Mixing
$f_n$ converge uniformly to $f$ then $mathrm{d}f_n(x_n)$ converges to $mathrm{d}f(x)$
$begingroup$
Let $f_n : mathbb{R}^p to mathbb{R}$ such that the $f_n$ are $C^1$ and such that the sequence $(f_n)_{n in mathbb{N}}$ converges uniformly to a function $f : mathbb{R}^p to mathbb{R}$ which is $C^1$. Then prove that for all $x in mathbb{R}^n$ there is a sequence $(x_n)_{n in mathbb{N}}$ which converge to $x$ such that $mathrm{d}f_n(x_n)$ converges to $mathrm{d}f(x)$.
I must say that I don't know at all how to do and don't have any intuition of what is really going on here. So we might look at the case qhere $p= 1$.
So we can write :
$$f(a+h) = f(a)+ f'(a)h +o(h)$$
$$forall n in mathbb{N}, f_n(a+h) = f_n(a) + f'_n(a)h +o(h)$$
Hence we have :
$$mid f'(a)h - f'_n(a)h mid leq mid f(a+h)-f(a) mid +mid f(a)-f_n(a) mid + mid o(h) mid$$
Since the function $f_n$ converge uniformly to $f$, we have :
$$mid f'(a)h - f'_{infty}(a) mid leq mid o(h) mid$$
And now using we let $h to 0$ so that :
$$mid f'(a) -f_infty'(a) mid = 0 $$
I don't know if this works, but it feels strange to me since in the case the sequence $x_n$ is just the constant sequence... and moreover if this is correct I don't see at all how to generalise to higher dimensions.
Thank you !
real-analysis calculus sequences-and-series multivariable-calculus uniform-convergence
asked Jan 28 at 23:30
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
Let $f_n : mathbb{R}^p to mathbb{R}$ such that the $f_n$ are $C^1$ and such that the sequence $(f_n)_{n in mathbb{N}}$ converges uniformly to a function $f : mathbb{R}^p to mathbb{R}$ which is $C^1$. Then prove that for all $x in mathbb{R}^n$ there is a sequence $(x_n)_{n in mathbb{N}}$ which converge to $x$ such that $mathrm{d}f_n(x_n)$ converges to $mathrm{d}f(x)$.
I must say that I don't know at all how to do and don't have any intuition of what is really going on here. So we might look at the case qhere $p= 1$.
So we can write :
$$f(a+h) = f(a)+ f'(a)h +o(h)$$
$$forall n in mathbb{N}, f_n(a+h) = f_n(a) + f'_n(a)h +o(h)$$
Hence we have :
$$mid f'(a)h - f'_n(a)h mid leq mid f(a+h)-f(a) mid +mid f(a)-f_n(a) mid + mid o(h) mid$$
Since the function $f_n$ converge uniformly to $f$, we have :
$$mid f'(a)h - f'_{infty}(a) mid leq mid o(h) mid$$
And now using we let $h to 0$ so that :
$$mid f'(a) -f_infty'(a) mid = 0 $$
I don't know if this works, but it feels strange to me since in the case the sequence $x_n$ is just the constant sequence... and moreover if this is correct I don't see at all how to generalise to higher dimensions.
Thank you !
real-analysis calculus sequences-and-series multivariable-calculus uniform-convergence
asked Jan 28 at 23:30
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
$begingroup$
Your demonstration is not correct, since you don’t know if $lim_{n to infty} f’_n(a)$ converges, hence you can note it like $f’_{infty}(a)$, that’s why you always need to be careful when dealing with double limit
$endgroup$
– Thinking
Jan 28 at 23:42
3
$begingroup$
For complex or vector valued functions the claim is wrong. Consider the sequence $f_n(x):={1over n}e^{inx}$, which converges uniformly to $0$, but $|f_n'(x)|=1$ for all $x$ and $n$.
$endgroup$
– Christian Blatter
Jan 31 at 10:34
add a comment |
$begingroup$
Let $f_n : mathbb{R}^p to mathbb{R}$ such that the $f_n$ are $C^1$ and such that the sequence $(f_n)_{n in mathbb{N}}$ converges uniformly to a function $f : mathbb{R}^p to mathbb{R}$ which is $C^1$. Then prove that for all $x in mathbb{R}^n$ there is a sequence $(x_n)_{n in mathbb{N}}$ which converge to $x$ such that $mathrm{d}f_n(x_n)$ converges to $mathrm{d}f(x)$.
I must say that I don't know at all how to do and don't have any intuition of what is really going on here. So we might look at the case qhere $p= 1$.
