Non-zero continuous function $f(x)$ such that $int_0^1x^k (1-x)^{n-k} f(x)dx=0$ for every $k=0,1,…,n$ and...
$begingroup$
Does there exists a non-zero continuous function $f$ such that $displaystyle int_0^1 x^k (1-x)^{n-k} f(x)dx=0$ for every $k=0,1,2,...,n$ where $n$ is a non-negative integer.
EDIT: The problem is related to Weierstrass approximation using Bernstein approximation of certain function. As you can see the term $x^k (1-x)^{n-k}$ comes from Bernstein polynomials. I want my integral to vanish for non-zero $f$.
Also, the result should hold for every $n$ and for every $k$ running from $0$ to $n$. For example, for $n=2$, we shall have three integrals corresponding to $k=0,1,2$. In this way, it should hold for any $n$.
real-analysis calculus integration analysis lebesgue-integral
asked Jan 20 at 0:36
ershersh
408113
$endgroup$
|
show 10 more comments
$begingroup$
Does there exists a non-zero continuous function $f$ such that $displaystyle int_0^1 x^k (1-x)^{n-k} f(x)dx=0$ for every $k=0,1,2,...,n$ where $n$ is a non-negative integer.
EDIT: The problem is related to Weierstrass approximation using Bernstein approximation of certain function. As you can see the term $x^k (1-x)^{n-k}$ comes from Bernstein polynomials. I want my integral to vanish for non-zero $f$.
Also, the result should hold for every $n$ and for every $k$ running from $0$ to $n$. For example, for $n=2$, we shall have three integrals corresponding to $k=0,1,2$. In this way, it should hold for any $n$.
real-analysis calculus integration analysis lebesgue-integral
asked Jan 20 at 0:36
ershersh
408113
$endgroup$
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:45
$begingroup$
Do you mean everywhere nonzero or not identically 0?
$endgroup$
– Van Latimer
Jan 20 at 0:48
2
$begingroup$
The set $$ V_n = left{ f in L^2([0,1]) : int_{0}^{1} x^k(1-x)^{n-k}f(x) , mathrm{d}x = 0 text{ for every } k = 0, cdots, n right} $$ is the orthogonal complement of the set of all polynomials of degree $leq n$, and in particular, is not only non-empty but in fact an infinite-dimensional subspace of $L^{2}([0,1])$. One can find an explicit orthogonal basis of $V_n$, for instance, using the shifted Legendre polynomials, which are of course continuous.
$endgroup$
– Sangchul Lee
Jan 20 at 1:10
1
$begingroup$
Ah, I thought that you are asking examples of $f$ for each given $n$. If the condition should hold for all $n$, then $f$ must be identically zero, by the standard approximation technique, i.e., choose a sequence of polynomials $p_n$ that approximate $f$ in $C([0,1])$ to see that $$ int_{0}^{1} f(x)^2 , mathrm{d}x = lim_{ntoinfty}int_{0}^{1} f(x)p_n(x) , mathrm{d}x stackrel{text{(by assumption)}}= 0. $$ This is enough to conclude $f equiv 0$. The conclusion continues to hold (modulo measure-zero modification) for larger class of functions.
$endgroup$
– Sangchul Lee
Jan 20 at 2:50
1
$begingroup$
If $f$ is $L^1$, then one can show that its Fourier transform $hat{f}(xi) = int_{0}^{1}f(x)e^{-2pi ixi x} , mathrm{d}x$ is identically zero, which is enough to show that $f$ is zero a.e.
$endgroup$
– Sangchul Lee
Jan 20 at 5:06
|
show 10 more comments
$begingroup$
Does there exists a non-zero continuous function $f$ such that $displaystyle int_0^1 x^k (1-x)^{n-k} f(x)dx=0$ for every $k=0,1,2,...,n$ where $n$ is a non-negative integer.
EDIT: The problem is related to Weierstrass approximation using Bernstein approximation of certain function. As you can see the term $x^k (1-x)^{n-k}$ comes from Bernstein polynomials. I want my integral to vanish for non-zero $f$.
