Fake proof of differentiability
$begingroup$
It's a theorem that if $fcolon UsubsetBbb R^nto Bbb R^m$ has the property that each of the partial derivatives $partial_if_j$ exist and are continuous $pin U$, then $f$ is differentiable at $p$. When I was trying to prove this, I came up with the following "proof" which doesn't use the continuity hypothesis. Can someone tell me what's wrong with this proof?
Since $f_j$ is differentiable at $p$, we can write
$$
f_j(p+v) = f_j(p) + sum_i partial_if_j(p)v_i + R_j(v),
$$
where $|R_j(v)|/|v| to 0$ as $vto 0$. Hence, we can write
begin{align*}
f(p+v) &= f(p) + big(sum_i partial_if_1(p)v_i + R_1(v),dots,sum_i partial_if_m(p)v_i + R_m(v)big) \
&= f(p) + sum_jbig(sum_ipartial_if_j(p)v_ibig)e_j + R_j(v)e_j \
&= f(p) + [Df_p][v] + (R_1,dots,R_m)(v),
end{align*}
where $[Df_p] = [partial_if_j(p)]$ is the usual Jacobian matrix, and $[v]$ is the column vector $[v_1 dotsb v_n]^T$. Now,
$$
frac{|(R_1,dots,R_m)(v)|^2}{|v|^2} = frac{R_1(v)^2 + dots + R_m(v)^2}{|v|^2} to 0,
$$
where the last expression goes to $0$ as $vto 0$ since it is a sum of finitely many terms, each of which goes to $0$. Hence we have written $f(p+v)$ as a sum of a constant term, a linear part, and a sublinear piece, so $f$ is differentiable at $p$. At no point did I explicitly use the continuity hypothesis, so what exactly is wrong with this proof? Best.
real-analysis multivariable-calculus derivatives proof-verification
asked Dec 26 '18 at 23:03
Alex OrtizAlex Ortiz
10.7k21441
$endgroup$
add a comment |
$begingroup$
It's a theorem that if $fcolon UsubsetBbb R^nto Bbb R^m$ has the property that each of the partial derivatives $partial_if_j$ exist and are continuous $pin U$, then $f$ is differentiable at $p$. When I was trying to prove this, I came up with the following "proof" which doesn't use the continuity hypothesis. Can someone tell me what's wrong with this proof?
Since $f_j$ is differentiable at $p$, we can write
$$
f_j(p+v) = f_j(p) + sum_i partial_if_j(p)v_i + R_j(v),
$$
where $|R_j(v)|/|v| to 0$ as $vto 0$. Hence, we can write
begin{align*}
f(p+v) &= f(p) + big(sum_i partial_if_1(p)v_i + R_1(v),dots,sum_i partial_if_m(p)v_i + R_m(v)big) \
&= f(p) + sum_jbig(sum_ipartial_if_j(p)v_ibig)e_j + R_j(v)e_j \
&= f(p) + [Df_p][v] + (R_1,dots,R_m)(v),
end{align*}
where $[Df_p] = [partial_if_j(p)]$ is the usual Jacobian matrix, and $[v]$ is the column vector $[v_1 dotsb v_n]^T$. Now,
$$
frac{|(R_1,dots,R_m)(v)|^2}{|v|^2} = frac{R_1(v)^2 + dots + R_m(v)^2}{|v|^2} to 0,
$$
where the last expression goes to $0$ as $vto 0$ since it is a sum of finitely many terms, each of which goes to $0$. Hence we have written $f(p+v)$ as a sum of a constant term, a linear part, and a sublinear piece, so $f$ is differentiable at $p$. At no point did I explicitly use the continuity hypothesis, so what exactly is wrong with this proof? Best.
real-analysis multivariable-calculus derivatives proof-verification
asked Dec 26 '18 at 23:03
Alex OrtizAlex Ortiz
10.7k21441
$endgroup$
add a comment |
$begingroup$
It's a theorem that if $fcolon UsubsetBbb R^nto Bbb R^m$ has the property that each of the partial derivatives $partial_if_j$ exist and are continuous $pin U$, then $f$ is differentiable at $p$. When I was trying to prove this, I came up with the following "proof" which doesn't use the continuity hypothesis. Can someone tell me what's wrong with this proof?
Since $f_j$ is differentiable at $p$, we can write
$$
f_j(p+v) = f_j(p) + sum_i partial_if_j(p)v_i + R_j(v),
$$
where $|R_j(v)|/|v| to 0$ as $vto 0$. Hence, we can write
begin{align*}
f(p+v) &= f(p) + big(sum_i partial_if_1(p)v_i + R_1(v),dots,sum_i partial_if_m(p)v_i + R_m(v)big) \
&= f(p) + sum_jbig(sum_ipartial_if_j(p)v_ibig)e_j + R_j(v)e_j \
&= f(p) + [Df_p][v] + (R_1,dots,R_m)(v),
end{align*}
where $[Df_p] = [partial_if_j(p)]$ is the usual Jacobian matrix, and $[v]$ is the column vector $[v_1 dotsb v_n]^T$. Now,
$$
frac{|(R_1,dots,R_m)(v)|^2}{|v|^2} = frac{R_1(v)^2 + dots + R_m(v)^2}{|v|^2} to 0,
$$
where the last expression goes to $0$ as $vto 0$ since it is a sum of finitely many terms, each of which goes to $0$. Hence we have written $f(p+v)$ as a sum of a constant term, a linear part, and a sublinear piece, so $f$ is differentiable at $p$. At no point did I explicitly use the continuity hypothesis, so what exactly is wrong with this proof? Best.