So we can write :
$$f(a+h) = f(a)+ f'(a)h +o(h)$$
$$forall n in mathbb{N}, f_n(a+h) = f_n(a) + f'_n(a)h +o(h)$$
Hence we have :
$$mid f'(a)h - f'_n(a)h mid leq mid f(a+h)-f(a) mid +mid f(a)-f_n(a) mid + mid o(h) mid$$
Since the function $f_n$ converge uniformly to $f$, we have :
$$mid f'(a)h - f'_{infty}(a) mid leq mid o(h) mid$$
And now using we let $h to 0$ so that :
$$mid f'(a) -f_infty'(a) mid = 0 $$
I don't know if this works, but it feels strange to me since in the case the sequence $x_n$ is just the constant sequence... and moreover if this is correct I don't see at all how to generalise to higher dimensions.
Thank you !
real-analysis calculus sequences-and-series multivariable-calculus uniform-convergence
asked Jan 28 at 23:30
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
Let $f_n : mathbb{R}^p to mathbb{R}$ such that the $f_n$ are $C^1$ and such that the sequence $(f_n)_{n in mathbb{N}}$ converges uniformly to a function $f : mathbb{R}^p to mathbb{R}$ which is $C^1$. Then prove that for all $x in mathbb{R}^n$ there is a sequence $(x_n)_{n in mathbb{N}}$ which converge to $x$ such that $mathrm{d}f_n(x_n)$ converges to $mathrm{d}f(x)$.
I must say that I don't know at all how to do and don't have any intuition of what is really going on here. So we might look at the case qhere $p= 1$.
So we can write :
$$f(a+h) = f(a)+ f'(a)h +o(h)$$
$$forall n in mathbb{N}, f_n(a+h) = f_n(a) + f'_n(a)h +o(h)$$
Hence we have :
$$mid f'(a)h - f'_n(a)h mid leq mid f(a+h)-f(a) mid +mid f(a)-f_n(a) mid + mid o(h) mid$$
Since the function $f_n$ converge uniformly to $f$, we have :
$$mid f'(a)h - f'_{infty}(a) mid leq mid o(h) mid$$
And now using we let $h to 0$ so that :
$$mid f'(a) -f_infty'(a) mid = 0 $$
I don't know if this works, but it feels strange to me since in the case the sequence $x_n$ is just the constant sequence... and moreover if this is correct I don't see at all how to generalise to higher dimensions.
Thank you !
real-analysis calculus sequences-and-series multivariable-calculus uniform-convergence
real-analysis calculus sequences-and-series multivariable-calculus uniform-convergence
asked Jan 28 at 23:30
dghkgfzyukzdghkgfzyukz
16612
asked Jan 28 at 23:30
dghkgfzyukzdghkgfzyukz
16612
asked Jan 28 at 23:30
dghkgfzyukzdghkgfzyukz
16612
asked Jan 28 at 23:30
dghkgfzyukzdghkgfzyukz
16612
asked Jan 28 at 23:30
dghkgfzyukzdghkgfzyukz
16612
16612
$begingroup$
Your demonstration is not correct, since you don’t know if $lim_{n to infty} f’_n(a)$ converges, hence you can note it like $f’_{infty}(a)$, that’s why you always need to be careful when dealing with double limit
$endgroup$
– Thinking
Jan 28 at 23:42
3
$begingroup$
For complex or vector valued functions the claim is wrong. Consider the sequence $f_n(x):={1over n}e^{inx}$, which converges uniformly to $0$, but $|f_n'(x)|=1$ for all $x$ and $n$.
$endgroup$
– Christian Blatter
Jan 31 at 10:34
add a comment |
$begingroup$
Your demonstration is not correct, since you don’t know if $lim_{n to infty} f’_n(a)$ converges, hence you can note it like $f’_{infty}(a)$, that’s why you always need to be careful when dealing with double limit
$endgroup$
– Thinking
Jan 28 at 23:42
3
$begingroup$
For complex or vector valued functions the claim is wrong. Consider the sequence $f_n(x):={1over n}e^{inx}$, which converges uniformly to $0$, but $|f_n'(x)|=1$ for all $x$ and $n$.