Also, the result should hold for every $n$ and for every $k$ running from $0$ to $n$. For example, for $n=2$, we shall have three integrals corresponding to $k=0,1,2$. In this way, it should hold for any $n$.
real-analysis calculus integration analysis lebesgue-integral
asked Jan 20 at 0:36
ershersh
408113
$endgroup$
Does there exists a non-zero continuous function $f$ such that $displaystyle int_0^1 x^k (1-x)^{n-k} f(x)dx=0$ for every $k=0,1,2,...,n$ where $n$ is a non-negative integer.
EDIT: The problem is related to Weierstrass approximation using Bernstein approximation of certain function. As you can see the term $x^k (1-x)^{n-k}$ comes from Bernstein polynomials. I want my integral to vanish for non-zero $f$.
Also, the result should hold for every $n$ and for every $k$ running from $0$ to $n$. For example, for $n=2$, we shall have three integrals corresponding to $k=0,1,2$. In this way, it should hold for any $n$.
real-analysis calculus integration analysis lebesgue-integral
real-analysis calculus integration analysis lebesgue-integral
asked Jan 20 at 0:36
ershersh
408113
asked Jan 20 at 0:36
ershersh
408113
edited Jan 20 at 1:03
ersh
asked Jan 20 at 0:36
ershersh
408113
asked Jan 20 at 0:36
ershersh
408113
asked Jan 20 at 0:36
ershersh
408113
408113
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:45
$begingroup$
Do you mean everywhere nonzero or not identically 0?
$endgroup$
– Van Latimer
Jan 20 at 0:48
2
$begingroup$
The set $$ V_n = left{ f in L^2([0,1]) : int_{0}^{1} x^k(1-x)^{n-k}f(x) , mathrm{d}x = 0 text{ for every } k = 0, cdots, n right} $$ is the orthogonal complement of the set of all polynomials of degree $leq n$, and in particular, is not only non-empty but in fact an infinite-dimensional subspace of $L^{2}([0,1])$. One can find an explicit orthogonal basis of $V_n$, for instance, using the shifted Legendre polynomials, which are of course continuous.
$endgroup$
– Sangchul Lee
Jan 20 at 1:10
1
$begingroup$
Ah, I thought that you are asking examples of $f$ for each given $n$. If the condition should hold for all $n$, then $f$ must be identically zero, by the standard approximation technique, i.e., choose a sequence of polynomials $p_n$ that approximate $f$ in $C([0,1])$ to see that $$ int_{0}^{1} f(x)^2 , mathrm{d}x = lim_{ntoinfty}int_{0}^{1} f(x)p_n(x) , mathrm{d}x stackrel{text{(by assumption)}}= 0. $$ This is enough to conclude $f equiv 0$. The conclusion continues to hold (modulo measure-zero modification) for larger class of functions.
$endgroup$
– Sangchul Lee
Jan 20 at 2:50
1
$begingroup$
If $f$ is $L^1$, then one can show that its Fourier transform $hat{f}(xi) = int_{0}^{1}f(x)e^{-2pi ixi x} , mathrm{d}x$ is identically zero, which is enough to show that $f$ is zero a.e.
$endgroup$
– Sangchul Lee
Jan 20 at 5:06
|
show 10 more comments
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:45
$begingroup$
Do you mean everywhere nonzero or not identically 0?
$endgroup$
– Van Latimer
Jan 20 at 0:48
2
$begingroup$
The set $$ V_n = left{ f in L^2([0,1]) : int_{0}^{1} x^k(1-x)^{n-k}f(x) , mathrm{d}x = 0 text{ for every } k = 0, cdots, n right} $$ is the orthogonal complement of the set of all polynomials of degree $leq n$, and in particular, is not only non-empty but in fact an infinite-dimensional subspace of $L^{2}([0,1])$. One can find an explicit orthogonal basis of $V_n$, for instance, using the shifted Legendre polynomials, which are of course continuous.
