Mixing
Uniform convergence and partial derivative in $R^d$
$begingroup$
I have a question about uniform convergences of functions in $mathbb{R}^d$. In $mathbb{R}$, we know that if ${f_n}_{n in mathbb{N}} subset mathbb{R}$, $f'_n$ converges uniformly towards $h$ and it exists $x_0 in mathbb{R}$ such that $f_n(x_0)$ converges, then $f_n$ converges uniformly.
But now, my question is : in $mathbb{R}^d$, if we have the uniform convergence of the partial derivatives, i.e $forall i in {1, ..., d} quad (frac{partial f_n}{partial x_i})_n$ converges uniformly, and if it exists $x_0 in mathbb{R}^d$ s.t $f_n(x_0)$ converges, then can we conclude that $f_n$ converges uniformly ?
Thank you !
real-analysis analysis partial-derivative
asked Jan 7 at 11:17
ChocoSavourChocoSavour
31418
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add a comment |
$begingroup$
I have a question about uniform convergences of functions in $mathbb{R}^d$. In $mathbb{R}$, we know that if ${f_n}_{n in mathbb{N}} subset mathbb{R}$, $f'_n$ converges uniformly towards $h$ and it exists $x_0 in mathbb{R}$ such that $f_n(x_0)$ converges, then $f_n$ converges uniformly.
But now, my question is : in $mathbb{R}^d$, if we have the uniform convergence of the partial derivatives, i.e $forall i in {1, ..., d} quad (frac{partial f_n}{partial x_i})_n$ converges uniformly, and if it exists $x_0 in mathbb{R}^d$ s.t $f_n(x_0)$ converges, then can we conclude that $f_n$ converges uniformly ?
Thank you !
real-analysis analysis partial-derivative
asked Jan 7 at 11:17
ChocoSavourChocoSavour
31418
$endgroup$
add a comment |
$begingroup$
I have a question about uniform convergences of functions in $mathbb{R}^d$. In $mathbb{R}$, we know that if ${f_n}_{n in mathbb{N}} subset mathbb{R}$, $f'_n$ converges uniformly towards $h$ and it exists $x_0 in mathbb{R}$ such that $f_n(x_0)$ converges, then $f_n$ converges uniformly.
But now, my question is : in $mathbb{R}^d$, if we have the uniform convergence of the partial derivatives, i.e $forall i in {1, ..., d} quad (frac{partial f_n}{partial x_i})_n$ converges uniformly, and if it exists $x_0 in mathbb{R}^d$ s.t $f_n(x_0)$ converges, then can we conclude that $f_n$ converges uniformly ?
Thank you !
real-analysis analysis partial-derivative
asked Jan 7 at 11:17
ChocoSavourChocoSavour
31418
$endgroup$
I have a question about uniform convergences of functions in $mathbb{R}^d$. In $mathbb{R}$, we know that if ${f_n}_{n in mathbb{N}} subset mathbb{R}$, $f'_n$ converges uniformly towards $h$ and it exists $x_0 in mathbb{R}$ such that $f_n(x_0)$ converges, then $f_n$ converges uniformly.
But now, my question is : in $mathbb{R}^d$, if we have the uniform convergence of the partial derivatives, i.e $forall i in {1, ..., d} quad (frac{partial f_n}{partial x_i})_n$ converges uniformly, and if it exists $x_0 in mathbb{R}^d$ s.t $f_n(x_0)$ converges, then can we conclude that $f_n$ converges uniformly ?
Thank you !
real-analysis analysis partial-derivative
real-analysis analysis partial-derivative
asked Jan 7 at 11:17
ChocoSavourChocoSavour
31418
asked Jan 7 at 11:17
ChocoSavourChocoSavour
31418
asked Jan 7 at 11:17
ChocoSavourChocoSavour
31418
asked Jan 7 at 11:17
ChocoSavourChocoSavour
31418
asked Jan 7 at 11:17
ChocoSavourChocoSavour
31418
31418
add a comment |
add a comment |
1 Answer
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$begingroup$
The result you say we know is false. First, ${f_n}subsetBbb R$ is not what you meant. That says $(f_n)$ is a sequence of real numbers, not a sequence of functions. You meant to say $f_n:Bbb RtoBbb R$.
Let $f_n(t)=t/n$, $h(t)=0$, $x_0=0$. Then $f_n'to h$ uniformly, $f(x_0)$ converges, but $(f_n)$ is not uniformly convergent.
I could tell you what the correct version is; better would be for you to look up the result wherever you learned it from and get it straight yourself.
answered Jan 7 at 15:14
David C. UllrichDavid C. Ullrich
60k43994
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1 Answer
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1 Answer
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$begingroup$
The result you say we know is false. First, ${f_n}subsetBbb R$ is not what you meant. That says $(f_n)$ is a sequence of real numbers, not a sequence of functions. You meant to say $f_n:Bbb RtoBbb R$.
Let $f_n(t)=t/n$, $h(t)=0$, $x_0=0$. Then $f_n'to h$ uniformly, $f(x_0)$ converges, but $(f_n)$ is not uniformly convergent.
I could tell you what the correct version is; better would be for you to look up the result wherever you learned it from and get it straight yourself.
answered Jan 7 at 15:14
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
add a comment |
$begingroup$
The result you say we know is false. First, ${f_n}subsetBbb R$ is not what you meant. That says $(f_n)$ is a sequence of real numbers, not a sequence of functions. You meant to say $f_n:Bbb RtoBbb R$.
Let $f_n(t)=t/n$, $h(t)=0$, $x_0=0$. Then $f_n'to h$ uniformly, $f(x_0)$ converges, but $(f_n)$ is not uniformly convergent.
I could tell you what the correct version is; better would be for you to look up the result wherever you learned it from and get it straight yourself.
answered Jan 7 at 15:14
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
add a comment |
$begingroup$
The result you say we know is false. First, ${f_n}subsetBbb R$ is not what you meant. That says $(f_n)$ is a sequence of real numbers, not a sequence of functions. You meant to say $f_n:Bbb RtoBbb R$.
Let $f_n(t)=t/n$, $h(t)=0$, $x_0=0$. Then $f_n'to h$ uniformly, $f(x_0)$ converges, but $(f_n)$ is not uniformly convergent.
I could tell you what the correct version is; better would be for you to look up the result wherever you learned it from and get it straight yourself.
answered Jan 7 at 15:14
David C. UllrichDavid C. Ullrich
60k43994
$endgroup$
The result you say we know is false. First, ${f_n}subsetBbb R$ is not what you meant. That says $(f_n)$ is a sequence of real numbers, not a sequence of functions. You meant to say $f_n:Bbb RtoBbb R$.
Let $f_n(t)=t/n$, $h(t)=0$, $x_0=0$. Then $f_n'to h$ uniformly, $f(x_0)$ converges, but $(f_n)$ is not uniformly convergent.
I could tell you what the correct version is; better would be for you to look up the result wherever you learned it from and get it straight yourself.
answered Jan 7 at 15:14
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 7 at 15:14
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 7 at 15:14
David C. UllrichDavid C. Ullrich
60k43994
answered Jan 7 at 15:14
David C. UllrichDavid C. Ullrich
60k43994
60k43994
add a comment |
add a comment |
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