Partial derivative being $0$ implies non-dependence on that coordinate for $f: U rightarrow mathbb{R}$ and $U...
$begingroup$
Let $U subset mathbb{R}^n$ be open and convex and $f : U rightarrow mathbb{R}$ differentiable such that $partial_1f(x)=0$ for all $x in U.$
Show that the value of $f(x)$ for $x = (x_1,...,x_n) in U$ does not depend
on $x_1.$
$text{Proof:}$
Suppose for a contradiction that $f$ depends on the first coordinate. Let $x,y in U$ be such that $x_i=y_i$ except for $i=1.$ Since $U$ is open and convex we can use the Mean Value Theorem. Hence there is $c= t_0x+(1-t_0)y,$ for some $t_0 in [0,1]$ such that
begin{align*}
f(x)-f(y)
&=nabla f(c)cdotlangle x-y rangle\
&=langle 0,partial_2f(c),...,partial_nf(c) rangle cdot langle x_1-y_1,0,...,0 rangle \
&=0.
end{align*}
Therefore, we have that $f(x)=f(y),$ which is a contraction, since $f$ is not constant. Thus, $f$ does not depend on the first coordinate.
Is this the correct approach to this problem? Any feedback is much appreciated. Thank you in advance.
calculus real-analysis multivariable-calculus proof-verification
asked Nov 4 '18 at 9:17
Gaby AlfonsoGaby Alfonso
839316
$endgroup$
add a comment |
$begingroup$
Let $U subset mathbb{R}^n$ be open and convex and $f : U rightarrow mathbb{R}$ differentiable such that $partial_1f(x)=0$ for all $x in U.$
Show that the value of $f(x)$ for $x = (x_1,...,x_n) in U$ does not depend
on $x_1.$
$text{Proof:}$
Suppose for a contradiction that $f$ depends on the first coordinate. Let $x,y in U$ be such that $x_i=y_i$ except for $i=1.$ Since $U$ is open and convex we can use the Mean Value Theorem. Hence there is $c= t_0x+(1-t_0)y,$ for some $t_0 in [0,1]$ such that
begin{align*}
f(x)-f(y)
&=nabla f(c)cdotlangle x-y rangle\
&=langle 0,partial_2f(c),...,partial_nf(c) rangle cdot langle x_1-y_1,0,...,0 rangle \
&=0.
end{align*}
Therefore, we have that $f(x)=f(y),$ which is a contraction, since $f$ is not constant. Thus, $f$ does not depend on the first coordinate.
Is this the correct approach to this problem? Any feedback is much appreciated. Thank you in advance.
calculus real-analysis multivariable-calculus proof-verification
asked Nov 4 '18 at 9:17
Gaby AlfonsoGaby Alfonso
839316
$endgroup$
1
$begingroup$
Yes, this looks good. You can drop the "If $f$ is constant..." sentence, by the way.
$endgroup$
– Lukas Geyer
Nov 5 '18 at 2:48
$begingroup$
@LukasGeyer Thank you for the feedback. I edited the proof as suggested.
$endgroup$
– Gaby Alfonso
Nov 5 '18 at 2:50
add a comment |
$begingroup$
Let $U subset mathbb{R}^n$ be open and convex and $f : U rightarrow mathbb{R}$ differentiable such that $partial_1f(x)=0$ for all $x in U.$
Show that the value of $f(x)$ for $x = (x_1,...,x_n) in U$ does not depend
on $x_1.$
$text{Proof:}$
Suppose for a contradiction that $f$ depends on the first coordinate. Let $x,y in U$ be such that $x_i=y_i$ except for $i=1.$ Since $U$ is open and convex we can use the Mean Value Theorem. Hence there is $c= t_0x+(1-t_0)y,$ for some $t_0 in [0,1]$ such that
begin{align*}
f(x)-f(y)
&=nabla f(c)cdotlangle x-y rangle\
&=langle 0,partial_2f(c),...,partial_nf(c) rangle cdot langle x_1-y_1,0,...,0 rangle \
&=0.
end{align*}
Therefore, we have that $f(x)=f(y),$ which is a contraction, since $f$ is not constant. Thus, $f$ does not depend on the first coordinate.
Is this the correct approach to this problem? Any feedback is much appreciated. Thank you in advance.
calculus real-analysis multivariable-calculus proof-verification
asked Nov 4 '18 at 9:17
Gaby AlfonsoGaby Alfonso
839316
$endgroup$
Let $U subset mathbb{R}^n$ be open and convex and $f : U rightarrow mathbb{R}$ differentiable such that $partial_1f(x)=0$ for all $x in U.$
Show that the value of $f(x)$ for $x = (x_1,...,x_n) in U$ does not depend
on $x_1.$
$text{Proof:}$
Suppose for a contradiction that $f$ depends on the first coordinate. Let $x,y in U$ be such that $x_i=y_i$ except for $i=1.$ Since $U$ is open and convex we can use the Mean Value Theorem. Hence there is $c= t_0x+(1-t_0)y,$ for some $t_0 in [0,1]$ such that
begin{align*}
f(x)-f(y)
&=nabla f(c)cdotlangle x-y rangle\
&=langle 0,partial_2f(c),...,partial_nf(c) rangle cdot langle x_1-y_1,0,...,0 rangle \
&=0.
end{align*}
Therefore, we have that $f(x)=f(y),$ which is a contraction, since $f$ is not constant. Thus, $f$ does not depend on the first coordinate.