$endgroup$
– Christian Blatter
Jan 31 at 10:34
$begingroup$
Your demonstration is not correct, since you don’t know if $lim_{n to infty} f’_n(a)$ converges, hence you can note it like $f’_{infty}(a)$, that’s why you always need to be careful when dealing with double limit
$endgroup$
– Thinking
Jan 28 at 23:42
$begingroup$
Your demonstration is not correct, since you don’t know if $lim_{n to infty} f’_n(a)$ converges, hence you can note it like $f’_{infty}(a)$, that’s why you always need to be careful when dealing with double limit
$endgroup$
– Thinking
Jan 28 at 23:42
3
3
$begingroup$
For complex or vector valued functions the claim is wrong. Consider the sequence $f_n(x):={1over n}e^{inx}$, which converges uniformly to $0$, but $|f_n'(x)|=1$ for all $x$ and $n$.
$endgroup$
– Christian Blatter
Jan 31 at 10:34
$begingroup$
For complex or vector valued functions the claim is wrong. Consider the sequence $f_n(x):={1over n}e^{inx}$, which converges uniformly to $0$, but $|f_n'(x)|=1$ for all $x$ and $n$.
$endgroup$
– Christian Blatter
Jan 31 at 10:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We first prove the result in the case where there is a local maximum. At a maximum, the differential vanishes, so the claim is here:
Lemma. If $(g_n)_n$ is a sequence of $C^1$ functions $mathbb{R}^pto mathbb{R}$ converging uniformly to a $C^1$ function $g$ having a local (strict) maximum at $y$, that is $g(x)<g(y)$ for all $xneq y$ in a ball neigborhood $B(y,r)$ of $y$, then there exist a sequence $x_n$ such that $lim x_n=y$ and $dg_n(x_n)=0$ for sufficiently large $n$.
Proof of the Lemma: Pick $N$ sufficiently large so that $forall ngeq N$,
$$sup_{ |x-y|=r} g_n(x)<g_n(y).$$
The existence of such an $N$ follows from the fact the corresponding inequality is true for $g$ by hypothesis, and the uniform convergence of $g_n$.
For any $ngeq N$, pick $x_n$ to be a maximum of $g_n$ on $B(y,r)$. Because of the previous inequality, $x_n$ is in the interior of the ball $B(y,r)$. So the derivative satisfies $dg_n(x_n)=0$.
Let $x$ be a limit point of a subsequence of $(x_n)$. Since $g_n(x_n)geq g_n(y)$ by definition of $x_n$, taking the limit we have $g(x)geq g(y)$, and of course $x_n$ is still in the closed ball $B(y,r)$. So necessarily $x=y$ since $y$ is a local strict maximum of $g$ on $B(y,r)$. This concludes the proof of the Lemma.
Now, to deal with the general case, pick a point $y$ and define
$$g_n(x)=f_n(x)-f(x)-|x-y|^2.$$
Clearly, this sequence of $C^1$ functions converges to $g(x)=-|x-y|^2.$
We now apply the Lemma, so there is a sequence $(x_n)_n$ such that $lim x_n=y$, and
$dg_n(x_n)=0$. But since
$$dg_n(x_n).h=df_n(x_n).h-df(x_n).h-2langle x_n-y, h rangle,$$
so
$$df_n(x_n)=df(x_n)+ 2langle x_n-y, . rangle.$$
The linear forms $hmapsto 2langle x_n-y, h rangle$ converges to zero because of Cauchy-Scharwz inequality, and $df(x_n)$ converges to $df(y)$ since $f$ is $C^1$. This concludes the proof.
answered Jan 31 at 18:17
user120527user120527
1,910315
$endgroup$
$begingroup$
This is clever ! I am wondering why it's actually simpler to prove the problem when there is a local minima/maxima. I guess it's because in this case we know what is the value $mathrm{d}f$ at this point. I'll deliver the bounty tonight. Thank you !
$endgroup$
– dghkgfzyukz
Jan 31 at 18:59
add a comment |
$begingroup$
In one dimension we don't need uniform convergence; pointwise converge will do. Also we only need $f$ and $f_n, n=1,2,dots$ differentiable everywhere, not necessarily $C^1.$
WLOG we can assume $fequiv 0$ because $f_n(x)-f(x)to 0$ everywhere and $f_n-f$ is differentiable everywhere.
Fix $xin mathbb R.$ Let $delta > 0.$ Then for $nin mathbb N$ the MVT shows there exists $c(n,delta)in (x,x+delta)$ such that
$$f_n(x+delta)- f_n(x) = f_n'(c(n,delta))delta.$$
Since the left side $to 0$ as $nto infty,$ we can make the right side as small as we like by taking $n$ large. We can thus find $N = N_delta$ such that $nge N_delta$ implies $|f_n'(c(n,delta))| < delta.$
Now think of $delta_k = 1/k, k=1,2,dots .$ Then from the above there exist integers $0<N_1<N_2 < cdots$ such that $N_kle n < N_{k+1}$ implies $|f_n'(c(n,1/k))| < 1/k.$ If we then define
$$x_n = c(n,1/k),,, N_kle n N_{k+1},$$
we have $x_nto 0$ and $f'(x_n)to 0.$ (We can let $x_n$ be anything for $1le n <N_1.$)
answered Feb 5 at 18:12
zhw.zhw.
