Mixing
If $D_{x_{0}}f$ is an isomorphism, there is $V_{x_{0}}$ s.t. $f(x) neq f(x_{0})$ for $x in...
$begingroup$
Let $f: U subset mathbb{R}^{n} to mathbb{R}^{n}$ be a continuous function, where $U$ is open. Suppose that $f$ is differentiable in $x_{0} in U$ and $D_{x_{0}}f$ is an isomorphism. Show that there is a neighborhood $V_{x_{0}} subset U$ of $x_{0}$ such that $f(x) neq f(x_{0})$ for any $x in V_{x_{0}}setminus{x_{0}}$.
My attempt My idea is suppose that in each neighborhood $V_{x_{0}}$ of $x_{0}$ there is $x in V_{x_{0}}setminus{x_{0}}$ such that $f(x) = f(x_{0})$ for get a contradiction.
We can take a sequence $(x_{n})$ with $x_{n} in B_{1/n}(x_{0})$ such that $x_{n} to x_{0}$ and $f(x_{n}) = f(x_{0})$. Write $x_{n} = x_{0} + t_{n}h_{n}$ where $h_{n}$ is a unity vector. Note that
$$Vert D_{x_{0}}f Vert = leftVert frac{f(x_{0}+t_{n}h_{n})-f(x_{0})}{t_{n}} - D_{x_{0}}frightVert.$$
But, given $epsilon > 0$,
$$leftVert frac{f(x_{0}+t_{n}h_{n})-f(x_{0})}{t_{n}} - D_{x_{0}}frightVert < epsilon$$
whenever $|t_{n}| < delta$. Since $x_{n} to x_{0}$, we can take $delta>0$ such that $|t_{n}| < delta$ for all $n > n_{0}$ for some $n_{0} in mathbb{N}$. For this $delta$, we get
$$Vert D_{x_{0}}f Vert < epsilon,$$
then $D_{x_{0}}f$ cannot be an isomorphism.
Is this correct? The idea seems good to me, but I'm not sure about the details of the proof. Also, if its correct, I couldnt see where total continuity was necessary.
real-analysis derivatives
asked Jan 24 at 18:00
Lucas CorrêaLucas Corrêa
1,5951421
$endgroup$
add a comment |
$begingroup$
Let $f: U subset mathbb{R}^{n} to mathbb{R}^{n}$ be a continuous function, where $U$ is open. Suppose that $f$ is differentiable in $x_{0} in U$ and $D_{x_{0}}f$ is an isomorphism. Show that there is a neighborhood $V_{x_{0}} subset U$ of $x_{0}$ such that $f(x) neq f(x_{0})$ for any $x in V_{x_{0}}setminus{x_{0}}$.
My attempt My idea is suppose that in each neighborhood $V_{x_{0}}$ of $x_{0}$ there is $x in V_{x_{0}}setminus{x_{0}}$ such that $f(x) = f(x_{0})$ for get a contradiction.
We can take a sequence $(x_{n})$ with $x_{n} in B_{1/n}(x_{0})$ such that $x_{n} to x_{0}$ and $f(x_{n}) = f(x_{0})$. Write $x_{n} = x_{0} + t_{n}h_{n}$ where $h_{n}$ is a unity vector. Note that
$$Vert D_{x_{0}}f Vert = leftVert frac{f(x_{0}+t_{n}h_{n})-f(x_{0})}{t_{n}} - D_{x_{0}}frightVert.$$
But, given $epsilon > 0$,
$$leftVert frac{f(x_{0}+t_{n}h_{n})-f(x_{0})}{t_{n}} - D_{x_{0}}frightVert < epsilon$$
whenever $|t_{n}| < delta$. Since $x_{n} to x_{0}$, we can take $delta>0$ such that $|t_{n}| < delta$ for all $n > n_{0}$ for some $n_{0} in mathbb{N}$. For this $delta$, we get
$$Vert D_{x_{0}}f Vert < epsilon,$$
then $D_{x_{0}}f$ cannot be an isomorphism.