real-analysis multivariable-calculus derivatives proof-verification
asked Dec 26 '18 at 23:03
Alex OrtizAlex Ortiz
10.7k21441
$endgroup$
It's a theorem that if $fcolon UsubsetBbb R^nto Bbb R^m$ has the property that each of the partial derivatives $partial_if_j$ exist and are continuous $pin U$, then $f$ is differentiable at $p$. When I was trying to prove this, I came up with the following "proof" which doesn't use the continuity hypothesis. Can someone tell me what's wrong with this proof?
Since $f_j$ is differentiable at $p$, we can write
$$
f_j(p+v) = f_j(p) + sum_i partial_if_j(p)v_i + R_j(v),
$$
where $|R_j(v)|/|v| to 0$ as $vto 0$. Hence, we can write
begin{align*}
f(p+v) &= f(p) + big(sum_i partial_if_1(p)v_i + R_1(v),dots,sum_i partial_if_m(p)v_i + R_m(v)big) \
&= f(p) + sum_jbig(sum_ipartial_if_j(p)v_ibig)e_j + R_j(v)e_j \
&= f(p) + [Df_p][v] + (R_1,dots,R_m)(v),
end{align*}
where $[Df_p] = [partial_if_j(p)]$ is the usual Jacobian matrix, and $[v]$ is the column vector $[v_1 dotsb v_n]^T$. Now,
$$
frac{|(R_1,dots,R_m)(v)|^2}{|v|^2} = frac{R_1(v)^2 + dots + R_m(v)^2}{|v|^2} to 0,
$$
where the last expression goes to $0$ as $vto 0$ since it is a sum of finitely many terms, each of which goes to $0$. Hence we have written $f(p+v)$ as a sum of a constant term, a linear part, and a sublinear piece, so $f$ is differentiable at $p$. At no point did I explicitly use the continuity hypothesis, so what exactly is wrong with this proof? Best.
real-analysis multivariable-calculus derivatives proof-verification
real-analysis multivariable-calculus derivatives proof-verification
asked Dec 26 '18 at 23:03
Alex OrtizAlex Ortiz
10.7k21441
asked Dec 26 '18 at 23:03
Alex OrtizAlex Ortiz
10.7k21441
asked Dec 26 '18 at 23:03
Alex OrtizAlex Ortiz
10.7k21441
asked Dec 26 '18 at 23:03
Alex OrtizAlex Ortiz
10.7k21441
asked Dec 26 '18 at 23:03
Alex OrtizAlex Ortiz
10.7k21441
10.7k21441
add a comment |
add a comment |
2 Answers
2
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oldest
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In short, if you don't assume that the $partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.
You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).
answered Dec 26 '18 at 23:12
0x5390x539
1,403518
$endgroup$
add a comment |
$begingroup$
The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $mathbb R^{n}$.
answered Dec 26 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
In short, if you don't assume that the $partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.
You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).
answered Dec 26 '18 at 23:12
0x5390x539
1,403518
$endgroup$
add a comment |
$begingroup$
In short, if you don't assume that the $partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.
You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).
answered Dec 26 '18 at 23:12
0x5390x539
1,403518
$endgroup$
add a comment |
$begingroup$
In short, if you don't assume that the $partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.
You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).
answered Dec 26 '18 at 23:12
0x5390x539
1,403518
$endgroup$
In short, if you don't assume that the $partial_i f_j$ are continuous then you can't assume that $f_j$ is differentiable at $p$.
You only know that all partial derivatives of $f_j$ exist, but you need continuity to guarantee that $f_j$ is actually differentiable (that's the $m=1$ case of the theorem you talk about).
answered Dec 26 '18 at 23:12
0x5390x539
1,403518
answered Dec 26 '18 at 23:12
0x5390x539
1,403518
answered Dec 26 '18 at 23:12
0x5390x539
1,403518
answered Dec 26 '18 at 23:12
0x5390x539
1,403518
1,403518
add a comment |
add a comment |
$begingroup$
The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $mathbb R^{n}$.
answered Dec 26 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
add a comment |
$begingroup$
The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $mathbb R^{n}$.
answered Dec 26 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
add a comment |
$begingroup$
The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $mathbb R^{n}$.
answered Dec 26 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
The very first step is wrong. You are only given that partial derivatives exist and are continuous, not that $f_j$ is a differentiable function on $mathbb R^{n}$.
answered Dec 26 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
answered Dec 26 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
answered Dec 26 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
answered Dec 26 '18 at 23:14
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
60.3k42161
add a comment |
add a comment |
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