74.8k43175
$endgroup$
$begingroup$
You can't say WLOG $f = 0$ since at the beginning you are assuming $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 19:02
$begingroup$
@Thinking I don't understand your comment.
$endgroup$
– zhw.
Feb 5 at 21:45
$begingroup$
It's not so clear for me that you are not loosing generality by saying $f = 0$, since $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 22:18
$begingroup$
@Thinking Like I wrote, I am assuming $f$ and $f_n, n=1,2,dots$ are differentiable everywhere. Where do you think the proof goes wrong?
$endgroup$
– zhw.
Feb 5 at 22:22
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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$begingroup$
We first prove the result in the case where there is a local maximum. At a maximum, the differential vanishes, so the claim is here:
Lemma. If $(g_n)_n$ is a sequence of $C^1$ functions $mathbb{R}^pto mathbb{R}$ converging uniformly to a $C^1$ function $g$ having a local (strict) maximum at $y$, that is $g(x)<g(y)$ for all $xneq y$ in a ball neigborhood $B(y,r)$ of $y$, then there exist a sequence $x_n$ such that $lim x_n=y$ and $dg_n(x_n)=0$ for sufficiently large $n$.
Proof of the Lemma: Pick $N$ sufficiently large so that $forall ngeq N$,
$$sup_{ |x-y|=r} g_n(x)<g_n(y).$$
The existence of such an $N$ follows from the fact the corresponding inequality is true for $g$ by hypothesis, and the uniform convergence of $g_n$.
For any $ngeq N$, pick $x_n$ to be a maximum of $g_n$ on $B(y,r)$. Because of the previous inequality, $x_n$ is in the interior of the ball $B(y,r)$. So the derivative satisfies $dg_n(x_n)=0$.
Let $x$ be a limit point of a subsequence of $(x_n)$. Since $g_n(x_n)geq g_n(y)$ by definition of $x_n$, taking the limit we have $g(x)geq g(y)$, and of course $x_n$ is still in the closed ball $B(y,r)$. So necessarily $x=y$ since $y$ is a local strict maximum of $g$ on $B(y,r)$. This concludes the proof of the Lemma.
Now, to deal with the general case, pick a point $y$ and define
$$g_n(x)=f_n(x)-f(x)-|x-y|^2.$$
Clearly, this sequence of $C^1$ functions converges to $g(x)=-|x-y|^2.$
We now apply the Lemma, so there is a sequence $(x_n)_n$ such that $lim x_n=y$, and
$dg_n(x_n)=0$. But since
$$dg_n(x_n).h=df_n(x_n).h-df(x_n).h-2langle x_n-y, h rangle,$$
so
$$df_n(x_n)=df(x_n)+ 2langle x_n-y, . rangle.$$
The linear forms $hmapsto 2langle x_n-y, h rangle$ converges to zero because of Cauchy-Scharwz inequality, and $df(x_n)$ converges to $df(y)$ since $f$ is $C^1$. This concludes the proof.
answered Jan 31 at 18:17
user120527user120527
1,910315
$endgroup$
$begingroup$
This is clever ! I am wondering why it's actually simpler to prove the problem when there is a local minima/maxima. I guess it's because in this case we know what is the value $mathrm{d}f$ at this point. I'll deliver the bounty tonight. Thank you !
$endgroup$
– dghkgfzyukz
Jan 31 at 18:59
add a comment |
$begingroup$
We first prove the result in the case where there is a local maximum. At a maximum, the differential vanishes, so the claim is here:
Lemma. If $(g_n)_n$ is a sequence of $C^1$ functions $mathbb{R}^pto mathbb{R}$ converging uniformly to a $C^1$ function $g$ having a local (strict) maximum at $y$, that is $g(x)<g(y)$ for all $xneq y$ in a ball neigborhood $B(y,r)$ of $y$, then there exist a sequence $x_n$ such that $lim x_n=y$ and $dg_n(x_n)=0$ for sufficiently large $n$.
Proof of the Lemma: Pick $N$ sufficiently large so that $forall ngeq N$,
$$sup_{ |x-y|=r} g_n(x)<g_n(y).$$
The existence of such an $N$ follows from the fact the corresponding inequality is true for $g$ by hypothesis, and the uniform convergence of $g_n$.