$endgroup$
– Sangchul Lee
Jan 20 at 1:10
1
$begingroup$
Ah, I thought that you are asking examples of $f$ for each given $n$. If the condition should hold for all $n$, then $f$ must be identically zero, by the standard approximation technique, i.e., choose a sequence of polynomials $p_n$ that approximate $f$ in $C([0,1])$ to see that $$ int_{0}^{1} f(x)^2 , mathrm{d}x = lim_{ntoinfty}int_{0}^{1} f(x)p_n(x) , mathrm{d}x stackrel{text{(by assumption)}}= 0. $$ This is enough to conclude $f equiv 0$. The conclusion continues to hold (modulo measure-zero modification) for larger class of functions.
$endgroup$
– Sangchul Lee
Jan 20 at 2:50
1
$begingroup$
If $f$ is $L^1$, then one can show that its Fourier transform $hat{f}(xi) = int_{0}^{1}f(x)e^{-2pi ixi x} , mathrm{d}x$ is identically zero, which is enough to show that $f$ is zero a.e.
$endgroup$
– Sangchul Lee
Jan 20 at 5:06
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:45
$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:45
$begingroup$
Do you mean everywhere nonzero or not identically 0?
$endgroup$
– Van Latimer
Jan 20 at 0:48
$begingroup$
Do you mean everywhere nonzero or not identically 0?
$endgroup$
– Van Latimer
Jan 20 at 0:48
2
2
$begingroup$
The set $$ V_n = left{ f in L^2([0,1]) : int_{0}^{1} x^k(1-x)^{n-k}f(x) , mathrm{d}x = 0 text{ for every } k = 0, cdots, n right} $$ is the orthogonal complement of the set of all polynomials of degree $leq n$, and in particular, is not only non-empty but in fact an infinite-dimensional subspace of $L^{2}([0,1])$. One can find an explicit orthogonal basis of $V_n$, for instance, using the shifted Legendre polynomials, which are of course continuous.
$endgroup$
– Sangchul Lee
Jan 20 at 1:10
$begingroup$
The set $$ V_n = left{ f in L^2([0,1]) : int_{0}^{1} x^k(1-x)^{n-k}f(x) , mathrm{d}x = 0 text{ for every } k = 0, cdots, n right} $$ is the orthogonal complement of the set of all polynomials of degree $leq n$, and in particular, is not only non-empty but in fact an infinite-dimensional subspace of $L^{2}([0,1])$. One can find an explicit orthogonal basis of $V_n$, for instance, using the shifted Legendre polynomials, which are of course continuous.
$endgroup$
– Sangchul Lee
Jan 20 at 1:10
1
1
$begingroup$
Ah, I thought that you are asking examples of $f$ for each given $n$. If the condition should hold for all $n$, then $f$ must be identically zero, by the standard approximation technique, i.e., choose a sequence of polynomials $p_n$ that approximate $f$ in $C([0,1])$ to see that $$ int_{0}^{1} f(x)^2 , mathrm{d}x = lim_{ntoinfty}int_{0}^{1} f(x)p_n(x) , mathrm{d}x stackrel{text{(by assumption)}}= 0. $$ This is enough to conclude $f equiv 0$. The conclusion continues to hold (modulo measure-zero modification) for larger class of functions.
$endgroup$
– Sangchul Lee
Jan 20 at 2:50
$begingroup$
Ah, I thought that you are asking examples of $f$ for each given $n$. If the condition should hold for all $n$, then $f$ must be identically zero, by the standard approximation technique, i.e., choose a sequence of polynomials $p_n$ that approximate $f$ in $C([0,1])$ to see that $$ int_{0}^{1} f(x)^2 , mathrm{d}x = lim_{ntoinfty}int_{0}^{1} f(x)p_n(x) , mathrm{d}x stackrel{text{(by assumption)}}= 0. $$ This is enough to conclude $f equiv 0$. The conclusion continues to hold (modulo measure-zero modification) for larger class of functions.