Is this correct? The idea seems good to me, but I'm not sure about the details of the proof. Also, if its correct, I couldnt see where total continuity was necessary.
real-analysis derivatives
asked Jan 24 at 18:00
Lucas CorrêaLucas Corrêa
1,5951421
$endgroup$
add a comment |
$begingroup$
Let $f: U subset mathbb{R}^{n} to mathbb{R}^{n}$ be a continuous function, where $U$ is open. Suppose that $f$ is differentiable in $x_{0} in U$ and $D_{x_{0}}f$ is an isomorphism. Show that there is a neighborhood $V_{x_{0}} subset U$ of $x_{0}$ such that $f(x) neq f(x_{0})$ for any $x in V_{x_{0}}setminus{x_{0}}$.
My attempt My idea is suppose that in each neighborhood $V_{x_{0}}$ of $x_{0}$ there is $x in V_{x_{0}}setminus{x_{0}}$ such that $f(x) = f(x_{0})$ for get a contradiction.
We can take a sequence $(x_{n})$ with $x_{n} in B_{1/n}(x_{0})$ such that $x_{n} to x_{0}$ and $f(x_{n}) = f(x_{0})$. Write $x_{n} = x_{0} + t_{n}h_{n}$ where $h_{n}$ is a unity vector. Note that
$$Vert D_{x_{0}}f Vert = leftVert frac{f(x_{0}+t_{n}h_{n})-f(x_{0})}{t_{n}} - D_{x_{0}}frightVert.$$
But, given $epsilon > 0$,
$$leftVert frac{f(x_{0}+t_{n}h_{n})-f(x_{0})}{t_{n}} - D_{x_{0}}frightVert < epsilon$$
whenever $|t_{n}| < delta$. Since $x_{n} to x_{0}$, we can take $delta>0$ such that $|t_{n}| < delta$ for all $n > n_{0}$ for some $n_{0} in mathbb{N}$. For this $delta$, we get
$$Vert D_{x_{0}}f Vert < epsilon,$$
then $D_{x_{0}}f$ cannot be an isomorphism.
Is this correct? The idea seems good to me, but I'm not sure about the details of the proof. Also, if its correct, I couldnt see where total continuity was necessary.
real-analysis derivatives
asked Jan 24 at 18:00
Lucas CorrêaLucas Corrêa
1,5951421
$endgroup$
Let $f: U subset mathbb{R}^{n} to mathbb{R}^{n}$ be a continuous function, where $U$ is open. Suppose that $f$ is differentiable in $x_{0} in U$ and $D_{x_{0}}f$ is an isomorphism. Show that there is a neighborhood $V_{x_{0}} subset U$ of $x_{0}$ such that $f(x) neq f(x_{0})$ for any $x in V_{x_{0}}setminus{x_{0}}$.
My attempt My idea is suppose that in each neighborhood $V_{x_{0}}$ of $x_{0}$ there is $x in V_{x_{0}}setminus{x_{0}}$ such that $f(x) = f(x_{0})$ for get a contradiction.
We can take a sequence $(x_{n})$ with $x_{n} in B_{1/n}(x_{0})$ such that $x_{n} to x_{0}$ and $f(x_{n}) = f(x_{0})$. Write $x_{n} = x_{0} + t_{n}h_{n}$ where $h_{n}$ is a unity vector. Note that
$$Vert D_{x_{0}}f Vert = leftVert frac{f(x_{0}+t_{n}h_{n})-f(x_{0})}{t_{n}} - D_{x_{0}}frightVert.$$
But, given $epsilon > 0$,
$$leftVert frac{f(x_{0}+t_{n}h_{n})-f(x_{0})}{t_{n}} - D_{x_{0}}frightVert < epsilon$$
whenever $|t_{n}| < delta$. Since $x_{n} to x_{0}$, we can take $delta>0$ such that $|t_{n}| < delta$ for all $n > n_{0}$ for some $n_{0} in mathbb{N}$. For this $delta$, we get
$$Vert D_{x_{0}}f Vert < epsilon,$$
then $D_{x_{0}}f$ cannot be an isomorphism.