For any $ngeq N$, pick $x_n$ to be a maximum of $g_n$ on $B(y,r)$. Because of the previous inequality, $x_n$ is in the interior of the ball $B(y,r)$. So the derivative satisfies $dg_n(x_n)=0$.
Let $x$ be a limit point of a subsequence of $(x_n)$. Since $g_n(x_n)geq g_n(y)$ by definition of $x_n$, taking the limit we have $g(x)geq g(y)$, and of course $x_n$ is still in the closed ball $B(y,r)$. So necessarily $x=y$ since $y$ is a local strict maximum of $g$ on $B(y,r)$. This concludes the proof of the Lemma.
Now, to deal with the general case, pick a point $y$ and define
$$g_n(x)=f_n(x)-f(x)-|x-y|^2.$$
Clearly, this sequence of $C^1$ functions converges to $g(x)=-|x-y|^2.$
We now apply the Lemma, so there is a sequence $(x_n)_n$ such that $lim x_n=y$, and
$dg_n(x_n)=0$. But since
$$dg_n(x_n).h=df_n(x_n).h-df(x_n).h-2langle x_n-y, h rangle,$$
so
$$df_n(x_n)=df(x_n)+ 2langle x_n-y, . rangle.$$
The linear forms $hmapsto 2langle x_n-y, h rangle$ converges to zero because of Cauchy-Scharwz inequality, and $df(x_n)$ converges to $df(y)$ since $f$ is $C^1$. This concludes the proof.
answered Jan 31 at 18:17
user120527user120527
1,910315
$endgroup$
$begingroup$
This is clever ! I am wondering why it's actually simpler to prove the problem when there is a local minima/maxima. I guess it's because in this case we know what is the value $mathrm{d}f$ at this point. I'll deliver the bounty tonight. Thank you !
$endgroup$
– dghkgfzyukz
Jan 31 at 18:59
add a comment |
$begingroup$
We first prove the result in the case where there is a local maximum. At a maximum, the differential vanishes, so the claim is here:
Lemma. If $(g_n)_n$ is a sequence of $C^1$ functions $mathbb{R}^pto mathbb{R}$ converging uniformly to a $C^1$ function $g$ having a local (strict) maximum at $y$, that is $g(x)<g(y)$ for all $xneq y$ in a ball neigborhood $B(y,r)$ of $y$, then there exist a sequence $x_n$ such that $lim x_n=y$ and $dg_n(x_n)=0$ for sufficiently large $n$.
Proof of the Lemma: Pick $N$ sufficiently large so that $forall ngeq N$,
$$sup_{ |x-y|=r} g_n(x)<g_n(y).$$
The existence of such an $N$ follows from the fact the corresponding inequality is true for $g$ by hypothesis, and the uniform convergence of $g_n$.
For any $ngeq N$, pick $x_n$ to be a maximum of $g_n$ on $B(y,r)$. Because of the previous inequality, $x_n$ is in the interior of the ball $B(y,r)$. So the derivative satisfies $dg_n(x_n)=0$.
Let $x$ be a limit point of a subsequence of $(x_n)$. Since $g_n(x_n)geq g_n(y)$ by definition of $x_n$, taking the limit we have $g(x)geq g(y)$, and of course $x_n$ is still in the closed ball $B(y,r)$. So necessarily $x=y$ since $y$ is a local strict maximum of $g$ on $B(y,r)$. This concludes the proof of the Lemma.
Now, to deal with the general case, pick a point $y$ and define
$$g_n(x)=f_n(x)-f(x)-|x-y|^2.$$
Clearly, this sequence of $C^1$ functions converges to $g(x)=-|x-y|^2.$
We now apply the Lemma, so there is a sequence $(x_n)_n$ such that $lim x_n=y$, and
$dg_n(x_n)=0$. But since
$$dg_n(x_n).h=df_n(x_n).h-df(x_n).h-2langle x_n-y, h rangle,$$
so
$$df_n(x_n)=df(x_n)+ 2langle x_n-y, . rangle.$$
The linear forms $hmapsto 2langle x_n-y, h rangle$ converges to zero because of Cauchy-Scharwz inequality, and $df(x_n)$ converges to $df(y)$ since $f$ is $C^1$. This concludes the proof.
answered Jan 31 at 18:17
user120527user120527
1,910315
$endgroup$
We first prove the result in the case where there is a local maximum. At a maximum, the differential vanishes, so the claim is here:
Lemma. If $(g_n)_n$ is a sequence of $C^1$ functions $mathbb{R}^pto mathbb{R}$ converging uniformly to a $C^1$ function $g$ having a local (strict) maximum at $y$, that is $g(x)<g(y)$ for all $xneq y$ in a ball neigborhood $B(y,r)$ of $y$, then there exist a sequence $x_n$ such that $lim x_n=y$ and $dg_n(x_n)=0$ for sufficiently large $n$.
