L2 and Scalar Product
$begingroup$
I struggle to show that for $, f,gin L^2(X,A,lambda)$ this is a scalar product:
$
langle f,grangle := int fg , dlambda
$
It wasn't hard to show that the length is positive, but I struggle with the symmetry and the linearity.
calculus measure-theory lebesgue-measure inner-product-space
asked Jan 7 at 13:55
KingDingelingKingDingeling
1857
$endgroup$
add a comment |
$begingroup$
I struggle to show that for $, f,gin L^2(X,A,lambda)$ this is a scalar product:
$
langle f,grangle := int fg , dlambda
$
It wasn't hard to show that the length is positive, but I struggle with the symmetry and the linearity.
calculus measure-theory lebesgue-measure inner-product-space
asked Jan 7 at 13:55
KingDingelingKingDingeling
1857
$endgroup$
add a comment |
$begingroup$
I struggle to show that for $, f,gin L^2(X,A,lambda)$ this is a scalar product:
$
langle f,grangle := int fg , dlambda
$
It wasn't hard to show that the length is positive, but I struggle with the symmetry and the linearity.
calculus measure-theory lebesgue-measure inner-product-space
asked Jan 7 at 13:55
KingDingelingKingDingeling
1857
$endgroup$
I struggle to show that for $, f,gin L^2(X,A,lambda)$ this is a scalar product:
$
langle f,grangle := int fg , dlambda
$
It wasn't hard to show that the length is positive, but I struggle with the symmetry and the linearity.
calculus measure-theory lebesgue-measure inner-product-space
calculus measure-theory lebesgue-measure inner-product-space
asked Jan 7 at 13:55
KingDingelingKingDingeling
1857
asked Jan 7 at 13:55
KingDingelingKingDingeling
1857
asked Jan 7 at 13:55
KingDingelingKingDingeling
1857
asked Jan 7 at 13:55
KingDingelingKingDingeling
1857
asked Jan 7 at 13:55
KingDingelingKingDingeling
1857
1857
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint:
Symmetry is trivial. $left<f,gright> = int fg dlambda = int gf dlambda=left<g,fright>$
because $fg(x)=f(x)g(x)=g(x)f(x)=gf(x)$.
For linearity you need to use the fact that
$$int (f+g) dlambda = int f dlambda + int g dlambda$$
answered Jan 7 at 13:57
YankoYanko
8,1282830
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Symmetry is trivial. $left<f,gright> = int fg dlambda = int gf dlambda=left<g,fright>$
because $fg(x)=f(x)g(x)=g(x)f(x)=gf(x)$.
For linearity you need to use the fact that
$$int (f+g) dlambda = int f dlambda + int g dlambda$$
answered Jan 7 at 13:57
YankoYanko
8,1282830
$endgroup$
add a comment |
$begingroup$
Hint:
Symmetry is trivial. $left<f,gright> = int fg dlambda = int gf dlambda=left<g,fright>$
because $fg(x)=f(x)g(x)=g(x)f(x)=gf(x)$.
For linearity you need to use the fact that
$$int (f+g) dlambda = int f dlambda + int g dlambda$$
answered Jan 7 at 13:57
YankoYanko
8,1282830
$endgroup$
add a comment |
$begingroup$
Hint:
Symmetry is trivial. $left<f,gright> = int fg dlambda = int gf dlambda=left<g,fright>$
because $fg(x)=f(x)g(x)=g(x)f(x)=gf(x)$.
For linearity you need to use the fact that
$$int (f+g) dlambda = int f dlambda + int g dlambda$$
answered Jan 7 at 13:57
YankoYanko
8,1282830
$endgroup$
Hint:
Symmetry is trivial. $left<f,gright> = int fg dlambda = int gf dlambda=left<g,fright>$
because $fg(x)=f(x)g(x)=g(x)f(x)=gf(x)$.
For linearity you need to use the fact that
$$int (f+g) dlambda = int f dlambda + int g dlambda$$
answered Jan 7 at 13:57
YankoYanko
8,1282830
answered Jan 7 at 13:57
YankoYanko
8,1282830
answered Jan 7 at 13:57
YankoYanko
8,1282830
answered Jan 7 at 13:57
YankoYanko
8,1282830
8,1282830
add a comment |
add a comment |
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