Is there a function $g$ such that $int_0^1 x^n g(x) , mathrm d x$ is $1$ if $n=0$ and $0$ for $n in mathbb...
up vote
2
down vote
favorite
Is there a function $g:[0,1]to mathbb R$ such that $$int_0^1 x^n g(x) , mathrm d x$$ is equal to $1$ if $n=0$ and equal to $0$ for $n=1,2,3, ldots$ ?
If there is, what would be an example of such a function? What if we require that $g$ be continuous?
I know I am expected to state what I have tried but I am honestly stuck. I wanted to integrate by parts but given that $g$ is not differentiable, this is rather useless, I think. Hints would be appreciated too.
real-analysis integration
add a comment |
up vote
2
down vote
favorite
Is there a function $g:[0,1]to mathbb R$ such that $$int_0^1 x^n g(x) , mathrm d x$$ is equal to $1$ if $n=0$ and equal to $0$ for $n=1,2,3, ldots$ ?
If there is, what would be an example of such a function? What if we require that $g$ be continuous?
I know I am expected to state what I have tried but I am honestly stuck. I wanted to integrate by parts but given that $g$ is not differentiable, this is rather useless, I think. Hints would be appreciated too.
real-analysis integration
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is there a function $g:[0,1]to mathbb R$ such that $$int_0^1 x^n g(x) , mathrm d x$$ is equal to $1$ if $n=0$ and equal to $0$ for $n=1,2,3, ldots$ ?
If there is, what would be an example of such a function? What if we require that $g$ be continuous?
I know I am expected to state what I have tried but I am honestly stuck. I wanted to integrate by parts but given that $g$ is not differentiable, this is rather useless, I think. Hints would be appreciated too.
real-analysis integration
Is there a function $g:[0,1]to mathbb R$ such that $$int_0^1 x^n g(x) , mathrm d x$$ is equal to $1$ if $n=0$ and equal to $0$ for $n=1,2,3, ldots$ ?
If there is, what would be an example of such a function? What if we require that $g$ be continuous?
I know I am expected to state what I have tried but I am honestly stuck. I wanted to integrate by parts but given that $g$ is not differentiable, this is rather useless, I think. Hints would be appreciated too.
real-analysis integration
real-analysis integration
edited Nov 18 at 4:36
asked Nov 18 at 4:13
Devilo
757
757
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Let $g$ be a Lebesgue-integrable function on $[0, 1]$. Assume that there exists $N geq 0$ such that
$$ forall n geq N : int_{0}^{1} x^n g(x) , dx = 0. $$
Then we prove the following claim.
Claim. $g$ vanishes almost everywhere. Consequently, $int_{0}^{1} x^n g(x) , dx = 0$ for all $n geq 0$.
Proof. For each $varphi$ in the set $C([0, 1])$ of all continuous functions on $[0, 1]$, Stone-Weierstrass theorem allows to find a sequence $p_n$ of polynomials such that $p_n(x) to varphi(x)$ uniformly on $[0, 1]$. This implies
$$ forall varphi in C([0, 1]) : int_{0}^{1} x^N varphi(x) g(x) , dx = 0. $$
Now for any $0 < a < b < 1$, we may choose $0 leq varphi_n(x) uparrow x^{-N} mathbf{1}_{[a, b]}(x)$, thus by the dominated convergence theorem, we have
$$ forall 0 < a < b < 1 : int_{a}^{b} g(x) , dx = 0. $$
This is enough to show that $g equiv 0$ a.e. on $[0, 1]$ as required.
add a comment |
up vote
1
down vote
There is no continuous function with this property: $int p(x) (xg(x)), dx =0$ for all polynomials $p$ so Weierstrass theorem tells you that $int (xg(x))^{2}, dx =0$ from which you get $g equiv 0$ . So the integral is $0$ for $n=0$ also. Actually, continuity is not required. Using the fact that $xg(x)$ can be approximated in $L^{1}$ norm by continuous functions (hence by polynomials) you can show, by a similar argument, that $xg(x)=0$ almost everywhere which forces the integral to be $0$ for $n=0$. Hence there is no such function as long as the integrals in he question exist for all $n geq 1$.
