Adjoint operator on $mathbb{L} ^2 _mathbb{R}$ of $frac{d}{dx}$
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If A is a linear operator defined on $mathbb{L} ^2 _mathbb{R}$ of $frac{d}{dx}$ then its adjoint must satisfy the property:
$$(A^*f,g) = (f,Ag) \ f,g in mathbb{L} ^2 _mathbb{R}$$$
Now if $A = frac{d}{dx}$, then how can I find $A^*$? I am unsure of this because first how is the scalar product defined on $mathbb{L} ^2 _mathbb{R}$? Do I just need to "plug in" $f,g$ in the scalar product and then work it out from there? I know the answer is $-frac{d}{dx}$.
linear-algebra vector-spaces adjoint-operators
asked Jan 15 at 16:49
daljit97daljit97
178111
$endgroup$
add a comment |
$begingroup$
If A is a linear operator defined on $mathbb{L} ^2 _mathbb{R}$ of $frac{d}{dx}$ then its adjoint must satisfy the property:
$$(A^*f,g) = (f,Ag) \ f,g in mathbb{L} ^2 _mathbb{R}$$$
Now if $A = frac{d}{dx}$, then how can I find $A^*$? I am unsure of this because first how is the scalar product defined on $mathbb{L} ^2 _mathbb{R}$? Do I just need to "plug in" $f,g$ in the scalar product and then work it out from there? I know the answer is $-frac{d}{dx}$.
linear-algebra vector-spaces adjoint-operators
asked Jan 15 at 16:49
daljit97daljit97
178111
$endgroup$
1
$begingroup$
In $L^2,$ the usual inner product is $langle f,grangle=int_{mathbb{R}}f(x) g(x),dmu,$ a Lebesgue integral. However, if your functions are differentiable, then they're continuous, and hence Riemann integrable.
$endgroup$
– Adrian Keister
Jan 15 at 16:51
1
$begingroup$
I believe integration by parts is useful here.
$endgroup$
– Math1000
Jan 15 at 17:27
$begingroup$
@Math1000 yeah I figured that would be helpful as inside the integral you would have something like $u dv$.
$endgroup$
– daljit97
Jan 15 at 17:38
add a comment |
$begingroup$
If A is a linear operator defined on $mathbb{L} ^2 _mathbb{R}$ of $frac{d}{dx}$ then its adjoint must satisfy the property:
$$(A^*f,g) = (f,Ag) \ f,g in mathbb{L} ^2 _mathbb{R}$$$
Now if $A = frac{d}{dx}$, then how can I find $A^*$? I am unsure of this because first how is the scalar product defined on $mathbb{L} ^2 _mathbb{R}$? Do I just need to "plug in" $f,g$ in the scalar product and then work it out from there? I know the answer is $-frac{d}{dx}$.
linear-algebra vector-spaces adjoint-operators
asked Jan 15 at 16:49
daljit97daljit97
178111
$endgroup$
If A is a linear operator defined on $mathbb{L} ^2 _mathbb{R}$ of $frac{d}{dx}$ then its adjoint must satisfy the property:
$$(A^*f,g) = (f,Ag) \ f,g in mathbb{L} ^2 _mathbb{R}$$$
Now if $A = frac{d}{dx}$, then how can I find $A^*$? I am unsure of this because first how is the scalar product defined on $mathbb{L} ^2 _mathbb{R}$? Do I just need to "plug in" $f,g$ in the scalar product and then work it out from there? I know the answer is $-frac{d}{dx}$.
linear-algebra vector-spaces adjoint-operators
linear-algebra vector-spaces adjoint-operators
asked Jan 15 at 16:49
daljit97daljit97
178111
asked Jan 15 at 16:49
daljit97daljit97
178111
asked Jan 15 at 16:49
daljit97daljit97
178111
asked Jan 15 at 16:49
daljit97daljit97
178111
asked Jan 15 at 16:49
daljit97daljit97
178111
178111
1
$begingroup$
In $L^2,$ the usual inner product is $langle f,grangle=int_{mathbb{R}}f(x) g(x),dmu,$ a Lebesgue integral. However, if your functions are differentiable, then they're continuous, and hence Riemann integrable.
$endgroup$
– Adrian Keister
Jan 15 at 16:51
1
$begingroup$
I believe integration by parts is useful here.
$endgroup$
– Math1000
Jan 15 at 17:27
$begingroup$
@Math1000 yeah I figured that would be helpful as inside the integral you would have something like $u dv$.
$endgroup$
– daljit97
Jan 15 at 17:38
add a comment |
1
$begingroup$
In $L^2,$ the usual inner product is $langle f,grangle=int_{mathbb{R}}f(x) g(x),dmu,$ a Lebesgue integral. However, if your functions are differentiable, then they're continuous, and hence Riemann integrable.
$endgroup$
– Adrian Keister
Jan 15 at 16:51
1
$begingroup$
I believe integration by parts is useful here.
$endgroup$
– Math1000
Jan 15 at 17:27
$begingroup$
@Math1000 yeah I figured that would be helpful as inside the integral you would have something like $u dv$.
$endgroup$
– daljit97
Jan 15 at 17:38
1
1
$begingroup$
In $L^2,$ the usual inner product is $langle f,grangle=int_{mathbb{R}}f(x) g(x),dmu,$ a Lebesgue integral. However, if your functions are differentiable, then they're continuous, and hence Riemann integrable.
$endgroup$
– Adrian Keister
Jan 15 at 16:51
$begingroup$
In $L^2,$ the usual inner product is $langle f,grangle=int_{mathbb{R}}f(x) g(x),dmu,$ a Lebesgue integral. However, if your functions are differentiable, then they're continuous, and hence Riemann integrable.
$endgroup$
– Adrian Keister
Jan 15 at 16:51
1
1
$begingroup$
I believe integration by parts is useful here.
$endgroup$
– Math1000
Jan 15 at 17:27
$begingroup$
I believe integration by parts is useful here.
$endgroup$
– Math1000
Jan 15 at 17:27
$begingroup$
@Math1000 yeah I figured that would be helpful as inside the integral you would have something like $u dv$.
$endgroup$
– daljit97
Jan 15 at 17:38
$begingroup$
@Math1000 yeah I figured that would be helpful as inside the integral you would have something like $u dv$.
$endgroup$
– daljit97
Jan 15 at 17:38
add a comment |
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1
$begingroup$
In $L^2,$ the usual inner product is $langle f,grangle=int_{mathbb{R}}f(x) g(x),dmu,$ a Lebesgue integral. However, if your functions are differentiable, then they're continuous, and hence Riemann integrable.
$endgroup$
– Adrian Keister
Jan 15 at 16:51
1
$begingroup$
I believe integration by parts is useful here.
$endgroup$
– Math1000
Jan 15 at 17:27
$begingroup$
@Math1000 yeah I figured that would be helpful as inside the integral you would have something like $u dv$.
$endgroup$
– daljit97
Jan 15 at 17:38