What is the maximum number $N$ of strict local minima that a degree 4 polynomial $p:\Bbb R^2\to \Bbb R$ can have?

One dimension: For a single variable quartic polynomial $q:\Bbb R \to \Bbb R$ the answer to this question would be easy: The polynomial has at most two local minima. Indeed, the derivative $q'$ is a cubic polynomial, so by Fundamental Theorem of Analysis $q$ has at most 3 critical points. However, when $q$ has 3 critical points, then all of them are simple roots, and so two consecutive critical points of $q$ can not be both local minima (local minima/maxima are alternating).

Along the lines: My first thought was to study $p$ along the lines $(x,y)=(a+bt,c+dt)$ as one would do to prove that a quadratic polynomial $p:\Bbb R^2\to \Bbb R$ has no more than one strict local minimum. However, this argument fails in the attempt to show that a quartic polynomial $p$ has at most two strict local minima. The problem is that unlike two points, three points in $\Bbb R^2$ can not in general be interpolated by a line. Three points in $\Bbb R^2$ can be interpolated by a quadratic curve like $(x(t),y(t))=(t,at^2+bt+c)$, but then $t\mapsto p(x(t),y(t))$ could be a polynomial of degree $8$.

At lest 4: Since the polynomial $q(t)=(t^2-1)^2$ has minima at the two points $t=\pm 1$, the polynomial $p(x,y) = q(x) + q(y)$ has minima at the four points points $(x,y)=(\pm 1,\pm 1)$. Thus $N\geq 4$.

At most 9: As commented by @GerryMyerson, the partial derivatives $p_x,p_y$ of $p$ are quadratic polynomials, and so by Bézout's Theorem the curves $p_x(x,y)=0$ and $p_y(x,y)=0$ intersect at no more than $3\times 3$ isolated points. Hence, $p$ has no more than 9 critical points. The question then is how many critical points $p$ needs to have in order to have $N$ strict local minima:

• For the case of $N=2$ and a general function $f$ this was answered in the post: If a two variable smooth function has two global minima, will it necessarily have a third critical point?
• I asked the related question for a degree 4 polynomial and $N\geq 2:$ Is there a quartic polynomial in two variables that have multiple local minima and no other critical points?

If you could not determine the exact value of $N$, could you at least give some lower or upper bound, like that $N>4$ or that $N$ is finite?

What is the maximum number $N$ of strict local minima that a degree 4 polynomial $p:\Bbb R^2\to \Bbb R$ can have?

One dimension: For a single variable quartic polynomial $q:\Bbb R \to \Bbb R$ the answer to this question would be easy: The polynomial has at most two local minima. Indeed, the derivative $q'$ is a cubic polynomial, so by Fundamental Theorem of Analysis $q$ has at most 3 critical points. However, when $q$ has 3 critical points, then all of them are simple roots, and so two consecutive critical points of $q$ can not be both local minima (local minima/maxima are alternating).

Along the lines: My first thought was to study $p$ along the lines $(x,y)=(a+bt,c+dt)$ as one would do to prove that a quadratic polynomial $p:\Bbb R^2\to \Bbb R$ has no more than one strict local minimum. However, this argument fails in the attempt to show that a quartic polynomial $p$ has at most two strict local minima. The problem is that unlike two points, three points in $\Bbb R^2$ can not in general be interpolated by a line. Three points in $\Bbb R^2$ can be interpolated by a quadratic curve like $(x(t),y(t))=(t,at^2+bt+c)$, but then $t\mapsto p(x(t),y(t))$ could be a polynomial of degree $8$.

At lest 4: Since the polynomial $q(t)=(t^2-1)^2$ has minima at the two points $t=\pm 1$, the polynomial $p(x,y) = q(x) + q(y)$ has minima at the four points points $(x,y)=(\pm 1,\pm 1)$. Thus $N\geq 4$.

