Find the differential of the following map.
Given $f:mathbb{R}^nto mathbb{R}$ and $Ain (mathbb{M_n(R}))$ such that $f(x) = x^{t}Ax.$ I want to find the derivative of this map and show that it is $C^{1}$. This problem has been partially answered before assuming that $A$ is symmetric, but note that in this case that $A$ is not necessarily symmetric. Therefore the derivative
$$df_x(h) = x^tAh+h^tAx.$$
The map is clearly linear but to show it is continuous, I am not sure whether I am using the right inequalities:
$$|df_x(h)|=|x^tAh|+|h^tAx|leq 2||x||cdot ||A||cdot ||h||$$
This shows that $df_x$ is continuous. To show that it is $C^{1}$ we do the following:
$$|df_x(h)-df_y(h)|leq 2||h||cdot ||A||cdot ||x-y||.$$
I am not sure whether these inequalities are correct, but I am using them because I have seen some exercise corrections similar to this problem. Any hints/suggestions will be much appreciated.
real-analysis
asked Nov 26 '18 at 20:22
Hello_World
4,02821630
add a comment |
Given $f:mathbb{R}^nto mathbb{R}$ and $Ain (mathbb{M_n(R}))$ such that $f(x) = x^{t}Ax.$ I want to find the derivative of this map and show that it is $C^{1}$. This problem has been partially answered before assuming that $A$ is symmetric, but note that in this case that $A$ is not necessarily symmetric. Therefore the derivative
$$df_x(h) = x^tAh+h^tAx.$$
The map is clearly linear but to show it is continuous, I am not sure whether I am using the right inequalities:
$$|df_x(h)|=|x^tAh|+|h^tAx|leq 2||x||cdot ||A||cdot ||h||$$
This shows that $df_x$ is continuous. To show that it is $C^{1}$ we do the following:
$$|df_x(h)-df_y(h)|leq 2||h||cdot ||A||cdot ||x-y||.$$
I am not sure whether these inequalities are correct, but I am using them because I have seen some exercise corrections similar to this problem. Any hints/suggestions will be much appreciated.
real-analysis
asked Nov 26 '18 at 20:22
Hello_World
4,02821630
$df_x$ is always continuous since it is, by definition, a linear map
– user25959
Nov 26 '18 at 20:34
What you have done (correctly) in the last part is to show that the map $xmapsto df_x$ is a continuous map. This is a map from $mathbb{R} rightarrow text{End}(mathbb{R}^n,mathbb{R})$, where "End(U,V)" denotes the linear maps from U to V.
– user25959
Nov 26 '18 at 20:36
@user25959 it is continuous because the dimension of the space is finite?
– Hello_World
Nov 26 '18 at 20:45
Yes that's a good point, in infinite dimensional spaces linear can be unbounded/discontinuous
– user25959
Nov 26 '18 at 20:46
add a comment |
Given $f:mathbb{R}^nto mathbb{R}$ and $Ain (mathbb{M_n(R}))$ such that $f(x) = x^{t}Ax.$ I want to find the derivative of this map and show that it is $C^{1}$. This problem has been partially answered before assuming that $A$ is symmetric, but note that in this case that $A$ is not necessarily symmetric. Therefore the derivative
$$df_x(h) = x^tAh+h^tAx.$$
The map is clearly linear but to show it is continuous, I am not sure whether I am using the right inequalities:
$$|df_x(h)|=|x^tAh|+|h^tAx|leq 2||x||cdot ||A||cdot ||h||$$
This shows that $df_x$ is continuous. To show that it is $C^{1}$ we do the following:
$$|df_x(h)-df_y(h)|leq 2||h||cdot ||A||cdot ||x-y||.$$
I am not sure whether these inequalities are correct, but I am using them because I have seen some exercise corrections similar to this problem. Any hints/suggestions will be much appreciated.