Proof of the Lemma: Pick $N$ sufficiently large so that $forall ngeq N$,
$$sup_{ |x-y|=r} g_n(x)<g_n(y).$$
The existence of such an $N$ follows from the fact the corresponding inequality is true for $g$ by hypothesis, and the uniform convergence of $g_n$.
For any $ngeq N$, pick $x_n$ to be a maximum of $g_n$ on $B(y,r)$. Because of the previous inequality, $x_n$ is in the interior of the ball $B(y,r)$. So the derivative satisfies $dg_n(x_n)=0$.
Let $x$ be a limit point of a subsequence of $(x_n)$. Since $g_n(x_n)geq g_n(y)$ by definition of $x_n$, taking the limit we have $g(x)geq g(y)$, and of course $x_n$ is still in the closed ball $B(y,r)$. So necessarily $x=y$ since $y$ is a local strict maximum of $g$ on $B(y,r)$. This concludes the proof of the Lemma.
Now, to deal with the general case, pick a point $y$ and define
$$g_n(x)=f_n(x)-f(x)-|x-y|^2.$$
Clearly, this sequence of $C^1$ functions converges to $g(x)=-|x-y|^2.$
We now apply the Lemma, so there is a sequence $(x_n)_n$ such that $lim x_n=y$, and
$dg_n(x_n)=0$. But since
$$dg_n(x_n).h=df_n(x_n).h-df(x_n).h-2langle x_n-y, h rangle,$$
so
$$df_n(x_n)=df(x_n)+ 2langle x_n-y, . rangle.$$
The linear forms $hmapsto 2langle x_n-y, h rangle$ converges to zero because of Cauchy-Scharwz inequality, and $df(x_n)$ converges to $df(y)$ since $f$ is $C^1$. This concludes the proof.
answered Jan 31 at 18:17
user120527user120527
1,910315
answered Jan 31 at 18:17
user120527user120527
1,910315
answered Jan 31 at 18:17
user120527user120527
1,910315
answered Jan 31 at 18:17
user120527user120527
1,910315
1,910315
$begingroup$
This is clever ! I am wondering why it's actually simpler to prove the problem when there is a local minima/maxima. I guess it's because in this case we know what is the value $mathrm{d}f$ at this point. I'll deliver the bounty tonight. Thank you !
$endgroup$
– dghkgfzyukz
Jan 31 at 18:59
add a comment |
$begingroup$
This is clever ! I am wondering why it's actually simpler to prove the problem when there is a local minima/maxima. I guess it's because in this case we know what is the value $mathrm{d}f$ at this point. I'll deliver the bounty tonight. Thank you !
$endgroup$
– dghkgfzyukz
Jan 31 at 18:59
$begingroup$
This is clever ! I am wondering why it's actually simpler to prove the problem when there is a local minima/maxima. I guess it's because in this case we know what is the value $mathrm{d}f$ at this point. I'll deliver the bounty tonight. Thank you !
$endgroup$
– dghkgfzyukz
Jan 31 at 18:59
$begingroup$
This is clever ! I am wondering why it's actually simpler to prove the problem when there is a local minima/maxima. I guess it's because in this case we know what is the value $mathrm{d}f$ at this point. I'll deliver the bounty tonight. Thank you !
$endgroup$
– dghkgfzyukz
Jan 31 at 18:59
add a comment |
$begingroup$
In one dimension we don't need uniform convergence; pointwise converge will do. Also we only need $f$ and $f_n, n=1,2,dots$ differentiable everywhere, not necessarily $C^1.$
WLOG we can assume $fequiv 0$ because $f_n(x)-f(x)to 0$ everywhere and $f_n-f$ is differentiable everywhere.