At most 9: As commented by @GerryMyerson, the partial derivatives $p_x,p_y$ of $p$ are cubic polynomials, and so by Bézout's Theorem the curves $p_x(x,y)=0$ and $p_y(x,y)=0$ intersect at no more than $3\times 3$ isolated points. Hence, $p$ has no more than 9 critical points. The question then is how many critical points $p$ needs to have in order to have $N$ strict local minima:

• For the case of $N=2$ and a general function $f$ this was answered in the post: If a two variable smooth function has two global minima, will it necessarily have a third critical point?
• I asked the related question for a degree 4 polynomial and $N\geq 2:$ Is there a quartic polynomial in two variables that have multiple local minima and no other critical points?

If you could not determine the exact value of $N$, could you at least give some lower or upper bound, like that $N>4$ or that $N$ is finite?

What is the maximum number $N$ of isolated local minima that a 4-degree (quartic) polynomial $p:\Bbb R^2\to \Bbb R$ might have?

One dimension: For a single variable quartic polynomial $q:\Bbb R \to \Bbb R$ the answer to this question would be easy: The polynomial has at most two local minima. Indeed, the derivative $q'$ is a cubic polynomial, so by Fundamental Theorem of Analysis $q$ has at most 3 critical points. However, when $q$ has 3 critical points, then all of them are simple roots, and so two consecutive critical points of $q$ can not be both local minima (local minima/maxima are alternating).

Along the lines: Back to the two dimensional case: My first idea was to study $p$ along lines $(x,y)=(a+bt,c+dt)$ as one would do to prove that a quadratic polynomial $p:\Bbb R^2\to \Bbb R$ has no more than one local minimum (unless there are infinitely many minima as in the case of the polynomial $p(x,y)=(x-y)^2$). However this argument fails in the attempt to show that a quartic or cubic polynomial $p$ has at most two local minima. The problem is that unlike two points, three points in $\Bbb R^2$ don't need to lie on a line. Three points in $\Bbb R^2$ can be interpolated by a quadratic curve like $(x(t),y(t))=(t,at^2+bt+c)$, but then $t\mapsto p(x(t),y(t))$ could be a polynomial of degree $8$.

At lest four: Since the polynomial $q(t)=(t^2-1)^2$ has minima at the two points $t=\pm 1$, the polynomial $p(x,y) = q(x) + q(y)$ has minima at the four points points $(x,y)=(\pm 1,\pm 1)$. Thus $N\geq 4$.

If you could not determine the exact value of $N$, could you at least give some lower or upper bound, like that $N>4$ or that $N$ is finite?

What is the maximum number $N$ of strict local minima that a degree 4 polynomial $p:\Bbb R^2\to \Bbb R$ can have?

One dimension: For a single variable quartic polynomial $q:\Bbb R \to \Bbb R$ the answer to this question would be easy: The polynomial has at most two local minima. Indeed, the derivative $q'$ is a cubic polynomial, so by Fundamental Theorem of Analysis $q$ has at most 3 critical points. However, when $q$ has 3 critical points, then all of them are simple roots, and so two consecutive critical points of $q$ can not be both local minima (local minima/maxima are alternating).

Along the lines: My first thought was to study $p$ along the lines $(x,y)=(a+bt,c+dt)$ as one would do to prove that a quadratic polynomial $p:\Bbb R^2\to \Bbb R$ has no more than one strict local minimum. However, this argument fails in the attempt to show that a quartic polynomial $p$ has at most two strict local minima. The problem is that unlike two points, three points in $\Bbb R^2$ can not in general be interpolated by a line. Three points in $\Bbb R^2$ can be interpolated by a quadratic curve like $(x(t),y(t))=(t,at^2+bt+c)$, but then $t\mapsto p(x(t),y(t))$ could be a polynomial of degree $8$.

At lest 4: Since the polynomial $q(t)=(t^2-1)^2$ has minima at the two points $t=\pm 1$, the polynomial $p(x,y) = q(x) + q(y)$ has minima at the four points points $(x,y)=(\pm 1,\pm 1)$. Thus $N\geq 4$.

At most 9: As commented by @GerryMyerson, the partial derivatives $p_x,p_y$ of $p$ are quadratic polynomials, and so by Bézout's Theorem the curves $p_x(x,y)=0$ and $p_y(x,y)=0$ intersect at no more than $3\times 3$ isolated points. Hence, $p$ has no more than 9 critical points. The question then is how many critical points $p$ needs to have in order to have $N$ strict local minima:

• For the case of $N=2$ and a general function $f$ this was answered in the post: If a two variable smooth function has two global minima, will it necessarily have a third critical point?
• I asked the related question for a degree 4 polynomial and $N\geq 2:$ Is there a quartic polynomial in two variables that have multiple local minima and no other critical points?

If you could not determine the exact value of $N$, could you at least give some lower or upper bound, like that $N>4$ or that $N$ is finite?

What is the maximum number $N$ of isolated local minima that a 4-degree (quartic) polynomial $p:\Bbb R^2\to \Bbb R$ might have?

One dimension: For a single variable quartic polynomial $q:\Bbb R \to \Bbb R$ the answer to this question would be easy: The polynomial has at most two local minima. Indeed, the derivative $q'$ is a cubic polynomial, so by Fundamental Theorem of Analysis $q$ has at most 3 critical points. However, when $q$ has 3 critical points, then all of them are simple roots, and so two consecutive critical points of $q$ can not be both local minima (local minima/maxima are alternating).

Along the lines: Back to the two dimensional case: My first idea was to study $p$ along lines $(x,y)=(a+bt,c+dt)$ as one would do to prove that a quadratic polynomial $p:\Bbb R^2\to \Bbb R$ has no more than one local minimum (unless there are infinitely many minima as in the case of the polynomial $p(x,y)=(x-y)^2$). However this argument fails in the attempt to show that a quartic polynomial $p$ has at most two local minima. The problem is that unlike two points, three points in $\Bbb R^2$ don't need to lie on a line. Three points in $\Bbb R^2$ can be interpolated by a quadratic curve like $(x(t),y(t))=(t,at^2+bt+c)$, but then $t\mapsto p(x(t),y(t))$ could be a polynomial of degree $8$.

At lest four: Since the polynomial $q(t)=(t^2-1)^2$ has minima at the two points $t=\pm 1$, the polynomial $p(x,y) = q(x) + q(y)$ has minima at the four points points $(x,y)=(\pm 1,\pm 1)$. Thus $N\geq 4$.

If you could not determine the exact value of $N$, could you at least give some lower or upper bound, like that $N>4$ or that $N$ is finite?

What is the maximum number $N$ of isolated local minima that a 4-degree (quartic) polynomial $p:\Bbb R^2\to \Bbb R$ might have?

One dimension: For a single variable quartic polynomial $q:\Bbb R \to \Bbb R$ the answer to this question would be easy: The polynomial has at most two local minima. Indeed, the derivative $q'$ is a cubic polynomial, so by Fundamental Theorem of Analysis $q$ has at most 3 critical points. However, when $q$ has 3 critical points, then all of them are simple roots, and so two consecutive critical points of $q$ can not be both local minima (local minima/maxima are alternating).

Along the lines: Back to the two dimensional case: My first idea was to study $p$ along lines $(x,y)=(a+bt,c+dt)$ as one would do to prove that a quadratic polynomial $p:\Bbb R^2\to \Bbb R$ has no more than one local minimum (unless there are infinitely many minima as in the case of the polynomial $p(x,y)=(x-y)^2$). However this argument fails in the attempt to show that a quartic or cubic polynomial $p$ has at most two local minima. The problem is that unlike two points, three points in $\Bbb R^2$ don't need to lie on a line. Three points in $\Bbb R^2$ can be interpolated by a quadratic curve like $(x(t),y(t))=(t,at^2+bt+c)$, but then $t\mapsto p(x(t),y(t))$ could be a polynomial of degree $8$.

At lest four: Since the polynomial $q(t)=(t^2-1)^2$ has minima at the two points $t=\pm 1$, the polynomial $p(x,y) = q(x) + q(y)$ has minima at the four points points $(x,y)=(\pm 1,\pm 1)$. Thus $N\geq 4$.

If you could not determine the exact value of $N$, could you at least give some lower or upper bound, like that $N>4$ or that $N$ is finite?