continuous function, existence of all its directional derivatives and positive in the unit ballFinding all...
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continuous function, existence of all its directional derivatives and positive in the unit ball
Finding all directional derivatives of a function involving absolute value.A continuous function with all directional derivatives but NOT differentiable??Is there a function that's continuous and has all directional derivatives as a linear function of direction, but still fails to be differentiable?Continuity of the directional derivatives implies continuity at the point ?Prove partial derivatives exist, but not all directional derivatives exists.Positive Directional derivativesExistence of Directional Derivatives Does Not Force Differentiability or ContinuityIf directional derivatives are bounded, then function is continuousPositive directional derivatives on sphereDifferentiability on a function from directional derivatives
$begingroup$
Let $f:mathbb{R}^{m} rightarrow mathbb{R}$ be a continuous function such that all the directional derivatives exist for every point in $mathbb{R}^{m}$. Suppose $frac{partial f}{partial u}(u)>0$ for all $u in S^{m-1}$. Prove there exists a point $a in mathbb{R}^{m}$ such that $frac{partial f}{partial v}(a)=0$ for all $v in mathbb{R}^{m}$.
So far I've given that I'll need to use that a continuous function attains its maximum and minimum value in the unit ball, but I don't know how to go from there; any books referenced or tips would be welcome.
real-analysis multivariable-calculus derivatives
edited Mar 10 at 5:05
rash
41412
asked Mar 10 at 4:42
ipreferpiipreferpi
348
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}^{m} rightarrow mathbb{R}$ be a continuous function such that all the directional derivatives exist for every point in $mathbb{R}^{m}$. Suppose $frac{partial f}{partial u}(u)>0$ for all $u in S^{m-1}$. Prove there exists a point $a in mathbb{R}^{m}$ such that $frac{partial f}{partial v}(a)=0$ for all $v in mathbb{R}^{m}$.
So far I've given that I'll need to use that a continuous function attains its maximum and minimum value in the unit ball, but I don't know how to go from there; any books referenced or tips would be welcome.
real-analysis multivariable-calculus derivatives
edited Mar 10 at 5:05
rash
41412
asked Mar 10 at 4:42
ipreferpiipreferpi
348
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}^{m} rightarrow mathbb{R}$ be a continuous function such that all the directional derivatives exist for every point in $mathbb{R}^{m}$. Suppose $frac{partial f}{partial u}(u)>0$ for all $u in S^{m-1}$. Prove there exists a point $a in mathbb{R}^{m}$ such that $frac{partial f}{partial v}(a)=0$ for all $v in mathbb{R}^{m}$.
So far I've given that I'll need to use that a continuous function attains its maximum and minimum value in the unit ball, but I don't know how to go from there; any books referenced or tips would be welcome.
real-analysis multivariable-calculus derivatives
edited Mar 10 at 5:05
rash
41412
asked Mar 10 at 4:42
ipreferpiipreferpi
348
$endgroup$
Let $f:mathbb{R}^{m} rightarrow mathbb{R}$ be a continuous function such that all the directional derivatives exist for every point in $mathbb{R}^{m}$. Suppose $frac{partial f}{partial u}(u)>0$ for all $u in S^{m-1}$. Prove there exists a point $a in mathbb{R}^{m}$ such that $frac{partial f}{partial v}(a)=0$ for all $v in mathbb{R}^{m}$.
So far I've given that I'll need to use that a continuous function attains its maximum and minimum value in the unit ball, but I don't know how to go from there; any books referenced or tips would be welcome.
real-analysis multivariable-calculus derivatives
real-analysis multivariable-calculus derivatives
edited Mar 10 at 5:05
rash
41412
asked Mar 10 at 4:42
ipreferpiipreferpi
348
edited Mar 10 at 5:05
rash
41412
asked Mar 10 at 4:42
ipreferpiipreferpi
348
edited Mar 10 at 5:05
rash
41412
edited Mar 10 at 5:05
rash
41412
edited Mar 10 at 5:05
rash
41412
41412
asked Mar 10 at 4:42
ipreferpiipreferpi
348
asked Mar 10 at 4:42
ipreferpiipreferpi
348
asked Mar 10 at 4:42
ipreferpiipreferpi
348
348
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your function, being continuous on the closed unit ball $B_1(0)={xin R^m: |x|le 1}$, should attain its minimum there (the closed ball is compact). It does not attain the min on the boundary of the ball $S^{m-1}$ for, suppose it does attain its min at $ain S^{m-1}$. Since at all those points the radial derivative $partial f/partial u>0$, $f(lambda a)<f(a)$ for $lambda<1$, sufficiently close to $1$. The min is thus achieved inside the unit ball. At that point, all directional derivatives have to be equal to zero (otherwise $f$ would decrease in some direction, a contradiction).
answered Mar 10 at 8:59
GReyesGReyes
1,95515
$endgroup$
$begingroup$
Sorry I got confused, when you say the radial derivative you mean $frac{partial f}{partial u}u$? Would you be so kind and refer to me any books that explain this kind of topic? Thanks.
