Can we tell if a function has a max or min by looking along specific directions?
Suppose we have a smooth function $f$ from $mathbb R^ntomathbb R$ such that $nabla f(0)=0$, and we want to check if $f$ has a local maximum at $0$ (as opposed to a local min or a saddle point).
For any vector $v$ in $mathbb R^n$, we can form a function $g_v:mathbb Rtomathbb R$ defined by $g_v(x)=f(vx)$. Intuitively, we are looking at the behavior of $f$ along the $v$ direction. If $f$ has a local max at zero, then of course $g_v$ also has a local max at zero. My question is, is the converse true? If the single-variable function $g_v$ has a local max at $0$ for every possible direction $v$, does $f$ also have a local max at $0$?
If the Hessian of $f$ at zero has full rank, then this question is easily answered as yes. By taking $v$ to be in turn each of the eigenvectors of the Hessian, the fact that $g_v$ has a local max means that the corresponding eigenvalues are all negative, and so $f$ has a local max at $0$. But if some of the eigenvalues are $0$, then I'm not sure how to analyze it.
As a first step in a proof, I might think: well, for any fixed direction, there is some radius $r$ such that as long as $v$ is within distance $r$ of the origin, then $f(v)le f(0)$. But since the $r$ depends on the direction, we can't necessarily find a single $r$ which works for all directions, so that might allow for a counter-example.
As a bonus, if it does hold for smooth functions, what about functions that are merely differentiable? Continuous? Or even all functions?
multivariable-calculus maxima-minima
asked Sep 16 '18 at 23:49
Carmeister
2,7842921
add a comment |
Suppose we have a smooth function $f$ from $mathbb R^ntomathbb R$ such that $nabla f(0)=0$, and we want to check if $f$ has a local maximum at $0$ (as opposed to a local min or a saddle point).
For any vector $v$ in $mathbb R^n$, we can form a function $g_v:mathbb Rtomathbb R$ defined by $g_v(x)=f(vx)$. Intuitively, we are looking at the behavior of $f$ along the $v$ direction. If $f$ has a local max at zero, then of course $g_v$ also has a local max at zero. My question is, is the converse true? If the single-variable function $g_v$ has a local max at $0$ for every possible direction $v$, does $f$ also have a local max at $0$?
If the Hessian of $f$ at zero has full rank, then this question is easily answered as yes. By taking $v$ to be in turn each of the eigenvectors of the Hessian, the fact that $g_v$ has a local max means that the corresponding eigenvalues are all negative, and so $f$ has a local max at $0$. But if some of the eigenvalues are $0$, then I'm not sure how to analyze it.
As a first step in a proof, I might think: well, for any fixed direction, there is some radius $r$ such that as long as $v$ is within distance $r$ of the origin, then $f(v)le f(0)$. But since the $r$ depends on the direction, we can't necessarily find a single $r$ which works for all directions, so that might allow for a counter-example.
As a bonus, if it does hold for smooth functions, what about functions that are merely differentiable? Continuous? Or even all functions?
multivariable-calculus maxima-minima
asked Sep 16 '18 at 23:49
Carmeister
2,7842921
I just had the same question myself. We can possibly apply compactness to get a minimum $r$, as long as we can choose $r$ as a continuous function of the direction, but I don't quite see a way to make this work.
– Misha Lavrov
2 days ago
add a comment |
Suppose we have a smooth function $f$ from $mathbb R^ntomathbb R$ such that $nabla f(0)=0$, and we want to check if $f$ has a local maximum at $0$ (as opposed to a local min or a saddle point).
For any vector $v$ in $mathbb R^n$, we can form a function $g_v:mathbb Rtomathbb R$ defined by $g_v(x)=f(vx)$. Intuitively, we are looking at the behavior of $f$ along the $v$ direction. If $f$ has a local max at zero, then of course $g_v$ also has a local max at zero. My question is, is the converse true? If the single-variable function $g_v$ has a local max at $0$ for every possible direction $v$, does $f$ also have a local max at $0$?