Is this correct? The idea seems good to me, but I'm not sure about the details of the proof. Also, if its correct, I couldnt see where total continuity was necessary.
real-analysis derivatives
real-analysis derivatives
asked Jan 24 at 18:00
Lucas CorrêaLucas Corrêa
1,5951421
asked Jan 24 at 18:00
Lucas CorrêaLucas Corrêa
1,5951421
asked Jan 24 at 18:00
Lucas CorrêaLucas Corrêa
1,5951421
asked Jan 24 at 18:00
Lucas CorrêaLucas Corrêa
1,5951421
asked Jan 24 at 18:00
Lucas CorrêaLucas Corrêa
1,5951421
1,5951421
add a comment |
add a comment |
1 Answer
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The equation
$$| D_{x_0}f | = left| frac{ f(x_0 + t_n h_n) - f(x_0)}{t_n} - D_{x_0}f right| $$
does not make sense to me.
The quotient $frac{f(x_0 + t_n h_n) - f(x_0)}{t_n}$ is a vector in $mathbb{R}^n$, but $D_{x_0}f$ is an $n times n$ matrix, so it is not meaningful to take their difference. Moreover, if the LHS refers to a matrix norm, there is no hope in proving that $ | D_{x_0}f |$ vanishes. (A singular matrix may still have large norm.) Instead, if you want to prove that $D_{x_0}f$ is not an isomorphism under your assumption, you should show that there is some vector $h in mathbb{R}^n backslash{0 }$ so that $(D_{x_0}f) h$ (i.e. the vector multiplied by the matrix $D_{x_0}f$) is zero.
I think you should replace $D_{x_0}f$ by $(D_{x_0}f) h_n$ in all of the centered equations. If you do so, your work shows that there is a sequence ${ h_n }$ of unit vectors so that $(D_{x_0}f) h_n to 0$. It follows by compactness of the unit ball that some subsequence ${ h_{n_k} }$ converges to a vector $h$. I claim that $(D_{x_0}f) h = 0$. Do you see why?
answered Jan 24 at 22:10
Jordan GreenJordan Green
1,133410
$endgroup$
$begingroup$
Perfect! The lack of $h_{n}$ was a notation error that I didn't realize, thank you for pointing it out! About the final conclusion, follows by continuity of $D_{x_{0}}f$?
$endgroup$
– Lucas Corrêa
Jan 25 at 0:09
1
$begingroup$
Continuity of $Df$ is not needed, because you are only looking at the derivative matrix at one point. Instead, I was thinking of writing something like $| (D_{x_0}f)h | leq | (D_{x_0}f (h - h_{n_k}) | + | (D_{x_0}f)h_{n_k} | leq | D_{x_0}f | | h - h_{n_k} | + | (D_{x_0} f )h_{n_k} |$. (Here, $ | D_{x_0}f |$ refers to the matrix norm.) The term $| (D_{x_0} f )h_{n_k} |$ can be made small by your work. The other term goes to zero because $h_{n_k} to h$.
$endgroup$
– Jordan Green
Jan 25 at 0:42
$begingroup$
I see. Nice attempt!
$endgroup$
– Lucas Corrêa
Jan 25 at 1:06
add a comment |
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$begingroup$
The equation
$$| D_{x_0}f | = left| frac{ f(x_0 + t_n h_n) - f(x_0)}{t_n} - D_{x_0}f right| $$
does not make sense to me.