$endgroup$
– Sangchul Lee
Jan 20 at 2:50
1
1
$begingroup$
If $f$ is $L^1$, then one can show that its Fourier transform $hat{f}(xi) = int_{0}^{1}f(x)e^{-2pi ixi x} , mathrm{d}x$ is identically zero, which is enough to show that $f$ is zero a.e.
$endgroup$
– Sangchul Lee
Jan 20 at 5:06
$begingroup$
If $f$ is $L^1$, then one can show that its Fourier transform $hat{f}(xi) = int_{0}^{1}f(x)e^{-2pi ixi x} , mathrm{d}x$ is identically zero, which is enough to show that $f$ is zero a.e.
$endgroup$
– Sangchul Lee
Jan 20 at 5:06
|
show 10 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Here’s the sketch: Since we require this for every $k$ and $n$, it amounts to saying that the integral of f against any $x^k (1-x)^j$ is zero. These are closed under multiplication, so their span is an algebra. By Stone-Weierstrass, find linear combinations $l_n$ of them approaching $f$ uniformly. By the assumption in the problem, $int l_n f$ is 0 for all $n$, so taking a limit we conclude $f^2$ integrates to 0. It however is positive, so it is identically 0. Therefore $f$ is identically 0.
answered Jan 20 at 2:13
Van LatimerVan Latimer
321110
$endgroup$
$begingroup$
@ Van Latimer, It really seems such $f$ must be zero. I am reading your proof.
$endgroup$
– ersh
Jan 20 at 2:41
$begingroup$
Right, sorry I didn’t state that explicitly till the end.
$endgroup$
– Van Latimer
Jan 20 at 3:20
add a comment |
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1 Answer
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$begingroup$
Here’s the sketch: Since we require this for every $k$ and $n$, it amounts to saying that the integral of f against any $x^k (1-x)^j$ is zero. These are closed under multiplication, so their span is an algebra. By Stone-Weierstrass, find linear combinations $l_n$ of them approaching $f$ uniformly. By the assumption in the problem, $int l_n f$ is 0 for all $n$, so taking a limit we conclude $f^2$ integrates to 0. It however is positive, so it is identically 0. Therefore $f$ is identically 0.
answered Jan 20 at 2:13
Van LatimerVan Latimer
321110
$endgroup$
$begingroup$
@ Van Latimer, It really seems such $f$ must be zero. I am reading your proof.
$endgroup$
– ersh
Jan 20 at 2:41
$begingroup$
Right, sorry I didn’t state that explicitly till the end.
$endgroup$
– Van Latimer
Jan 20 at 3:20
add a comment |
$begingroup$
Here’s the sketch: Since we require this for every $k$ and $n$, it amounts to saying that the integral of f against any $x^k (1-x)^j$ is zero. These are closed under multiplication, so their span is an algebra. By Stone-Weierstrass, find linear combinations $l_n$ of them approaching $f$ uniformly. By the assumption in the problem, $int l_n f$ is 0 for all $n$, so taking a limit we conclude $f^2$ integrates to 0. It however is positive, so it is identically 0. Therefore $f$ is identically 0.
answered Jan 20 at 2:13
Van LatimerVan Latimer
321110
$endgroup$
$begingroup$
@ Van Latimer, It really seems such $f$ must be zero. I am reading your proof.
$endgroup$
– ersh
Jan 20 at 2:41
$begingroup$
Right, sorry I didn’t state that explicitly till the end.