add a comment |
up vote
0
down vote
Assuming $gin L^2(0,1)$ we are allowed to write
$$ g(x) stackrel{L^2}{=} sum_{ngeq 0} c_n P_n(2x-1),qquad c_n=(2n+1)int_{0}^{1}g(x)P_n(2x-1),dx.$$
Our constraints give $c_0=1$ and
$$ c_n = (2n+1)int_{0}^{1}g(x)left[(-1)^n+x q_n(x)right],dx = (-1)^n (2n+1) $$
so, formally,
$$ g(x) stackrel{L^2}{=}sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1) $$
but the RHS of the last line is not a square-integrable function over $(0,1)$:
$$ int_{0}^{1}left[sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1)right]^2,dx = sum_{ngeq 0}frac{1}{2n+1}=+infty $$
so there are no solutions in $L^2(0,1)$. A fortiori, no continuous solutions.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let $g$ be a Lebesgue-integrable function on $[0, 1]$. Assume that there exists $N geq 0$ such that
$$ forall n geq N : int_{0}^{1} x^n g(x) , dx = 0. $$
Then we prove the following claim.
Claim. $g$ vanishes almost everywhere. Consequently, $int_{0}^{1} x^n g(x) , dx = 0$ for all $n geq 0$.
Proof. For each $varphi$ in the set $C([0, 1])$ of all continuous functions on $[0, 1]$, Stone-Weierstrass theorem allows to find a sequence $p_n$ of polynomials such that $p_n(x) to varphi(x)$ uniformly on $[0, 1]$. This implies
$$ forall varphi in C([0, 1]) : int_{0}^{1} x^N varphi(x) g(x) , dx = 0. $$
Now for any $0 < a < b < 1$, we may choose $0 leq varphi_n(x) uparrow x^{-N} mathbf{1}_{[a, b]}(x)$, thus by the dominated convergence theorem, we have
$$ forall 0 < a < b < 1 : int_{a}^{b} g(x) , dx = 0. $$
This is enough to show that $g equiv 0$ a.e. on $[0, 1]$ as required.
add a comment |
up vote
4
down vote
accepted
Let $g$ be a Lebesgue-integrable function on $[0, 1]$. Assume that there exists $N geq 0$ such that
$$ forall n geq N : int_{0}^{1} x^n g(x) , dx = 0. $$
Then we prove the following claim.
Claim. $g$ vanishes almost everywhere. Consequently, $int_{0}^{1} x^n g(x) , dx = 0$ for all $n geq 0$.
Proof. For each $varphi$ in the set $C([0, 1])$ of all continuous functions on $[0, 1]$, Stone-Weierstrass theorem allows to find a sequence $p_n$ of polynomials such that $p_n(x) to varphi(x)$ uniformly on $[0, 1]$. This implies
$$ forall varphi in C([0, 1]) : int_{0}^{1} x^N varphi(x) g(x) , dx = 0. $$
Now for any $0 < a < b < 1$, we may choose $0 leq varphi_n(x) uparrow x^{-N} mathbf{1}_{[a, b]}(x)$, thus by the dominated convergence theorem, we have
$$ forall 0 < a < b < 1 : int_{a}^{b} g(x) , dx = 0. $$
This is enough to show that $g equiv 0$ a.e. on $[0, 1]$ as required.
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let $g$ be a Lebesgue-integrable function on $[0, 1]$. Assume that there exists $N geq 0$ such that
$$ forall n geq N : int_{0}^{1} x^n g(x) , dx = 0. $$
Then we prove the following claim.
Claim. $g$ vanishes almost everywhere. Consequently, $int_{0}^{1} x^n g(x) , dx = 0$ for all $n geq 0$.