The quotient $frac{f(x_0 + t_n h_n) - f(x_0)}{t_n}$ is a vector in $mathbb{R}^n$, but $D_{x_0}f$ is an $n times n$ matrix, so it is not meaningful to take their difference. Moreover, if the LHS refers to a matrix norm, there is no hope in proving that $ | D_{x_0}f |$ vanishes. (A singular matrix may still have large norm.) Instead, if you want to prove that $D_{x_0}f$ is not an isomorphism under your assumption, you should show that there is some vector $h in mathbb{R}^n backslash{0 }$ so that $(D_{x_0}f) h$ (i.e. the vector multiplied by the matrix $D_{x_0}f$) is zero.
I think you should replace $D_{x_0}f$ by $(D_{x_0}f) h_n$ in all of the centered equations. If you do so, your work shows that there is a sequence ${ h_n }$ of unit vectors so that $(D_{x_0}f) h_n to 0$. It follows by compactness of the unit ball that some subsequence ${ h_{n_k} }$ converges to a vector $h$. I claim that $(D_{x_0}f) h = 0$. Do you see why?
answered Jan 24 at 22:10
Jordan GreenJordan Green
1,133410
$endgroup$
$begingroup$
Perfect! The lack of $h_{n}$ was a notation error that I didn't realize, thank you for pointing it out! About the final conclusion, follows by continuity of $D_{x_{0}}f$?
$endgroup$
– Lucas Corrêa
Jan 25 at 0:09
1
$begingroup$
Continuity of $Df$ is not needed, because you are only looking at the derivative matrix at one point. Instead, I was thinking of writing something like $| (D_{x_0}f)h | leq | (D_{x_0}f (h - h_{n_k}) | + | (D_{x_0}f)h_{n_k} | leq | D_{x_0}f | | h - h_{n_k} | + | (D_{x_0} f )h_{n_k} |$. (Here, $ | D_{x_0}f |$ refers to the matrix norm.) The term $| (D_{x_0} f )h_{n_k} |$ can be made small by your work. The other term goes to zero because $h_{n_k} to h$.
$endgroup$
– Jordan Green
Jan 25 at 0:42
$begingroup$
I see. Nice attempt!
$endgroup$
– Lucas Corrêa
Jan 25 at 1:06
add a comment |
$begingroup$
The equation
$$| D_{x_0}f | = left| frac{ f(x_0 + t_n h_n) - f(x_0)}{t_n} - D_{x_0}f right| $$
does not make sense to me.
The quotient $frac{f(x_0 + t_n h_n) - f(x_0)}{t_n}$ is a vector in $mathbb{R}^n$, but $D_{x_0}f$ is an $n times n$ matrix, so it is not meaningful to take their difference. Moreover, if the LHS refers to a matrix norm, there is no hope in proving that $ | D_{x_0}f |$ vanishes. (A singular matrix may still have large norm.) Instead, if you want to prove that $D_{x_0}f$ is not an isomorphism under your assumption, you should show that there is some vector $h in mathbb{R}^n backslash{0 }$ so that $(D_{x_0}f) h$ (i.e. the vector multiplied by the matrix $D_{x_0}f$) is zero.
I think you should replace $D_{x_0}f$ by $(D_{x_0}f) h_n$ in all of the centered equations. If you do so, your work shows that there is a sequence ${ h_n }$ of unit vectors so that $(D_{x_0}f) h_n to 0$. It follows by compactness of the unit ball that some subsequence ${ h_{n_k} }$ converges to a vector $h$. I claim that $(D_{x_0}f) h = 0$. Do you see why?
answered Jan 24 at 22:10
Jordan GreenJordan Green
1,133410
$endgroup$
$begingroup$
Perfect! The lack of $h_{n}$ was a notation error that I didn't realize, thank you for pointing it out! About the final conclusion, follows by continuity of $D_{x_{0}}f$?