Is this the correct approach to this problem? Any feedback is much appreciated. Thank you in advance.
calculus real-analysis multivariable-calculus proof-verification
calculus real-analysis multivariable-calculus proof-verification
asked Nov 4 '18 at 9:17
Gaby AlfonsoGaby Alfonso
839316
asked Nov 4 '18 at 9:17
Gaby AlfonsoGaby Alfonso
839316
edited Dec 25 '18 at 5:46
Gaby Alfonso
asked Nov 4 '18 at 9:17
Gaby AlfonsoGaby Alfonso
839316
asked Nov 4 '18 at 9:17
Gaby AlfonsoGaby Alfonso
839316
asked Nov 4 '18 at 9:17
Gaby AlfonsoGaby Alfonso
839316
839316
1
$begingroup$
Yes, this looks good. You can drop the "If $f$ is constant..." sentence, by the way.
$endgroup$
– Lukas Geyer
Nov 5 '18 at 2:48
$begingroup$
@LukasGeyer Thank you for the feedback. I edited the proof as suggested.
$endgroup$
– Gaby Alfonso
Nov 5 '18 at 2:50
add a comment |
1
$begingroup$
Yes, this looks good. You can drop the "If $f$ is constant..." sentence, by the way.
$endgroup$
– Lukas Geyer
Nov 5 '18 at 2:48
$begingroup$
@LukasGeyer Thank you for the feedback. I edited the proof as suggested.
$endgroup$
– Gaby Alfonso
Nov 5 '18 at 2:50
1
1
$begingroup$
Yes, this looks good. You can drop the "If $f$ is constant..." sentence, by the way.
$endgroup$
– Lukas Geyer
Nov 5 '18 at 2:48
$begingroup$
Yes, this looks good. You can drop the "If $f$ is constant..." sentence, by the way.
$endgroup$
– Lukas Geyer
Nov 5 '18 at 2:48
$begingroup$
@LukasGeyer Thank you for the feedback. I edited the proof as suggested.
$endgroup$
– Gaby Alfonso
Nov 5 '18 at 2:50
$begingroup$
@LukasGeyer Thank you for the feedback. I edited the proof as suggested.
$endgroup$
– Gaby Alfonso
Nov 5 '18 at 2:50
add a comment |
1 Answer
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Your proof is fine, but it is needless to phrase it as a proof by contradiction. The same computation that you did shows that $g(t) = f(t,x_2,dots,x_n)$ is a constant function, which is all you want. As a matter of style, direct proofs are preferred to proofs by contradiction.
answered Nov 5 '18 at 4:47
Alex OrtizAlex Ortiz
10.6k21441
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add a comment |
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Your proof is fine, but it is needless to phrase it as a proof by contradiction. The same computation that you did shows that $g(t) = f(t,x_2,dots,x_n)$ is a constant function, which is all you want. As a matter of style, direct proofs are preferred to proofs by contradiction.
answered Nov 5 '18 at 4:47
Alex OrtizAlex Ortiz
10.6k21441
$endgroup$
add a comment |
$begingroup$
Your proof is fine, but it is needless to phrase it as a proof by contradiction. The same computation that you did shows that $g(t) = f(t,x_2,dots,x_n)$ is a constant function, which is all you want. As a matter of style, direct proofs are preferred to proofs by contradiction.
answered Nov 5 '18 at 4:47
Alex OrtizAlex Ortiz
10.6k21441
$endgroup$
add a comment |
$begingroup$
Your proof is fine, but it is needless to phrase it as a proof by contradiction. The same computation that you did shows that $g(t) = f(t,x_2,dots,x_n)$ is a constant function, which is all you want. As a matter of style, direct proofs are preferred to proofs by contradiction.
answered Nov 5 '18 at 4:47
Alex OrtizAlex Ortiz
10.6k21441
$endgroup$
Your proof is fine, but it is needless to phrase it as a proof by contradiction. The same computation that you did shows that $g(t) = f(t,x_2,dots,x_n)$ is a constant function, which is all you want. As a matter of style, direct proofs are preferred to proofs by contradiction.
answered Nov 5 '18 at 4:47
Alex OrtizAlex Ortiz
10.6k21441
answered Nov 5 '18 at 4:47
Alex OrtizAlex Ortiz
10.6k21441
answered Nov 5 '18 at 4:47
Alex OrtizAlex Ortiz
10.6k21441
answered Nov 5 '18 at 4:47
Alex OrtizAlex Ortiz
10.6k21441
10.6k21441
add a comment |
add a comment |
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$begingroup$
Yes, this looks good. You can drop the "If $f$ is constant..." sentence, by the way.
$endgroup$
– Lukas Geyer
Nov 5 '18 at 2:48
$begingroup$
@LukasGeyer Thank you for the feedback. I edited the proof as suggested.
$endgroup$
– Gaby Alfonso
Nov 5 '18 at 2:50