Mixing
Differentiability of $f(x, y) = |xy|$ at $0$; is this a mistake by Munkres?
$begingroup$
I am reading Analysis on Manifolds by Munkres, and question $1$ on page $54$ says
Show that the function $f(x, y) = |xy|$ is differentiable at $0$, but it is not of class $C^1$ in any neighborhood of $0$.
I know that if a function $f$ is differentiable at a point, then all its partial derivatives exist at that point. However,
$$D_1f(x, y) = begin{cases}
|y|&text{if}, x>0 \
-|y| &text{if}, x<0
end{cases}
$$
and $D_1f(0, 0)$ does not exist. How can $f$ be differentiable at $0$?
real-analysis multivariable-calculus
asked Jan 21 at 1:26
OviOvi
12.4k1040113
$endgroup$
add a comment |
$begingroup$
I am reading Analysis on Manifolds by Munkres, and question $1$ on page $54$ says
Show that the function $f(x, y) = |xy|$ is differentiable at $0$, but it is not of class $C^1$ in any neighborhood of $0$.
I know that if a function $f$ is differentiable at a point, then all its partial derivatives exist at that point. However,
$$D_1f(x, y) = begin{cases}
|y|&text{if}, x>0 \
-|y| &text{if}, x<0
end{cases}
$$
and $D_1f(0, 0)$ does not exist. How can $f$ be differentiable at $0$?
real-analysis multivariable-calculus
asked Jan 21 at 1:26
OviOvi
12.4k1040113
$endgroup$
add a comment |
$begingroup$
I am reading Analysis on Manifolds by Munkres, and question $1$ on page $54$ says
Show that the function $f(x, y) = |xy|$ is differentiable at $0$, but it is not of class $C^1$ in any neighborhood of $0$.
I know that if a function $f$ is differentiable at a point, then all its partial derivatives exist at that point. However,
$$D_1f(x, y) = begin{cases}
|y|&text{if}, x>0 \
-|y| &text{if}, x<0
end{cases}
$$
and $D_1f(0, 0)$ does not exist. How can $f$ be differentiable at $0$?
real-analysis multivariable-calculus
asked Jan 21 at 1:26
OviOvi
12.4k1040113
$endgroup$
I am reading Analysis on Manifolds by Munkres, and question $1$ on page $54$ says
Show that the function $f(x, y) = |xy|$ is differentiable at $0$, but it is not of class $C^1$ in any neighborhood of $0$.
I know that if a function $f$ is differentiable at a point, then all its partial derivatives exist at that point. However,
$$D_1f(x, y) = begin{cases}
|y|&text{if}, x>0 \
-|y| &text{if}, x<0
end{cases}
$$
and $D_1f(0, 0)$ does not exist. How can $f$ be differentiable at $0$?
real-analysis multivariable-calculus
real-analysis multivariable-calculus
asked Jan 21 at 1:26
OviOvi
12.4k1040113
asked Jan 21 at 1:26
OviOvi
12.4k1040113
asked Jan 21 at 1:26
OviOvi
12.4k1040113
asked Jan 21 at 1:26
OviOvi
12.4k1040113
asked Jan 21 at 1:26
OviOvi
12.4k1040113
12.4k1040113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have that
$$frac{x^2y^2}{x^2+y^2}leq x^2+y^2,$$
so if $||(x,y)||=r$ then
$$frac{|xy|}{||(x,y)||}=sqrt{frac{x^2y^2}{x^2+y^2}}leq r,$$
and as such
$$lim_{x,yto 0} frac{|xy|}{||(x,y)||}=0$$
as $(x,y)to(0,0)$ implies that $||(x,y)||to 0$. The reason your statement does not imply nondifferentiability is that the fact that
$$D_1f(x, y) = begin{cases}
|y|&text{if}, x>0 \
-|y| &text{if}, x<0
end{cases}
$$
is only problematic when $y$ is away from $0$, otherwise $|y|approx-|y|$. The partials still very much exist at $f(0,0)$, as $|y|,-|y|to 0$. However, you can use your result to conclude that $D_1 f(x,y)$ does not exist if $yneq 0$.
answered Jan 21 at 1:35
Carl SchildkrautCarl Schildkraut
11.7k11443
$endgroup$
add a comment |
$begingroup$
How did you conclude that $D_1f(0,0)$ does not exist? Try using the definition directly. You are probably confusing differentiable and continuously differentiable.
answered Jan 21 at 1:31
KlausKlaus
2,12711
$endgroup$
1
$begingroup$
Ohhh I was thinking that when $y$ is fixed, we have the absolute value function, which is not differentiable at $0$. But of course, when we talk about $D_1f$ at $(0, 0)$, we're fixing $y$ to be zero, in which case we get the zero function.
$endgroup$
– Ovi
Jan 21 at 1:35
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
We have that
$$frac{x^2y^2}{x^2+y^2}leq x^2+y^2,$$
so if $||(x,y)||=r$ then
$$frac{|xy|}{||(x,y)||}=sqrt{frac{x^2y^2}{x^2+y^2}}leq r,$$
and as such
$$lim_{x,yto 0} frac{|xy|}{||(x,y)||}=0$$
as $(x,y)to(0,0)$ implies that $||(x,y)||to 0$. The reason your statement does not imply nondifferentiability is that the fact that
$$D_1f(x, y) = begin{cases}
|y|&text{if}, x>0 \
-|y| &text{if}, x<0
end{cases}
$$
is only problematic when $y$ is away from $0$, otherwise $|y|approx-|y|$. The partials still very much exist at $f(0,0)$, as $|y|,-|y|to 0$. However, you can use your result to conclude that $D_1 f(x,y)$ does not exist if $yneq 0$.
answered Jan 21 at 1:35
Carl SchildkrautCarl Schildkraut
11.7k11443
$endgroup$
add a comment |
$begingroup$
We have that
$$frac{x^2y^2}{x^2+y^2}leq x^2+y^2,$$
so if $||(x,y)||=r$ then
$$frac{|xy|}{||(x,y)||}=sqrt{frac{x^2y^2}{x^2+y^2}}leq r,$$
and as such
$$lim_{x,yto 0} frac{|xy|}{||(x,y)||}=0$$
as $(x,y)to(0,0)$ implies that $||(x,y)||to 0$. The reason your statement does not imply nondifferentiability is that the fact that
$$D_1f(x, y) = begin{cases}
|y|&text{if}, x>0 \
-|y| &text{if}, x<0
end{cases}
$$
is only problematic when $y$ is away from $0$, otherwise $|y|approx-|y|$. The partials still very much exist at $f(0,0)$, as $|y|,-|y|to 0$. However, you can use your result to conclude that $D_1 f(x,y)$ does not exist if $yneq 0$.
answered Jan 21 at 1:35
Carl SchildkrautCarl Schildkraut
11.7k11443
$endgroup$
add a comment |
$begingroup$
We have that
$$frac{x^2y^2}{x^2+y^2}leq x^2+y^2,$$
so if $||(x,y)||=r$ then
$$frac{|xy|}{||(x,y)||}=sqrt{frac{x^2y^2}{x^2+y^2}}leq r,$$
and as such
$$lim_{x,yto 0} frac{|xy|}{||(x,y)||}=0$$
as $(x,y)to(0,0)$ implies that $||(x,y)||to 0$. The reason your statement does not imply nondifferentiability is that the fact that
$$D_1f(x, y) = begin{cases}
|y|&text{if}, x>0 \
-|y| &text{if}, x<0
end{cases}
$$
is only problematic when $y$ is away from $0$, otherwise $|y|approx-|y|$. The partials still very much exist at $f(0,0)$, as $|y|,-|y|to 0$. However, you can use your result to conclude that $D_1 f(x,y)$ does not exist if $yneq 0$.
answered Jan 21 at 1:35
Carl SchildkrautCarl Schildkraut
11.7k11443
$endgroup$
We have that
$$frac{x^2y^2}{x^2+y^2}leq x^2+y^2,$$
so if $||(x,y)||=r$ then
$$frac{|xy|}{||(x,y)||}=sqrt{frac{x^2y^2}{x^2+y^2}}leq r,$$
and as such
$$lim_{x,yto 0} frac{|xy|}{||(x,y)||}=0$$
as $(x,y)to(0,0)$ implies that $||(x,y)||to 0$. The reason your statement does not imply nondifferentiability is that the fact that
$$D_1f(x, y) = begin{cases}
|y|&text{if}, x>0 \
-|y| &text{if}, x<0
end{cases}
$$
is only problematic when $y$ is away from $0$, otherwise $|y|approx-|y|$. The partials still very much exist at $f(0,0)$, as $|y|,-|y|to 0$. However, you can use your result to conclude that $D_1 f(x,y)$ does not exist if $yneq 0$.
answered Jan 21 at 1:35
Carl SchildkrautCarl Schildkraut
11.7k11443
answered Jan 21 at 1:35
Carl SchildkrautCarl Schildkraut
11.7k11443
answered Jan 21 at 1:35
Carl SchildkrautCarl Schildkraut
11.7k11443
answered Jan 21 at 1:35
Carl SchildkrautCarl Schildkraut
11.7k11443
11.7k11443
add a comment |
add a comment |
$begingroup$
How did you conclude that $D_1f(0,0)$ does not exist? Try using the definition directly. You are probably confusing differentiable and continuously differentiable.
answered Jan 21 at 1:31
KlausKlaus
2,12711
$endgroup$
1
$begingroup$
Ohhh I was thinking that when $y$ is fixed, we have the absolute value function, which is not differentiable at $0$. But of course, when we talk about $D_1f$ at $(0, 0)$, we're fixing $y$ to be zero, in which case we get the zero function.
$endgroup$
– Ovi
Jan 21 at 1:35
add a comment |
$begingroup$
How did you conclude that $D_1f(0,0)$ does not exist? Try using the definition directly. You are probably confusing differentiable and continuously differentiable.
answered Jan 21 at 1:31
KlausKlaus
2,12711
$endgroup$
1
$begingroup$
Ohhh I was thinking that when $y$ is fixed, we have the absolute value function, which is not differentiable at $0$. But of course, when we talk about $D_1f$ at $(0, 0)$, we're fixing $y$ to be zero, in which case we get the zero function.
$endgroup$
– Ovi
Jan 21 at 1:35
add a comment |
$begingroup$
How did you conclude that $D_1f(0,0)$ does not exist? Try using the definition directly. You are probably confusing differentiable and continuously differentiable.
answered Jan 21 at 1:31
KlausKlaus
2,12711
$endgroup$
How did you conclude that $D_1f(0,0)$ does not exist? Try using the definition directly. You are probably confusing differentiable and continuously differentiable.
answered Jan 21 at 1:31
KlausKlaus
2,12711
answered Jan 21 at 1:31
KlausKlaus
2,12711
answered Jan 21 at 1:31
KlausKlaus
2,12711
answered Jan 21 at 1:31
KlausKlaus
2,12711
2,12711
1
$begingroup$
Ohhh I was thinking that when $y$ is fixed, we have the absolute value function, which is not differentiable at $0$. But of course, when we talk about $D_1f$ at $(0, 0)$, we're fixing $y$ to be zero, in which case we get the zero function.
$endgroup$
– Ovi
Jan 21 at 1:35
add a comment |
1
$begingroup$
Ohhh I was thinking that when $y$ is fixed, we have the absolute value function, which is not differentiable at $0$. But of course, when we talk about $D_1f$ at $(0, 0)$, we're fixing $y$ to be zero, in which case we get the zero function.
$endgroup$
– Ovi
Jan 21 at 1:35
1
1
$begingroup$
Ohhh I was thinking that when $y$ is fixed, we have the absolute value function, which is not differentiable at $0$. But of course, when we talk about $D_1f$ at $(0, 0)$, we're fixing $y$ to be zero, in which case we get the zero function.
$endgroup$
– Ovi
Jan 21 at 1:35
$begingroup$
Ohhh I was thinking that when $y$ is fixed, we have the absolute value function, which is not differentiable at $0$. But of course, when we talk about $D_1f$ at $(0, 0)$, we're fixing $y$ to be zero, in which case we get the zero function.
$endgroup$
– Ovi
Jan 21 at 1:35
add a comment |
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