$endgroup$
– Van Latimer
Jan 20 at 3:20
add a comment |
$begingroup$
Here’s the sketch: Since we require this for every $k$ and $n$, it amounts to saying that the integral of f against any $x^k (1-x)^j$ is zero. These are closed under multiplication, so their span is an algebra. By Stone-Weierstrass, find linear combinations $l_n$ of them approaching $f$ uniformly. By the assumption in the problem, $int l_n f$ is 0 for all $n$, so taking a limit we conclude $f^2$ integrates to 0. It however is positive, so it is identically 0. Therefore $f$ is identically 0.
answered Jan 20 at 2:13
Van LatimerVan Latimer
321110
$endgroup$
Here’s the sketch: Since we require this for every $k$ and $n$, it amounts to saying that the integral of f against any $x^k (1-x)^j$ is zero. These are closed under multiplication, so their span is an algebra. By Stone-Weierstrass, find linear combinations $l_n$ of them approaching $f$ uniformly. By the assumption in the problem, $int l_n f$ is 0 for all $n$, so taking a limit we conclude $f^2$ integrates to 0. It however is positive, so it is identically 0. Therefore $f$ is identically 0.
answered Jan 20 at 2:13
Van LatimerVan Latimer
321110
answered Jan 20 at 2:13
Van LatimerVan Latimer
321110
answered Jan 20 at 2:13
Van LatimerVan Latimer
321110
answered Jan 20 at 2:13
Van LatimerVan Latimer
321110
321110
$begingroup$
@ Van Latimer, It really seems such $f$ must be zero. I am reading your proof.
$endgroup$
– ersh
Jan 20 at 2:41
$begingroup$
Right, sorry I didn’t state that explicitly till the end.
$endgroup$
– Van Latimer
Jan 20 at 3:20
add a comment |
$begingroup$
@ Van Latimer, It really seems such $f$ must be zero. I am reading your proof.
$endgroup$
– ersh
Jan 20 at 2:41
$begingroup$
Right, sorry I didn’t state that explicitly till the end.
$endgroup$
– Van Latimer
Jan 20 at 3:20
$begingroup$
@ Van Latimer, It really seems such $f$ must be zero. I am reading your proof.
$endgroup$
– ersh
Jan 20 at 2:41
$begingroup$
@ Van Latimer, It really seems such $f$ must be zero. I am reading your proof.
$endgroup$
– ersh
Jan 20 at 2:41
$begingroup$
Right, sorry I didn’t state that explicitly till the end.
$endgroup$
– Van Latimer
Jan 20 at 3:20
$begingroup$
Right, sorry I didn’t state that explicitly till the end.
$endgroup$
– Van Latimer
Jan 20 at 3:20
add a comment |
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$begingroup$
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:45
$begingroup$
Do you mean everywhere nonzero or not identically 0?
$endgroup$
– Van Latimer
Jan 20 at 0:48
2
$begingroup$
The set $$ V_n = left{ f in L^2([0,1]) : int_{0}^{1} x^k(1-x)^{n-k}f(x) , mathrm{d}x = 0 text{ for every } k = 0, cdots, n right} $$ is the orthogonal complement of the set of all polynomials of degree $leq n$, and in particular, is not only non-empty but in fact an infinite-dimensional subspace of $L^{2}([0,1])$. One can find an explicit orthogonal basis of $V_n$, for instance, using the shifted Legendre polynomials, which are of course continuous.
$endgroup$
– Sangchul Lee
Jan 20 at 1:10
1
$begingroup$
Ah, I thought that you are asking examples of $f$ for each given $n$. If the condition should hold for all $n$, then $f$ must be identically zero, by the standard approximation technique, i.e., choose a sequence of polynomials $p_n$ that approximate $f$ in $C([0,1])$ to see that $$ int_{0}^{1} f(x)^2 , mathrm{d}x = lim_{ntoinfty}int_{0}^{1} f(x)p_n(x) , mathrm{d}x stackrel{text{(by assumption)}}= 0. $$ This is enough to conclude $f equiv 0$. The conclusion continues to hold (modulo measure-zero modification) for larger class of functions.
$endgroup$
– Sangchul Lee
Jan 20 at 2:50
1
$begingroup$
If $f$ is $L^1$, then one can show that its Fourier transform $hat{f}(xi) = int_{0}^{1}f(x)e^{-2pi ixi x} , mathrm{d}x$ is identically zero, which is enough to show that $f$ is zero a.e.
$endgroup$
– Sangchul Lee
Jan 20 at 5:06