Proof. For each $varphi$ in the set $C([0, 1])$ of all continuous functions on $[0, 1]$, Stone-Weierstrass theorem allows to find a sequence $p_n$ of polynomials such that $p_n(x) to varphi(x)$ uniformly on $[0, 1]$. This implies
$$ forall varphi in C([0, 1]) : int_{0}^{1} x^N varphi(x) g(x) , dx = 0. $$
Now for any $0 < a < b < 1$, we may choose $0 leq varphi_n(x) uparrow x^{-N} mathbf{1}_{[a, b]}(x)$, thus by the dominated convergence theorem, we have
$$ forall 0 < a < b < 1 : int_{a}^{b} g(x) , dx = 0. $$
This is enough to show that $g equiv 0$ a.e. on $[0, 1]$ as required.
Let $g$ be a Lebesgue-integrable function on $[0, 1]$. Assume that there exists $N geq 0$ such that
$$ forall n geq N : int_{0}^{1} x^n g(x) , dx = 0. $$
Then we prove the following claim.
Claim. $g$ vanishes almost everywhere. Consequently, $int_{0}^{1} x^n g(x) , dx = 0$ for all $n geq 0$.
Proof. For each $varphi$ in the set $C([0, 1])$ of all continuous functions on $[0, 1]$, Stone-Weierstrass theorem allows to find a sequence $p_n$ of polynomials such that $p_n(x) to varphi(x)$ uniformly on $[0, 1]$. This implies
$$ forall varphi in C([0, 1]) : int_{0}^{1} x^N varphi(x) g(x) , dx = 0. $$
Now for any $0 < a < b < 1$, we may choose $0 leq varphi_n(x) uparrow x^{-N} mathbf{1}_{[a, b]}(x)$, thus by the dominated convergence theorem, we have
$$ forall 0 < a < b < 1 : int_{a}^{b} g(x) , dx = 0. $$
This is enough to show that $g equiv 0$ a.e. on $[0, 1]$ as required.
answered Nov 18 at 5:35
Sangchul Lee
89.9k12162262
89.9k12162262
add a comment |
add a comment |
up vote
1
down vote
There is no continuous function with this property: $int p(x) (xg(x)), dx =0$ for all polynomials $p$ so Weierstrass theorem tells you that $int (xg(x))^{2}, dx =0$ from which you get $g equiv 0$ . So the integral is $0$ for $n=0$ also. Actually, continuity is not required. Using the fact that $xg(x)$ can be approximated in $L^{1}$ norm by continuous functions (hence by polynomials) you can show, by a similar argument, that $xg(x)=0$ almost everywhere which forces the integral to be $0$ for $n=0$. Hence there is no such function as long as the integrals in he question exist for all $n geq 1$.
add a comment |
up vote
1
down vote
There is no continuous function with this property: $int p(x) (xg(x)), dx =0$ for all polynomials $p$ so Weierstrass theorem tells you that $int (xg(x))^{2}, dx =0$ from which you get $g equiv 0$ . So the integral is $0$ for $n=0$ also. Actually, continuity is not required. Using the fact that $xg(x)$ can be approximated in $L^{1}$ norm by continuous functions (hence by polynomials) you can show, by a similar argument, that $xg(x)=0$ almost everywhere which forces the integral to be $0$ for $n=0$. Hence there is no such function as long as the integrals in he question exist for all $n geq 1$.
add a comment |
up vote
1
down vote
up vote
1
down vote
There is no continuous function with this property: $int p(x) (xg(x)), dx =0$ for all polynomials $p$ so Weierstrass theorem tells you that $int (xg(x))^{2}, dx =0$ from which you get $g equiv 0$ . So the integral is $0$ for $n=0$ also. Actually, continuity is not required. Using the fact that $xg(x)$ can be approximated in $L^{1}$ norm by continuous functions (hence by polynomials) you can show, by a similar argument, that $xg(x)=0$ almost everywhere which forces the integral to be $0$ for $n=0$. Hence there is no such function as long as the integrals in he question exist for all $n geq 1$.