$endgroup$
– Lucas Corrêa
Jan 25 at 0:09
1
$begingroup$
Continuity of $Df$ is not needed, because you are only looking at the derivative matrix at one point. Instead, I was thinking of writing something like $| (D_{x_0}f)h | leq | (D_{x_0}f (h - h_{n_k}) | + | (D_{x_0}f)h_{n_k} | leq | D_{x_0}f | | h - h_{n_k} | + | (D_{x_0} f )h_{n_k} |$. (Here, $ | D_{x_0}f |$ refers to the matrix norm.) The term $| (D_{x_0} f )h_{n_k} |$ can be made small by your work. The other term goes to zero because $h_{n_k} to h$.
$endgroup$
– Jordan Green
Jan 25 at 0:42
$begingroup$
I see. Nice attempt!
$endgroup$
– Lucas Corrêa
Jan 25 at 1:06
add a comment |
$begingroup$
The equation
$$| D_{x_0}f | = left| frac{ f(x_0 + t_n h_n) - f(x_0)}{t_n} - D_{x_0}f right| $$
does not make sense to me.
The quotient $frac{f(x_0 + t_n h_n) - f(x_0)}{t_n}$ is a vector in $mathbb{R}^n$, but $D_{x_0}f$ is an $n times n$ matrix, so it is not meaningful to take their difference. Moreover, if the LHS refers to a matrix norm, there is no hope in proving that $ | D_{x_0}f |$ vanishes. (A singular matrix may still have large norm.) Instead, if you want to prove that $D_{x_0}f$ is not an isomorphism under your assumption, you should show that there is some vector $h in mathbb{R}^n backslash{0 }$ so that $(D_{x_0}f) h$ (i.e. the vector multiplied by the matrix $D_{x_0}f$) is zero.
I think you should replace $D_{x_0}f$ by $(D_{x_0}f) h_n$ in all of the centered equations. If you do so, your work shows that there is a sequence ${ h_n }$ of unit vectors so that $(D_{x_0}f) h_n to 0$. It follows by compactness of the unit ball that some subsequence ${ h_{n_k} }$ converges to a vector $h$. I claim that $(D_{x_0}f) h = 0$. Do you see why?
answered Jan 24 at 22:10
Jordan GreenJordan Green
1,133410
$endgroup$
The equation
$$| D_{x_0}f | = left| frac{ f(x_0 + t_n h_n) - f(x_0)}{t_n} - D_{x_0}f right| $$
does not make sense to me.
The quotient $frac{f(x_0 + t_n h_n) - f(x_0)}{t_n}$ is a vector in $mathbb{R}^n$, but $D_{x_0}f$ is an $n times n$ matrix, so it is not meaningful to take their difference. Moreover, if the LHS refers to a matrix norm, there is no hope in proving that $ | D_{x_0}f |$ vanishes. (A singular matrix may still have large norm.) Instead, if you want to prove that $D_{x_0}f$ is not an isomorphism under your assumption, you should show that there is some vector $h in mathbb{R}^n backslash{0 }$ so that $(D_{x_0}f) h$ (i.e. the vector multiplied by the matrix $D_{x_0}f$) is zero.
I think you should replace $D_{x_0}f$ by $(D_{x_0}f) h_n$ in all of the centered equations. If you do so, your work shows that there is a sequence ${ h_n }$ of unit vectors so that $(D_{x_0}f) h_n to 0$. It follows by compactness of the unit ball that some subsequence ${ h_{n_k} }$ converges to a vector $h$. I claim that $(D_{x_0}f) h = 0$. Do you see why?
answered Jan 24 at 22:10
Jordan GreenJordan Green
1,133410
answered Jan 24 at 22:10
Jordan GreenJordan Green
1,133410
answered Jan 24 at 22:10
Jordan GreenJordan Green
1,133410
answered Jan 24 at 22:10
Jordan GreenJordan Green
1,133410
1,133410
$begingroup$
Perfect! The lack of $h_{n}$ was a notation error that I didn't realize, thank you for pointing it out! About the final conclusion, follows by continuity of $D_{x_{0}}f$?