If the Hessian of $f$ at zero has full rank, then this question is easily answered as yes. By taking $v$ to be in turn each of the eigenvectors of the Hessian, the fact that $g_v$ has a local max means that the corresponding eigenvalues are all negative, and so $f$ has a local max at $0$. But if some of the eigenvalues are $0$, then I'm not sure how to analyze it.
As a first step in a proof, I might think: well, for any fixed direction, there is some radius $r$ such that as long as $v$ is within distance $r$ of the origin, then $f(v)le f(0)$. But since the $r$ depends on the direction, we can't necessarily find a single $r$ which works for all directions, so that might allow for a counter-example.
As a bonus, if it does hold for smooth functions, what about functions that are merely differentiable? Continuous? Or even all functions?
multivariable-calculus maxima-minima
asked Sep 16 '18 at 23:49
Carmeister
2,7842921
Suppose we have a smooth function $f$ from $mathbb R^ntomathbb R$ such that $nabla f(0)=0$, and we want to check if $f$ has a local maximum at $0$ (as opposed to a local min or a saddle point).
For any vector $v$ in $mathbb R^n$, we can form a function $g_v:mathbb Rtomathbb R$ defined by $g_v(x)=f(vx)$. Intuitively, we are looking at the behavior of $f$ along the $v$ direction. If $f$ has a local max at zero, then of course $g_v$ also has a local max at zero. My question is, is the converse true? If the single-variable function $g_v$ has a local max at $0$ for every possible direction $v$, does $f$ also have a local max at $0$?
If the Hessian of $f$ at zero has full rank, then this question is easily answered as yes. By taking $v$ to be in turn each of the eigenvectors of the Hessian, the fact that $g_v$ has a local max means that the corresponding eigenvalues are all negative, and so $f$ has a local max at $0$. But if some of the eigenvalues are $0$, then I'm not sure how to analyze it.
As a first step in a proof, I might think: well, for any fixed direction, there is some radius $r$ such that as long as $v$ is within distance $r$ of the origin, then $f(v)le f(0)$. But since the $r$ depends on the direction, we can't necessarily find a single $r$ which works for all directions, so that might allow for a counter-example.
As a bonus, if it does hold for smooth functions, what about functions that are merely differentiable? Continuous? Or even all functions?
multivariable-calculus maxima-minima
multivariable-calculus maxima-minima
asked Sep 16 '18 at 23:49
Carmeister
2,7842921
asked Sep 16 '18 at 23:49
Carmeister
2,7842921
asked Sep 16 '18 at 23:49
Carmeister
2,7842921
asked Sep 16 '18 at 23:49
Carmeister
2,7842921
asked Sep 16 '18 at 23:49
Carmeister
2,7842921
2,7842921
I just had the same question myself. We can possibly apply compactness to get a minimum $r$, as long as we can choose $r$ as a continuous function of the direction, but I don't quite see a way to make this work.
– Misha Lavrov
2 days ago
add a comment |
I just had the same question myself. We can possibly apply compactness to get a minimum $r$, as long as we can choose $r$ as a continuous function of the direction, but I don't quite see a way to make this work.
– Misha Lavrov
2 days ago
I just had the same question myself. We can possibly apply compactness to get a minimum $r$, as long as we can choose $r$ as a continuous function of the direction, but I don't quite see a way to make this work.
– Misha Lavrov
2 days ago
I just had the same question myself. We can possibly apply compactness to get a minimum $r$, as long as we can choose $r$ as a continuous function of the direction, but I don't quite see a way to make this work.
– Misha Lavrov
2 days ago
add a comment |
1 Answer
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After thinking about it for a while I realized that the answer to this question is no, even for polynomial functions. Consider the following function, defined in polar coordinates:
$$f(r,theta)=r^4(r^2-cos^2theta)(r^2-2cos^2theta)$$
For any fixed value of $theta$, we have a local min at $0$: if $costheta=0$, then we just get $g(r)=r^8$, while for $costhetane 0$, we get $g(r)=2r^4cos^4theta-3r^6cos^2theta+r^8$. The leading term for small $r$ is $2r^4cos^4theta$, which also has a local min at $r=0$.
However, can find arbitrarily small $r$ with $f(r,theta)<0$. Given any (small) $r$, choose $theta$ so that $frac{r}{sqrt 2}<costheta<r$.
If we switch to rectangular coordinates, $x=rcostheta$, $y=rsintheta$, we get the degree $8$ polynomial:
$$2x^4-3x^6+x^8-6x^4y^2+4x^6y^2-3x^2y^4+6x^4y^4+4x^2y^6+y^8$$
So this function is, in fact, a polynomial, the nicest type of function around!
answered 2 days ago
Carmeister
2,7842921
Neat construction! To visualize what's going on, we can factor $f(x,y)$ as $(x^2+y^2+x)(x^2+y^2-x)(x^2+y^2+sqrt2 x)(x^2+y^2-sqrt2 x)$: the product of four circles passing through $(0,0)$, two small ones inside two larger ones. For this product to be negative, $(x,y)$ must be inside a large circle but outside a small one; here is a plot of this region. In any direction from $0$, we must go through a positive region before reaching a negative one.
– Misha Lavrov
2 days ago
1
I guess on that basis, we could take half the picture for another counterexample: say, $f(x,y) = (x^2+y^2-x)(x^2+y^2-2x)$ would also work.
– Misha Lavrov
2 days ago
@MishaLavrov Ah, good thinking! Looking at the regions where $f$ is positive and negative is a nice way to visualize it.
– Carmeister
2 days ago
add a comment |
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After thinking about it for a while I realized that the answer to this question is no, even for polynomial functions. Consider the following function, defined in polar coordinates:
$$f(r,theta)=r^4(r^2-cos^2theta)(r^2-2cos^2theta)$$
For any fixed value of $theta$, we have a local min at $0$: if $costheta=0$, then we just get $g(r)=r^8$, while for $costhetane 0$, we get $g(r)=2r^4cos^4theta-3r^6cos^2theta+r^8$. The leading term for small $r$ is $2r^4cos^4theta$, which also has a local min at $r=0$.
However, can find arbitrarily small $r$ with $f(r,theta)<0$. Given any (small) $r$, choose $theta$ so that $frac{r}{sqrt 2}<costheta<r$.
If we switch to rectangular coordinates, $x=rcostheta$, $y=rsintheta$, we get the degree $8$ polynomial:
$$2x^4-3x^6+x^8-6x^4y^2+4x^6y^2-3x^2y^4+6x^4y^4+4x^2y^6+y^8$$
So this function is, in fact, a polynomial, the nicest type of function around!
answered 2 days ago
Carmeister
2,7842921
Neat construction! To visualize what's going on, we can factor $f(x,y)$ as $(x^2+y^2+x)(x^2+y^2-x)(x^2+y^2+sqrt2 x)(x^2+y^2-sqrt2 x)$: the product of four circles passing through $(0,0)$, two small ones inside two larger ones. For this product to be negative, $(x,y)$ must be inside a large circle but outside a small one; here is a plot of this region. In any direction from $0$, we must go through a positive region before reaching a negative one.
– Misha Lavrov
2 days ago
1
I guess on that basis, we could take half the picture for another counterexample: say, $f(x,y) = (x^2+y^2-x)(x^2+y^2-2x)$ would also work.
– Misha Lavrov
2 days ago
@MishaLavrov Ah, good thinking! Looking at the regions where $f$ is positive and negative is a nice way to visualize it.
– Carmeister
2 days ago
add a comment |
After thinking about it for a while I realized that the answer to this question is no, even for polynomial functions. Consider the following function, defined in polar coordinates:
$$f(r,theta)=r^4(r^2-cos^2theta)(r^2-2cos^2theta)$$
For any fixed value of $theta$, we have a local min at $0$: if $costheta=0$, then we just get $g(r)=r^8$, while for $costhetane 0$, we get $g(r)=2r^4cos^4theta-3r^6cos^2theta+r^8$. The leading term for small $r$ is $2r^4cos^4theta$, which also has a local min at $r=0$.
However, can find arbitrarily small $r$ with $f(r,theta)<0$. Given any (small) $r$, choose $theta$ so that $frac{r}{sqrt 2}<costheta<r$.
If we switch to rectangular coordinates, $x=rcostheta$, $y=rsintheta$, we get the degree $8$ polynomial:
$$2x^4-3x^6+x^8-6x^4y^2+4x^6y^2-3x^2y^4+6x^4y^4+4x^2y^6+y^8$$
So this function is, in fact, a polynomial, the nicest type of function around!
answered 2 days ago
Carmeister
2,7842921
Neat construction! To visualize what's going on, we can factor $f(x,y)$ as $(x^2+y^2+x)(x^2+y^2-x)(x^2+y^2+sqrt2 x)(x^2+y^2-sqrt2 x)$: the product of four circles passing through $(0,0)$, two small ones inside two larger ones. For this product to be negative, $(x,y)$ must be inside a large circle but outside a small one; here is a plot of this region. In any direction from $0$, we must go through a positive region before reaching a negative one.
– Misha Lavrov
2 days ago
1
I guess on that basis, we could take half the picture for another counterexample: say, $f(x,y) = (x^2+y^2-x)(x^2+y^2-2x)$ would also work.
– Misha Lavrov
2 days ago
@MishaLavrov Ah, good thinking! Looking at the regions where $f$ is positive and negative is a nice way to visualize it.
– Carmeister
2 days ago
add a comment |
After thinking about it for a while I realized that the answer to this question is no, even for polynomial functions. Consider the following function, defined in polar coordinates:
$$f(r,theta)=r^4(r^2-cos^2theta)(r^2-2cos^2theta)$$
For any fixed value of $theta$, we have a local min at $0$: if $costheta=0$, then we just get $g(r)=r^8$, while for $costhetane 0$, we get $g(r)=2r^4cos^4theta-3r^6cos^2theta+r^8$. The leading term for small $r$ is $2r^4cos^4theta$, which also has a local min at $r=0$.
However, can find arbitrarily small $r$ with $f(r,theta)<0$. Given any (small) $r$, choose $theta$ so that $frac{r}{sqrt 2}<costheta<r$.
If we switch to rectangular coordinates, $x=rcostheta$, $y=rsintheta$, we get the degree $8$ polynomial:
$$2x^4-3x^6+x^8-6x^4y^2+4x^6y^2-3x^2y^4+6x^4y^4+4x^2y^6+y^8$$
So this function is, in fact, a polynomial, the nicest type of function around!
answered 2 days ago
Carmeister
2,7842921
After thinking about it for a while I realized that the answer to this question is no, even for polynomial functions. Consider the following function, defined in polar coordinates:
$$f(r,theta)=r^4(r^2-cos^2theta)(r^2-2cos^2theta)$$
For any fixed value of $theta$, we have a local min at $0$: if $costheta=0$, then we just get $g(r)=r^8$, while for $costhetane 0$, we get $g(r)=2r^4cos^4theta-3r^6cos^2theta+r^8$. The leading term for small $r$ is $2r^4cos^4theta$, which also has a local min at $r=0$.
However, can find arbitrarily small $r$ with $f(r,theta)<0$. Given any (small) $r$, choose $theta$ so that $frac{r}{sqrt 2}<costheta<r$.
If we switch to rectangular coordinates, $x=rcostheta$, $y=rsintheta$, we get the degree $8$ polynomial:
$$2x^4-3x^6+x^8-6x^4y^2+4x^6y^2-3x^2y^4+6x^4y^4+4x^2y^6+y^8$$
So this function is, in fact, a polynomial, the nicest type of function around!
answered 2 days ago
Carmeister
2,7842921
answered 2 days ago
Carmeister
2,7842921
answered 2 days ago
Carmeister
2,7842921
answered 2 days ago
Carmeister
2,7842921
2,7842921
Neat construction! To visualize what's going on, we can factor $f(x,y)$ as $(x^2+y^2+x)(x^2+y^2-x)(x^2+y^2+sqrt2 x)(x^2+y^2-sqrt2 x)$: the product of four circles passing through $(0,0)$, two small ones inside two larger ones. For this product to be negative, $(x,y)$ must be inside a large circle but outside a small one; here is a plot of this region. In any direction from $0$, we must go through a positive region before reaching a negative one.
– Misha Lavrov
2 days ago
1
I guess on that basis, we could take half the picture for another counterexample: say, $f(x,y) = (x^2+y^2-x)(x^2+y^2-2x)$ would also work.
– Misha Lavrov
2 days ago
@MishaLavrov Ah, good thinking! Looking at the regions where $f$ is positive and negative is a nice way to visualize it.
– Carmeister
2 days ago
add a comment |
Neat construction! To visualize what's going on, we can factor $f(x,y)$ as $(x^2+y^2+x)(x^2+y^2-x)(x^2+y^2+sqrt2 x)(x^2+y^2-sqrt2 x)$: the product of four circles passing through $(0,0)$, two small ones inside two larger ones. For this product to be negative, $(x,y)$ must be inside a large circle but outside a small one; here is a plot of this region. In any direction from $0$, we must go through a positive region before reaching a negative one.
– Misha Lavrov
2 days ago
1
I guess on that basis, we could take half the picture for another counterexample: say, $f(x,y) = (x^2+y^2-x)(x^2+y^2-2x)$ would also work.
– Misha Lavrov
2 days ago
@MishaLavrov Ah, good thinking! Looking at the regions where $f$ is positive and negative is a nice way to visualize it.
– Carmeister
2 days ago
Neat construction! To visualize what's going on, we can factor $f(x,y)$ as $(x^2+y^2+x)(x^2+y^2-x)(x^2+y^2+sqrt2 x)(x^2+y^2-sqrt2 x)$: the product of four circles passing through $(0,0)$, two small ones inside two larger ones. For this product to be negative, $(x,y)$ must be inside a large circle but outside a small one; here is a plot of this region. In any direction from $0$, we must go through a positive region before reaching a negative one.
– Misha Lavrov
2 days ago
Neat construction! To visualize what's going on, we can factor $f(x,y)$ as $(x^2+y^2+x)(x^2+y^2-x)(x^2+y^2+sqrt2 x)(x^2+y^2-sqrt2 x)$: the product of four circles passing through $(0,0)$, two small ones inside two larger ones. For this product to be negative, $(x,y)$ must be inside a large circle but outside a small one; here is a plot of this region. In any direction from $0$, we must go through a positive region before reaching a negative one.
– Misha Lavrov
2 days ago
1
1
I guess on that basis, we could take half the picture for another counterexample: say, $f(x,y) = (x^2+y^2-x)(x^2+y^2-2x)$ would also work.
– Misha Lavrov
2 days ago
I guess on that basis, we could take half the picture for another counterexample: say, $f(x,y) = (x^2+y^2-x)(x^2+y^2-2x)$ would also work.
– Misha Lavrov
2 days ago
@MishaLavrov Ah, good thinking! Looking at the regions where $f$ is positive and negative is a nice way to visualize it.
– Carmeister
2 days ago
@MishaLavrov Ah, good thinking! Looking at the regions where $f$ is positive and negative is a nice way to visualize it.
– Carmeister
2 days ago
add a comment |
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I just had the same question myself. We can possibly apply compactness to get a minimum $r$, as long as we can choose $r$ as a continuous function of the direction, but I don't quite see a way to make this work.
– Misha Lavrov
2 days ago