There is no continuous function with this property: $int p(x) (xg(x)), dx =0$ for all polynomials $p$ so Weierstrass theorem tells you that $int (xg(x))^{2}, dx =0$ from which you get $g equiv 0$ . So the integral is $0$ for $n=0$ also. Actually, continuity is not required. Using the fact that $xg(x)$ can be approximated in $L^{1}$ norm by continuous functions (hence by polynomials) you can show, by a similar argument, that $xg(x)=0$ almost everywhere which forces the integral to be $0$ for $n=0$. Hence there is no such function as long as the integrals in he question exist for all $n geq 1$.
edited Nov 18 at 6:02
answered Nov 18 at 5:33
Kavi Rama Murthy
42.4k31751
42.4k31751
add a comment |
add a comment |
up vote
0
down vote
Assuming $gin L^2(0,1)$ we are allowed to write
$$ g(x) stackrel{L^2}{=} sum_{ngeq 0} c_n P_n(2x-1),qquad c_n=(2n+1)int_{0}^{1}g(x)P_n(2x-1),dx.$$
Our constraints give $c_0=1$ and
$$ c_n = (2n+1)int_{0}^{1}g(x)left[(-1)^n+x q_n(x)right],dx = (-1)^n (2n+1) $$
so, formally,
$$ g(x) stackrel{L^2}{=}sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1) $$
but the RHS of the last line is not a square-integrable function over $(0,1)$:
$$ int_{0}^{1}left[sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1)right]^2,dx = sum_{ngeq 0}frac{1}{2n+1}=+infty $$
so there are no solutions in $L^2(0,1)$. A fortiori, no continuous solutions.
add a comment |
up vote
0
down vote
Assuming $gin L^2(0,1)$ we are allowed to write
$$ g(x) stackrel{L^2}{=} sum_{ngeq 0} c_n P_n(2x-1),qquad c_n=(2n+1)int_{0}^{1}g(x)P_n(2x-1),dx.$$
Our constraints give $c_0=1$ and
$$ c_n = (2n+1)int_{0}^{1}g(x)left[(-1)^n+x q_n(x)right],dx = (-1)^n (2n+1) $$
so, formally,
$$ g(x) stackrel{L^2}{=}sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1) $$
but the RHS of the last line is not a square-integrable function over $(0,1)$:
$$ int_{0}^{1}left[sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1)right]^2,dx = sum_{ngeq 0}frac{1}{2n+1}=+infty $$
so there are no solutions in $L^2(0,1)$. A fortiori, no continuous solutions.
add a comment |
up vote
0
down vote
up vote
0
down vote
Assuming $gin L^2(0,1)$ we are allowed to write
$$ g(x) stackrel{L^2}{=} sum_{ngeq 0} c_n P_n(2x-1),qquad c_n=(2n+1)int_{0}^{1}g(x)P_n(2x-1),dx.$$
Our constraints give $c_0=1$ and
$$ c_n = (2n+1)int_{0}^{1}g(x)left[(-1)^n+x q_n(x)right],dx = (-1)^n (2n+1) $$
so, formally,
$$ g(x) stackrel{L^2}{=}sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1) $$
but the RHS of the last line is not a square-integrable function over $(0,1)$:
$$ int_{0}^{1}left[sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1)right]^2,dx = sum_{ngeq 0}frac{1}{2n+1}=+infty $$
so there are no solutions in $L^2(0,1)$. A fortiori, no continuous solutions.
Assuming $gin L^2(0,1)$ we are allowed to write
$$ g(x) stackrel{L^2}{=} sum_{ngeq 0} c_n P_n(2x-1),qquad c_n=(2n+1)int_{0}^{1}g(x)P_n(2x-1),dx.$$
Our constraints give $c_0=1$ and
$$ c_n = (2n+1)int_{0}^{1}g(x)left[(-1)^n+x q_n(x)right],dx = (-1)^n (2n+1) $$
so, formally,
$$ g(x) stackrel{L^2}{=}sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1) $$
but the RHS of the last line is not a square-integrable function over $(0,1)$:
$$ int_{0}^{1}left[sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1)right]^2,dx = sum_{ngeq 0}frac{1}{2n+1}=+infty $$
so there are no solutions in $L^2(0,1)$. A fortiori, no continuous solutions.
answered Nov 18 at 16:46
Jack D'Aurizio
283k33275653
283k33275653
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003123%2fis-there-a-function-g-such-that-int-01-xn-gx-mathrm-d-x-is-1-if%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown