Show that the function $f(x, y) = |xy|$ is not of class $C^1$ in any neighborhood of $(0, 0)$
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Show that the function $f(x, y) = |xy|$ is not of class $C^1$ in any neighborhood of $(0, 0)$
This is my attempted proof:
Proof: Choose a neighborhood $U$ of $(0, 0) in mathbb{R}^2$. Since $U$ is a neighborhood of $(0, 0)$ it must contain points of the form $x = (0, b)$ for sufficiently small $b$.
But then
begin{align*}
D_1f(x) &= lim_{t to 0} frac{f(x+te_1) -f(x)}{t} \
&= lim_{t to 0} frac{f((0, b) + (t, 0)) -f(0, b)}{t} \
&= lim_{t to 0} frac{|tb|-|b|}{t} \
&= lim_{t to 0} frac{|t|}{t} |b| - frac{1}{t}|b|
end{align*}
But now consider the limit as positive $t$ approaches $0$, $$lim_{t^+ to 0} frac{|t|}{t} |b| - frac{1}{t}|b| = lim_{t^+ to 0}|b|-frac{1}{t}|b| (*)$$
and $|b| - frac{1}{t}|b| to -infty$ as $t^+ to 0$, so the above limit $(*)$ doesn't exist and thus $lim_{t to 0} frac{|t|}{t} |b| - frac{1}{t}|b|$ doesn't exist and so $D_1f(x)$ doesn't exist and thus $f$ cannot be of class $C^1$ in any neighborhood of $(0, 0)$. $square$.
Is the above proof correct and rigorous? (Note I have left out some details on why there exists such an $x$ for sufficiently small $b$ as it's not too hard to show using the definition of open sets in metric spaces).
multivariable-calculus proof-verification
asked Nov 24 at 17:00
Perturbative
3,93011449
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0
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Show that the function $f(x, y) = |xy|$ is not of class $C^1$ in any neighborhood of $(0, 0)$
This is my attempted proof:
Proof: Choose a neighborhood $U$ of $(0, 0) in mathbb{R}^2$. Since $U$ is a neighborhood of $(0, 0)$ it must contain points of the form $x = (0, b)$ for sufficiently small $b$.
But then
begin{align*}
D_1f(x) &= lim_{t to 0} frac{f(x+te_1) -f(x)}{t} \
&= lim_{t to 0} frac{f((0, b) + (t, 0)) -f(0, b)}{t} \
&= lim_{t to 0} frac{|tb|-|b|}{t} \
&= lim_{t to 0} frac{|t|}{t} |b| - frac{1}{t}|b|
end{align*}
But now consider the limit as positive $t$ approaches $0$, $$lim_{t^+ to 0} frac{|t|}{t} |b| - frac{1}{t}|b| = lim_{t^+ to 0}|b|-frac{1}{t}|b| (*)$$
and $|b| - frac{1}{t}|b| to -infty$ as $t^+ to 0$, so the above limit $(*)$ doesn't exist and thus $lim_{t to 0} frac{|t|}{t} |b| - frac{1}{t}|b|$ doesn't exist and so $D_1f(x)$ doesn't exist and thus $f$ cannot be of class $C^1$ in any neighborhood of $(0, 0)$. $square$.
Is the above proof correct and rigorous? (Note I have left out some details on why there exists such an $x$ for sufficiently small $b$ as it's not too hard to show using the definition of open sets in metric spaces).
multivariable-calculus proof-verification
asked Nov 24 at 17:00
Perturbative
3,93011449
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that the function $f(x, y) = |xy|$ is not of class $C^1$ in any neighborhood of $(0, 0)$
This is my attempted proof:
Proof: Choose a neighborhood $U$ of $(0, 0) in mathbb{R}^2$. Since $U$ is a neighborhood of $(0, 0)$ it must contain points of the form $x = (0, b)$ for sufficiently small $b$.
But then
begin{align*}
D_1f(x) &= lim_{t to 0} frac{f(x+te_1) -f(x)}{t} \
&= lim_{t to 0} frac{f((0, b) + (t, 0)) -f(0, b)}{t} \
&= lim_{t to 0} frac{|tb|-|b|}{t} \
&= lim_{t to 0} frac{|t|}{t} |b| - frac{1}{t}|b|
end{align*}
But now consider the limit as positive $t$ approaches $0$, $$lim_{t^+ to 0} frac{|t|}{t} |b| - frac{1}{t}|b| = lim_{t^+ to 0}|b|-frac{1}{t}|b| (*)$$
and $|b| - frac{1}{t}|b| to -infty$ as $t^+ to 0$, so the above limit $(*)$ doesn't exist and thus $lim_{t to 0} frac{|t|}{t} |b| - frac{1}{t}|b|$ doesn't exist and so $D_1f(x)$ doesn't exist and thus $f$ cannot be of class $C^1$ in any neighborhood of $(0, 0)$. $square$.
Is the above proof correct and rigorous? (Note I have left out some details on why there exists such an $x$ for sufficiently small $b$ as it's not too hard to show using the definition of open sets in metric spaces).
multivariable-calculus proof-verification
asked Nov 24 at 17:00
Perturbative
3,93011449
Show that the function $f(x, y) = |xy|$ is not of class $C^1$ in any neighborhood of $(0, 0)$
This is my attempted proof:
Proof: Choose a neighborhood $U$ of $(0, 0) in mathbb{R}^2$. Since $U$ is a neighborhood of $(0, 0)$ it must contain points of the form $x = (0, b)$ for sufficiently small $b$.
But then
begin{align*}
D_1f(x) &= lim_{t to 0} frac{f(x+te_1) -f(x)}{t} \
&= lim_{t to 0} frac{f((0, b) + (t, 0)) -f(0, b)}{t} \
&= lim_{t to 0} frac{|tb|-|b|}{t} \
&= lim_{t to 0} frac{|t|}{t} |b| - frac{1}{t}|b|
end{align*}
But now consider the limit as positive $t$ approaches $0$, $$lim_{t^+ to 0} frac{|t|}{t} |b| - frac{1}{t}|b| = lim_{t^+ to 0}|b|-frac{1}{t}|b| (*)$$
and $|b| - frac{1}{t}|b| to -infty$ as $t^+ to 0$, so the above limit $(*)$ doesn't exist and thus $lim_{t to 0} frac{|t|}{t} |b| - frac{1}{t}|b|$ doesn't exist and so $D_1f(x)$ doesn't exist and thus $f$ cannot be of class $C^1$ in any neighborhood of $(0, 0)$. $square$.
Is the above proof correct and rigorous? (Note I have left out some details on why there exists such an $x$ for sufficiently small $b$ as it's not too hard to show using the definition of open sets in metric spaces).
multivariable-calculus proof-verification
multivariable-calculus proof-verification
asked Nov 24 at 17:00
Perturbative
3,93011449
asked Nov 24 at 17:00
Perturbative
3,93011449
asked Nov 24 at 17:00
Perturbative
3,93011449
asked Nov 24 at 17:00
Perturbative
3,93011449
asked Nov 24 at 17:00
Perturbative
3,93011449
3,93011449
add a comment |
add a comment |
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