$$
answered Jan 31 at 16:44
cspruncsprun
1,04318
$endgroup$
add a comment |
$begingroup$
You dropped absolute values in your computation of $langle f,grangle$ and $||f||cdot||g||$, but when you put them back in, the dot product is not bilinear, which is a problem. And if you do leave them out, then it is not positive-definite:
$$||f|| = sqrt{langle f,f rangle} = sqrt{left(int_a^b f(x) , dxright)^2} = 0$$
does not imply $f=0$.
You might want
$$||f|| := int_a^b f(x)^2 , dx$$
and
$$langle f,g rangle := int_a^b f(x)g(x) , dx.$$
answered Jan 31 at 16:44
cspruncsprun
1,04318
$endgroup$
You dropped absolute values in your computation of $langle f,grangle$ and $||f||cdot||g||$, but when you put them back in, the dot product is not bilinear, which is a problem. And if you do leave them out, then it is not positive-definite:
$$||f|| = sqrt{langle f,f rangle} = sqrt{left(int_a^b f(x) , dxright)^2} = 0$$
does not imply $f=0$.
You might want
$$||f|| := int_a^b f(x)^2 , dx$$
and
$$langle f,g rangle := int_a^b f(x)g(x) , dx.$$
answered Jan 31 at 16:44
cspruncsprun
1,04318
edited Jan 31 at 16:50
answered Jan 31 at 16:44
cspruncsprun
1,04318
answered Jan 31 at 16:44
cspruncsprun
1,04318
answered Jan 31 at 16:44
cspruncsprun
1,04318
1,04318
add a comment |
add a comment |
$begingroup$
The scalar product you introduce is not a scalar product.
answered Jan 31 at 16:46
AlessioDVAlessioDV
2387
$endgroup$
$begingroup$
While true, this doesn't really provide any additional information beyond what is stated in the other two answers. @AlessioDV
$endgroup$
– Tyberius
Jan 31 at 20:13
add a comment |
$begingroup$
The scalar product you introduce is not a scalar product.
answered Jan 31 at 16:46
AlessioDVAlessioDV
2387
$endgroup$
$begingroup$
While true, this doesn't really provide any additional information beyond what is stated in the other two answers. @AlessioDV
$endgroup$
– Tyberius
Jan 31 at 20:13
add a comment |
$begingroup$
The scalar product you introduce is not a scalar product.
answered Jan 31 at 16:46
AlessioDVAlessioDV
2387
$endgroup$
The scalar product you introduce is not a scalar product.
answered Jan 31 at 16:46
AlessioDVAlessioDV
2387
answered Jan 31 at 16:46
AlessioDVAlessioDV
2387
answered Jan 31 at 16:46
AlessioDVAlessioDV
2387
answered Jan 31 at 16:46
AlessioDVAlessioDV
2387
2387
$begingroup$
While true, this doesn't really provide any additional information beyond what is stated in the other two answers. @AlessioDV
$endgroup$
– Tyberius
Jan 31 at 20:13
add a comment |
$begingroup$
While true, this doesn't really provide any additional information beyond what is stated in the other two answers. @AlessioDV
$endgroup$
– Tyberius
Jan 31 at 20:13
$begingroup$
While true, this doesn't really provide any additional information beyond what is stated in the other two answers. @AlessioDV
$endgroup$
– Tyberius
Jan 31 at 20:13
$begingroup$
While true, this doesn't really provide any additional information beyond what is stated in the other two answers. @AlessioDV
$endgroup$
– Tyberius
Jan 31 at 20:13
add a comment |
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