Fix $xin mathbb R.$ Let $delta > 0.$ Then for $nin mathbb N$ the MVT shows there exists $c(n,delta)in (x,x+delta)$ such that
$$f_n(x+delta)- f_n(x) = f_n'(c(n,delta))delta.$$
Since the left side $to 0$ as $nto infty,$ we can make the right side as small as we like by taking $n$ large. We can thus find $N = N_delta$ such that $nge N_delta$ implies $|f_n'(c(n,delta))| < delta.$
Now think of $delta_k = 1/k, k=1,2,dots .$ Then from the above there exist integers $0<N_1<N_2 < cdots$ such that $N_kle n < N_{k+1}$ implies $|f_n'(c(n,1/k))| < 1/k.$ If we then define
$$x_n = c(n,1/k),,, N_kle n N_{k+1},$$
we have $x_nto 0$ and $f'(x_n)to 0.$ (We can let $x_n$ be anything for $1le n <N_1.$)
answered Feb 5 at 18:12
zhw.zhw.
74.8k43175
$endgroup$
$begingroup$
You can't say WLOG $f = 0$ since at the beginning you are assuming $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 19:02
$begingroup$
@Thinking I don't understand your comment.
$endgroup$
– zhw.
Feb 5 at 21:45
$begingroup$
It's not so clear for me that you are not loosing generality by saying $f = 0$, since $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 22:18
$begingroup$
@Thinking Like I wrote, I am assuming $f$ and $f_n, n=1,2,dots$ are differentiable everywhere. Where do you think the proof goes wrong?
$endgroup$
– zhw.
Feb 5 at 22:22
add a comment |
$begingroup$
In one dimension we don't need uniform convergence; pointwise converge will do. Also we only need $f$ and $f_n, n=1,2,dots$ differentiable everywhere, not necessarily $C^1.$
WLOG we can assume $fequiv 0$ because $f_n(x)-f(x)to 0$ everywhere and $f_n-f$ is differentiable everywhere.
Fix $xin mathbb R.$ Let $delta > 0.$ Then for $nin mathbb N$ the MVT shows there exists $c(n,delta)in (x,x+delta)$ such that
$$f_n(x+delta)- f_n(x) = f_n'(c(n,delta))delta.$$
Since the left side $to 0$ as $nto infty,$ we can make the right side as small as we like by taking $n$ large. We can thus find $N = N_delta$ such that $nge N_delta$ implies $|f_n'(c(n,delta))| < delta.$
Now think of $delta_k = 1/k, k=1,2,dots .$ Then from the above there exist integers $0<N_1<N_2 < cdots$ such that $N_kle n < N_{k+1}$ implies $|f_n'(c(n,1/k))| < 1/k.$ If we then define
$$x_n = c(n,1/k),,, N_kle n N_{k+1},$$
we have $x_nto 0$ and $f'(x_n)to 0.$ (We can let $x_n$ be anything for $1le n <N_1.$)
answered Feb 5 at 18:12
zhw.zhw.
74.8k43175
$endgroup$
$begingroup$
You can't say WLOG $f = 0$ since at the beginning you are assuming $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 19:02
$begingroup$
@Thinking I don't understand your comment.
$endgroup$
– zhw.
Feb 5 at 21:45
$begingroup$
It's not so clear for me that you are not loosing generality by saying $f = 0$, since $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 22:18
$begingroup$
@Thinking Like I wrote, I am assuming $f$ and $f_n, n=1,2,dots$ are differentiable everywhere. Where do you think the proof goes wrong?
$endgroup$
– zhw.
Feb 5 at 22:22
add a comment |
$begingroup$
In one dimension we don't need uniform convergence; pointwise converge will do. Also we only need $f$ and $f_n, n=1,2,dots$ differentiable everywhere, not necessarily $C^1.$
WLOG we can assume $fequiv 0$ because $f_n(x)-f(x)to 0$ everywhere and $f_n-f$ is differentiable everywhere.
Fix $xin mathbb R.$ Let $delta > 0.$ Then for $nin mathbb N$ the MVT shows there exists $c(n,delta)in (x,x+delta)$ such that
$$f_n(x+delta)- f_n(x) = f_n'(c(n,delta))delta.$$
Since the left side $to 0$ as $nto infty,$ we can make the right side as small as we like by taking $n$ large. We can thus find $N = N_delta$ such that $nge N_delta$ implies $|f_n'(c(n,delta))| < delta.$
Now think of $delta_k = 1/k, k=1,2,dots .$ Then from the above there exist integers $0<N_1<N_2 < cdots$ such that $N_kle n < N_{k+1}$ implies $|f_n'(c(n,1/k))| < 1/k.$ If we then define
$$x_n = c(n,1/k),,, N_kle n N_{k+1},$$
we have $x_nto 0$ and $f'(x_n)to 0.$ (We can let $x_n$ be anything for $1le n <N_1.$)
answered Feb 5 at 18:12
zhw.zhw.