real-analysis
asked Nov 26 '18 at 20:22
Hello_World
4,02821630
Given $f:mathbb{R}^nto mathbb{R}$ and $Ain (mathbb{M_n(R}))$ such that $f(x) = x^{t}Ax.$ I want to find the derivative of this map and show that it is $C^{1}$. This problem has been partially answered before assuming that $A$ is symmetric, but note that in this case that $A$ is not necessarily symmetric. Therefore the derivative
$$df_x(h) = x^tAh+h^tAx.$$
The map is clearly linear but to show it is continuous, I am not sure whether I am using the right inequalities:
$$|df_x(h)|=|x^tAh|+|h^tAx|leq 2||x||cdot ||A||cdot ||h||$$
This shows that $df_x$ is continuous. To show that it is $C^{1}$ we do the following:
$$|df_x(h)-df_y(h)|leq 2||h||cdot ||A||cdot ||x-y||.$$
I am not sure whether these inequalities are correct, but I am using them because I have seen some exercise corrections similar to this problem. Any hints/suggestions will be much appreciated.
real-analysis
real-analysis
asked Nov 26 '18 at 20:22
Hello_World
4,02821630
asked Nov 26 '18 at 20:22
Hello_World
4,02821630
asked Nov 26 '18 at 20:22
Hello_World
4,02821630
asked Nov 26 '18 at 20:22
Hello_World
4,02821630
asked Nov 26 '18 at 20:22
Hello_World
4,02821630
4,02821630
$df_x$ is always continuous since it is, by definition, a linear map
– user25959
Nov 26 '18 at 20:34
What you have done (correctly) in the last part is to show that the map $xmapsto df_x$ is a continuous map. This is a map from $mathbb{R} rightarrow text{End}(mathbb{R}^n,mathbb{R})$, where "End(U,V)" denotes the linear maps from U to V.
– user25959
Nov 26 '18 at 20:36
@user25959 it is continuous because the dimension of the space is finite?
– Hello_World
Nov 26 '18 at 20:45
Yes that's a good point, in infinite dimensional spaces linear can be unbounded/discontinuous
– user25959
Nov 26 '18 at 20:46
add a comment |
$df_x$ is always continuous since it is, by definition, a linear map
– user25959
Nov 26 '18 at 20:34
What you have done (correctly) in the last part is to show that the map $xmapsto df_x$ is a continuous map. This is a map from $mathbb{R} rightarrow text{End}(mathbb{R}^n,mathbb{R})$, where "End(U,V)" denotes the linear maps from U to V.
– user25959
Nov 26 '18 at 20:36
@user25959 it is continuous because the dimension of the space is finite?
– Hello_World
Nov 26 '18 at 20:45
Yes that's a good point, in infinite dimensional spaces linear can be unbounded/discontinuous
– user25959
Nov 26 '18 at 20:46
$df_x$ is always continuous since it is, by definition, a linear map
– user25959
Nov 26 '18 at 20:34
$df_x$ is always continuous since it is, by definition, a linear map
– user25959
Nov 26 '18 at 20:34
What you have done (correctly) in the last part is to show that the map $xmapsto df_x$ is a continuous map. This is a map from $mathbb{R} rightarrow text{End}(mathbb{R}^n,mathbb{R})$, where "End(U,V)" denotes the linear maps from U to V.
– user25959
Nov 26 '18 at 20:36
What you have done (correctly) in the last part is to show that the map $xmapsto df_x$ is a continuous map. This is a map from $mathbb{R} rightarrow text{End}(mathbb{R}^n,mathbb{R})$, where "End(U,V)" denotes the linear maps from U to V.
– user25959
Nov 26 '18 at 20:36
@user25959 it is continuous because the dimension of the space is finite?
– Hello_World
Nov 26 '18 at 20:45
@user25959 it is continuous because the dimension of the space is finite?
– Hello_World
Nov 26 '18 at 20:45
Yes that's a good point, in infinite dimensional spaces linear can be unbounded/discontinuous
– user25959
Nov 26 '18 at 20:46
Yes that's a good point, in infinite dimensional spaces linear can be unbounded/discontinuous
– user25959
Nov 26 '18 at 20:46
add a comment |
1 Answer
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The directional derivative you have for $d_{x}f: mathbb{R}^{n} rightarrow mathbb{R}$ is correct, and as sort of stated in the comments that a linear map between finite-dimensional vector spaces is continuous iff it is linear (which is equivalent to boundedness with linear maps and finite-dimensional vector spaces).