$endgroup$
– ipreferpi
Mar 10 at 17:58
1
$begingroup$
@ipreferpi Since $u$ is a unit vector and your derivative is evaluated at $u$ itself, $partial f/partial u(u)$ is the rate of change of your function at points on the unit sphere in the direction of the unit vector joining the origin to $u$, that is, in the radial direction.Your function is increasing along the radius, that is at any point inside the unit ball its value is less than the value at the point you hit when you extend the segment $Ou$ up to the boundary. Any Calculus book liker Apostol's, or Spivak's books explain the concept. Also more elementary books like Stewart's textbook.
$endgroup$
– GReyes
Mar 10 at 23:55
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
Your function, being continuous on the closed unit ball $B_1(0)={xin R^m: |x|le 1}$, should attain its minimum there (the closed ball is compact). It does not attain the min on the boundary of the ball $S^{m-1}$ for, suppose it does attain its min at $ain S^{m-1}$. Since at all those points the radial derivative $partial f/partial u>0$, $f(lambda a)<f(a)$ for $lambda<1$, sufficiently close to $1$. The min is thus achieved inside the unit ball. At that point, all directional derivatives have to be equal to zero (otherwise $f$ would decrease in some direction, a contradiction).
answered Mar 10 at 8:59
GReyesGReyes
1,95515
$endgroup$
$begingroup$
Sorry I got confused, when you say the radial derivative you mean $frac{partial f}{partial u}u$? Would you be so kind and refer to me any books that explain this kind of topic? Thanks.
$endgroup$
– ipreferpi
Mar 10 at 17:58
1
$begingroup$
@ipreferpi Since $u$ is a unit vector and your derivative is evaluated at $u$ itself, $partial f/partial u(u)$ is the rate of change of your function at points on the unit sphere in the direction of the unit vector joining the origin to $u$, that is, in the radial direction.Your function is increasing along the radius, that is at any point inside the unit ball its value is less than the value at the point you hit when you extend the segment $Ou$ up to the boundary. Any Calculus book liker Apostol's, or Spivak's books explain the concept. Also more elementary books like Stewart's textbook.
$endgroup$
– GReyes
Mar 10 at 23:55
add a comment |
$begingroup$
Your function, being continuous on the closed unit ball $B_1(0)={xin R^m: |x|le 1}$, should attain its minimum there (the closed ball is compact). It does not attain the min on the boundary of the ball $S^{m-1}$ for, suppose it does attain its min at $ain S^{m-1}$. Since at all those points the radial derivative $partial f/partial u>0$, $f(lambda a)<f(a)$ for $lambda<1$, sufficiently close to $1$. The min is thus achieved inside the unit ball. At that point, all directional derivatives have to be equal to zero (otherwise $f$ would decrease in some direction, a contradiction).
answered Mar 10 at 8:59
GReyesGReyes
1,95515
$endgroup$
$begingroup$
Sorry I got confused, when you say the radial derivative you mean $frac{partial f}{partial u}u$? Would you be so kind and refer to me any books that explain this kind of topic? Thanks.
$endgroup$
– ipreferpi
Mar 10 at 17:58
1
$begingroup$
@ipreferpi Since $u$ is a unit vector and your derivative is evaluated at $u$ itself, $partial f/partial u(u)$ is the rate of change of your function at points on the unit sphere in the direction of the unit vector joining the origin to $u$, that is, in the radial direction.Your function is increasing along the radius, that is at any point inside the unit ball its value is less than the value at the point you hit when you extend the segment $Ou$ up to the boundary. Any Calculus book liker Apostol's, or Spivak's books explain the concept. Also more elementary books like Stewart's textbook.