74.8k43175
$endgroup$
In one dimension we don't need uniform convergence; pointwise converge will do. Also we only need $f$ and $f_n, n=1,2,dots$ differentiable everywhere, not necessarily $C^1.$
WLOG we can assume $fequiv 0$ because $f_n(x)-f(x)to 0$ everywhere and $f_n-f$ is differentiable everywhere.
Fix $xin mathbb R.$ Let $delta > 0.$ Then for $nin mathbb N$ the MVT shows there exists $c(n,delta)in (x,x+delta)$ such that
$$f_n(x+delta)- f_n(x) = f_n'(c(n,delta))delta.$$
Since the left side $to 0$ as $nto infty,$ we can make the right side as small as we like by taking $n$ large. We can thus find $N = N_delta$ such that $nge N_delta$ implies $|f_n'(c(n,delta))| < delta.$
Now think of $delta_k = 1/k, k=1,2,dots .$ Then from the above there exist integers $0<N_1<N_2 < cdots$ such that $N_kle n < N_{k+1}$ implies $|f_n'(c(n,1/k))| < 1/k.$ If we then define
$$x_n = c(n,1/k),,, N_kle n N_{k+1},$$
we have $x_nto 0$ and $f'(x_n)to 0.$ (We can let $x_n$ be anything for $1le n <N_1.$)
answered Feb 5 at 18:12
zhw.zhw.
74.8k43175
answered Feb 5 at 18:12
zhw.zhw.
74.8k43175
answered Feb 5 at 18:12
zhw.zhw.
74.8k43175
answered Feb 5 at 18:12
zhw.zhw.
74.8k43175
74.8k43175
$begingroup$
You can't say WLOG $f = 0$ since at the beginning you are assuming $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 19:02
$begingroup$
@Thinking I don't understand your comment.
$endgroup$
– zhw.
Feb 5 at 21:45
$begingroup$
It's not so clear for me that you are not loosing generality by saying $f = 0$, since $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 22:18
$begingroup$
@Thinking Like I wrote, I am assuming $f$ and $f_n, n=1,2,dots$ are differentiable everywhere. Where do you think the proof goes wrong?
$endgroup$
– zhw.
Feb 5 at 22:22
add a comment |
$begingroup$
You can't say WLOG $f = 0$ since at the beginning you are assuming $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 19:02
$begingroup$
@Thinking I don't understand your comment.
$endgroup$
– zhw.
Feb 5 at 21:45
$begingroup$
It's not so clear for me that you are not loosing generality by saying $f = 0$, since $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 22:18
$begingroup$
@Thinking Like I wrote, I am assuming $f$ and $f_n, n=1,2,dots$ are differentiable everywhere. Where do you think the proof goes wrong?
$endgroup$
– zhw.
Feb 5 at 22:22
$begingroup$
You can't say WLOG $f = 0$ since at the beginning you are assuming $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 19:02
$begingroup$
You can't say WLOG $f = 0$ since at the beginning you are assuming $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 19:02
$begingroup$
@Thinking I don't understand your comment.
$endgroup$
– zhw.
Feb 5 at 21:45
$begingroup$
@Thinking I don't understand your comment.
$endgroup$
– zhw.
Feb 5 at 21:45
$begingroup$
It's not so clear for me that you are not loosing generality by saying $f = 0$, since $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 22:18
$begingroup$
It's not so clear for me that you are not loosing generality by saying $f = 0$, since $f$ is not necessarily $C^1$.
$endgroup$
– Thinking
Feb 5 at 22:18
$begingroup$
@Thinking Like I wrote, I am assuming $f$ and $f_n, n=1,2,dots$ are differentiable everywhere. Where do you think the proof goes wrong?
$endgroup$
– zhw.
Feb 5 at 22:22
$begingroup$
@Thinking Like I wrote, I am assuming $f$ and $f_n, n=1,2,dots$ are differentiable everywhere. Where do you think the proof goes wrong?
$endgroup$
– zhw.
Feb 5 at 22:22
add a comment |
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Your demonstration is not correct, since you don’t know if $lim_{n to infty} f’_n(a)$ converges, hence you can note it like $f’_{infty}(a)$, that’s why you always need to be careful when dealing with double limit
$endgroup$
– Thinking
Jan 28 at 23:42
3
$begingroup$
For complex or vector valued functions the claim is wrong. Consider the sequence $f_n(x):={1over n}e^{inx}$, which converges uniformly to $0$, but $|f_n'(x)|=1$ for all $x$ and $n$.
$endgroup$
– Christian Blatter
Jan 31 at 10:34