More explicitly however, you can proceed as: for all $epsilon > 0$, suppose $y,z in mathbb{R}^{n}$ are such that $| y - z | < epsilon$. Then for a fixed $x in mathbb{R}^{n}$, and $A in M_{ntimes n}(mathbb{R})$,
$$
d_{x}f(y) - d_{x}f(z) = x^{T}Ay + y^{T}Ax - x^{T}Az - z^{T}Ax\
= x^{T}A(y-z) + (y-z)^{T} Ax
$$
and therefore with respect to any norm (all norms are equivalent on finite-dimensional vector spaces),
$$
| d_{x}f(y) - d_{x}f(z) | = | x^{T}A(y-z) + (y-z)^{T} Ax | leq 2|x| |A||y - z | < 2|x| |A| epsilon.
$$
So as both $xin mathbb{R}^{n}$ and $A in M_{n times n}(mathbb{R})$ are fixed, we can set $delta := 2| x | | A | epsilon$, then formally:
For any $epsilon > 0$ such that $| y - z | < epsilon$, there exists a $delta > 0$ such that
$$
|y - z|< epsilon implies | d_{x}f(y) - d_{x}f(z)| < delta
$$
answered Nov 26 '18 at 21:21
BenCWBrown
3807
add a comment |
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The directional derivative you have for $d_{x}f: mathbb{R}^{n} rightarrow mathbb{R}$ is correct, and as sort of stated in the comments that a linear map between finite-dimensional vector spaces is continuous iff it is linear (which is equivalent to boundedness with linear maps and finite-dimensional vector spaces).
More explicitly however, you can proceed as: for all $epsilon > 0$, suppose $y,z in mathbb{R}^{n}$ are such that $| y - z | < epsilon$. Then for a fixed $x in mathbb{R}^{n}$, and $A in M_{ntimes n}(mathbb{R})$,
$$
d_{x}f(y) - d_{x}f(z) = x^{T}Ay + y^{T}Ax - x^{T}Az - z^{T}Ax\
= x^{T}A(y-z) + (y-z)^{T} Ax
$$
and therefore with respect to any norm (all norms are equivalent on finite-dimensional vector spaces),
$$
| d_{x}f(y) - d_{x}f(z) | = | x^{T}A(y-z) + (y-z)^{T} Ax | leq 2|x| |A||y - z | < 2|x| |A| epsilon.
$$
So as both $xin mathbb{R}^{n}$ and $A in M_{n times n}(mathbb{R})$ are fixed, we can set $delta := 2| x | | A | epsilon$, then formally:
For any $epsilon > 0$ such that $| y - z | < epsilon$, there exists a $delta > 0$ such that
$$
|y - z|< epsilon implies | d_{x}f(y) - d_{x}f(z)| < delta
$$
answered Nov 26 '18 at 21:21
BenCWBrown
3807
add a comment |
The directional derivative you have for $d_{x}f: mathbb{R}^{n} rightarrow mathbb{R}$ is correct, and as sort of stated in the comments that a linear map between finite-dimensional vector spaces is continuous iff it is linear (which is equivalent to boundedness with linear maps and finite-dimensional vector spaces).
More explicitly however, you can proceed as: for all $epsilon > 0$, suppose $y,z in mathbb{R}^{n}$ are such that $| y - z | < epsilon$. Then for a fixed $x in mathbb{R}^{n}$, and $A in M_{ntimes n}(mathbb{R})$,
$$
d_{x}f(y) - d_{x}f(z) = x^{T}Ay + y^{T}Ax - x^{T}Az - z^{T}Ax\
= x^{T}A(y-z) + (y-z)^{T} Ax
$$
and therefore with respect to any norm (all norms are equivalent on finite-dimensional vector spaces),
$$
| d_{x}f(y) - d_{x}f(z) | = | x^{T}A(y-z) + (y-z)^{T} Ax | leq 2|x| |A||y - z | < 2|x| |A| epsilon.
$$
So as both $xin mathbb{R}^{n}$ and $A in M_{n times n}(mathbb{R})$ are fixed, we can set $delta := 2| x | | A | epsilon$, then formally:
For any $epsilon > 0$ such that $| y - z | < epsilon$, there exists a $delta > 0$ such that
$$
|y - z|< epsilon implies | d_{x}f(y) - d_{x}f(z)| < delta
$$
answered Nov 26 '18 at 21:21
BenCWBrown
3807
add a comment |
The directional derivative you have for $d_{x}f: mathbb{R}^{n} rightarrow mathbb{R}$ is correct, and as sort of stated in the comments that a linear map between finite-dimensional vector spaces is continuous iff it is linear (which is equivalent to boundedness with linear maps and finite-dimensional vector spaces).