$endgroup$
– GReyes
Mar 10 at 23:55
add a comment |
$begingroup$
Your function, being continuous on the closed unit ball $B_1(0)={xin R^m: |x|le 1}$, should attain its minimum there (the closed ball is compact). It does not attain the min on the boundary of the ball $S^{m-1}$ for, suppose it does attain its min at $ain S^{m-1}$. Since at all those points the radial derivative $partial f/partial u>0$, $f(lambda a)<f(a)$ for $lambda<1$, sufficiently close to $1$. The min is thus achieved inside the unit ball. At that point, all directional derivatives have to be equal to zero (otherwise $f$ would decrease in some direction, a contradiction).
answered Mar 10 at 8:59
GReyesGReyes
1,95515
$endgroup$
Your function, being continuous on the closed unit ball $B_1(0)={xin R^m: |x|le 1}$, should attain its minimum there (the closed ball is compact). It does not attain the min on the boundary of the ball $S^{m-1}$ for, suppose it does attain its min at $ain S^{m-1}$. Since at all those points the radial derivative $partial f/partial u>0$, $f(lambda a)<f(a)$ for $lambda<1$, sufficiently close to $1$. The min is thus achieved inside the unit ball. At that point, all directional derivatives have to be equal to zero (otherwise $f$ would decrease in some direction, a contradiction).
answered Mar 10 at 8:59
GReyesGReyes
1,95515
answered Mar 10 at 8:59
GReyesGReyes
1,95515
answered Mar 10 at 8:59
GReyesGReyes
1,95515
answered Mar 10 at 8:59
GReyesGReyes
1,95515
1,95515
$begingroup$
Sorry I got confused, when you say the radial derivative you mean $frac{partial f}{partial u}u$? Would you be so kind and refer to me any books that explain this kind of topic? Thanks.
$endgroup$
– ipreferpi
Mar 10 at 17:58
1
$begingroup$
@ipreferpi Since $u$ is a unit vector and your derivative is evaluated at $u$ itself, $partial f/partial u(u)$ is the rate of change of your function at points on the unit sphere in the direction of the unit vector joining the origin to $u$, that is, in the radial direction.Your function is increasing along the radius, that is at any point inside the unit ball its value is less than the value at the point you hit when you extend the segment $Ou$ up to the boundary. Any Calculus book liker Apostol's, or Spivak's books explain the concept. Also more elementary books like Stewart's textbook.
$endgroup$
– GReyes
Mar 10 at 23:55
add a comment |
$begingroup$
Sorry I got confused, when you say the radial derivative you mean $frac{partial f}{partial u}u$? Would you be so kind and refer to me any books that explain this kind of topic? Thanks.
$endgroup$
– ipreferpi
Mar 10 at 17:58
1
$begingroup$
@ipreferpi Since $u$ is a unit vector and your derivative is evaluated at $u$ itself, $partial f/partial u(u)$ is the rate of change of your function at points on the unit sphere in the direction of the unit vector joining the origin to $u$, that is, in the radial direction.Your function is increasing along the radius, that is at any point inside the unit ball its value is less than the value at the point you hit when you extend the segment $Ou$ up to the boundary. Any Calculus book liker Apostol's, or Spivak's books explain the concept. Also more elementary books like Stewart's textbook.
$endgroup$
– GReyes
Mar 10 at 23:55
$begingroup$
Sorry I got confused, when you say the radial derivative you mean $frac{partial f}{partial u}u$? Would you be so kind and refer to me any books that explain this kind of topic? Thanks.
$endgroup$
– ipreferpi
Mar 10 at 17:58
$begingroup$
Sorry I got confused, when you say the radial derivative you mean $frac{partial f}{partial u}u$? Would you be so kind and refer to me any books that explain this kind of topic? Thanks.
$endgroup$
– ipreferpi
Mar 10 at 17:58
1
1
$begingroup$
@ipreferpi Since $u$ is a unit vector and your derivative is evaluated at $u$ itself, $partial f/partial u(u)$ is the rate of change of your function at points on the unit sphere in the direction of the unit vector joining the origin to $u$, that is, in the radial direction.Your function is increasing along the radius, that is at any point inside the unit ball its value is less than the value at the point you hit when you extend the segment $Ou$ up to the boundary. Any Calculus book liker Apostol's, or Spivak's books explain the concept. Also more elementary books like Stewart's textbook.
$endgroup$
– GReyes
Mar 10 at 23:55
$begingroup$
@ipreferpi Since $u$ is a unit vector and your derivative is evaluated at $u$ itself, $partial f/partial u(u)$ is the rate of change of your function at points on the unit sphere in the direction of the unit vector joining the origin to $u$, that is, in the radial direction.Your function is increasing along the radius, that is at any point inside the unit ball its value is less than the value at the point you hit when you extend the segment $Ou$ up to the boundary. Any Calculus book liker Apostol's, or Spivak's books explain the concept. Also more elementary books like Stewart's textbook.
$endgroup$
– GReyes
Mar 10 at 23:55
add a comment |
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