$endgroup$
– Lucas Corrêa
Jan 25 at 0:09
1
$begingroup$
Continuity of $Df$ is not needed, because you are only looking at the derivative matrix at one point. Instead, I was thinking of writing something like $| (D_{x_0}f)h | leq | (D_{x_0}f (h - h_{n_k}) | + | (D_{x_0}f)h_{n_k} | leq | D_{x_0}f | | h - h_{n_k} | + | (D_{x_0} f )h_{n_k} |$. (Here, $ | D_{x_0}f |$ refers to the matrix norm.) The term $| (D_{x_0} f )h_{n_k} |$ can be made small by your work. The other term goes to zero because $h_{n_k} to h$.
$endgroup$
– Jordan Green
Jan 25 at 0:42
$begingroup$
I see. Nice attempt!
$endgroup$
– Lucas Corrêa
Jan 25 at 1:06
add a comment |
$begingroup$
Perfect! The lack of $h_{n}$ was a notation error that I didn't realize, thank you for pointing it out! About the final conclusion, follows by continuity of $D_{x_{0}}f$?
$endgroup$
– Lucas Corrêa
Jan 25 at 0:09
1
$begingroup$
Continuity of $Df$ is not needed, because you are only looking at the derivative matrix at one point. Instead, I was thinking of writing something like $| (D_{x_0}f)h | leq | (D_{x_0}f (h - h_{n_k}) | + | (D_{x_0}f)h_{n_k} | leq | D_{x_0}f | | h - h_{n_k} | + | (D_{x_0} f )h_{n_k} |$. (Here, $ | D_{x_0}f |$ refers to the matrix norm.) The term $| (D_{x_0} f )h_{n_k} |$ can be made small by your work. The other term goes to zero because $h_{n_k} to h$.
$endgroup$
– Jordan Green
Jan 25 at 0:42
$begingroup$
I see. Nice attempt!
$endgroup$
– Lucas Corrêa
Jan 25 at 1:06
$begingroup$
Perfect! The lack of $h_{n}$ was a notation error that I didn't realize, thank you for pointing it out! About the final conclusion, follows by continuity of $D_{x_{0}}f$?
$endgroup$
– Lucas Corrêa
Jan 25 at 0:09
$begingroup$
Perfect! The lack of $h_{n}$ was a notation error that I didn't realize, thank you for pointing it out! About the final conclusion, follows by continuity of $D_{x_{0}}f$?
$endgroup$
– Lucas Corrêa
Jan 25 at 0:09
1
1
$begingroup$
Continuity of $Df$ is not needed, because you are only looking at the derivative matrix at one point. Instead, I was thinking of writing something like $| (D_{x_0}f)h | leq | (D_{x_0}f (h - h_{n_k}) | + | (D_{x_0}f)h_{n_k} | leq | D_{x_0}f | | h - h_{n_k} | + | (D_{x_0} f )h_{n_k} |$. (Here, $ | D_{x_0}f |$ refers to the matrix norm.) The term $| (D_{x_0} f )h_{n_k} |$ can be made small by your work. The other term goes to zero because $h_{n_k} to h$.
$endgroup$
– Jordan Green
Jan 25 at 0:42
$begingroup$
Continuity of $Df$ is not needed, because you are only looking at the derivative matrix at one point. Instead, I was thinking of writing something like $| (D_{x_0}f)h | leq | (D_{x_0}f (h - h_{n_k}) | + | (D_{x_0}f)h_{n_k} | leq | D_{x_0}f | | h - h_{n_k} | + | (D_{x_0} f )h_{n_k} |$. (Here, $ | D_{x_0}f |$ refers to the matrix norm.) The term $| (D_{x_0} f )h_{n_k} |$ can be made small by your work. The other term goes to zero because $h_{n_k} to h$.
$endgroup$
– Jordan Green
Jan 25 at 0:42
$begingroup$
I see. Nice attempt!
$endgroup$
– Lucas Corrêa
Jan 25 at 1:06
$begingroup$
I see. Nice attempt!
$endgroup$
– Lucas Corrêa
Jan 25 at 1:06
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