More explicitly however, you can proceed as: for all $epsilon > 0$, suppose $y,z in mathbb{R}^{n}$ are such that $| y - z | < epsilon$. Then for a fixed $x in mathbb{R}^{n}$, and $A in M_{ntimes n}(mathbb{R})$,
$$
d_{x}f(y) - d_{x}f(z) = x^{T}Ay + y^{T}Ax - x^{T}Az - z^{T}Ax\
= x^{T}A(y-z) + (y-z)^{T} Ax
$$
and therefore with respect to any norm (all norms are equivalent on finite-dimensional vector spaces),
$$
| d_{x}f(y) - d_{x}f(z) | = | x^{T}A(y-z) + (y-z)^{T} Ax | leq 2|x| |A||y - z | < 2|x| |A| epsilon.
$$
So as both $xin mathbb{R}^{n}$ and $A in M_{n times n}(mathbb{R})$ are fixed, we can set $delta := 2| x | | A | epsilon$, then formally:
For any $epsilon > 0$ such that $| y - z | < epsilon$, there exists a $delta > 0$ such that
$$
|y - z|< epsilon implies | d_{x}f(y) - d_{x}f(z)| < delta
$$
answered Nov 26 '18 at 21:21
BenCWBrown
3807
The directional derivative you have for $d_{x}f: mathbb{R}^{n} rightarrow mathbb{R}$ is correct, and as sort of stated in the comments that a linear map between finite-dimensional vector spaces is continuous iff it is linear (which is equivalent to boundedness with linear maps and finite-dimensional vector spaces).
More explicitly however, you can proceed as: for all $epsilon > 0$, suppose $y,z in mathbb{R}^{n}$ are such that $| y - z | < epsilon$. Then for a fixed $x in mathbb{R}^{n}$, and $A in M_{ntimes n}(mathbb{R})$,
$$
d_{x}f(y) - d_{x}f(z) = x^{T}Ay + y^{T}Ax - x^{T}Az - z^{T}Ax\
= x^{T}A(y-z) + (y-z)^{T} Ax
$$
and therefore with respect to any norm (all norms are equivalent on finite-dimensional vector spaces),
$$
| d_{x}f(y) - d_{x}f(z) | = | x^{T}A(y-z) + (y-z)^{T} Ax | leq 2|x| |A||y - z | < 2|x| |A| epsilon.
$$
So as both $xin mathbb{R}^{n}$ and $A in M_{n times n}(mathbb{R})$ are fixed, we can set $delta := 2| x | | A | epsilon$, then formally:
For any $epsilon > 0$ such that $| y - z | < epsilon$, there exists a $delta > 0$ such that
$$
|y - z|< epsilon implies | d_{x}f(y) - d_{x}f(z)| < delta
$$
answered Nov 26 '18 at 21:21
BenCWBrown
3807
edited Nov 26 '18 at 21:32
answered Nov 26 '18 at 21:21
BenCWBrown
3807
answered Nov 26 '18 at 21:21
BenCWBrown
3807
answered Nov 26 '18 at 21:21
BenCWBrown
3807
3807
add a comment |
add a comment |
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$df_x$ is always continuous since it is, by definition, a linear map
– user25959
Nov 26 '18 at 20:34
What you have done (correctly) in the last part is to show that the map $xmapsto df_x$ is a continuous map. This is a map from $mathbb{R} rightarrow text{End}(mathbb{R}^n,mathbb{R})$, where "End(U,V)" denotes the linear maps from U to V.
– user25959
Nov 26 '18 at 20:36
@user25959 it is continuous because the dimension of the space is finite?
– Hello_World
Nov 26 '18 at 20:45
Yes that's a good point, in infinite dimensional spaces linear can be unbounded/discontinuous
– user25959
Nov 26 '18 at 20:46