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With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is $$ f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, . $$ $f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$.

If the gradient $$ \nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(e^y-x^2) \bigr) $$ is zero then $e^y =x^2$ and $x(x^2-1) = 0$. $x= 0 $ is not possible, so that the gradient is zero only if $x=\pm1$ and $y=0$, that is only at the global minima.

The construction is inspired by Does $f$ have a critical point if $f(x, y) \to +\infty$ on all horizontal lines and $f(x, y) \to -\infty$ on all vertical lines?. We have $f(x, y) = g(\phi(x, y))$ where:

• $g(u, v) = (u^2-1)^2 + v^2$ has two global minima, but also an additional critical point at $(0, 0)$, and
• $ \phi(x, y) = ( x , e^y-x^2)$ is a diffeomorphism from the plane onto the set $\{ (u, v) \mid v > -u^2 \}$. The image is chosen such that it contains the minima of the function $g$, but not its critical point.

With respect to the “connected lakes” approach: The level sets $$ L(z) = \{ (x, y) \mid f(x, y) \le z \} $$ connect the minima $(-1, 0)$ and $(1, 0)$ exactly if $z > 1$. The infimum of such levels is therefore $m=1$, but $L(1)$ does not connect the minima (it does not contain the y-axis). Therefore this approach does not lead to a candidate for a critical point.

The above approach can also be used to construct a counterexample with bounded derivatives. Set $f(x, y) = g(\phi(x, y))$ with

• $g(u, v) = \frac{(u^2-1)^2}{1+u^4} + \frac{v^2}{1+v^2}$, which has two global minima at $(\pm 1, 0)$, one critical point at $(0, 0)$, and bounded derivatives.
• $\phi(x, y) = (x, \log(1+e^y) +1 -\sqrt{1+x^2} )$, which is a diffeomorphism from $\Bbb R^2$ with bounded derivatives onto the set $\{ (u, v) \mid v > 1- \sqrt{1+v^2} \}$, which contains the points $(\pm 1, 0)$ but not the point $(0, 0)$.

With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is $$ f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, . $$ $f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$.

If the gradient $$ \nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(e^y-x^2) \bigr) $$ is zero then $e^y =x^2$ and $x(x^2-1) = 0$. $x= 0 $ is not possible, so that the gradient is zero only if $x=\pm1$ and $y=0$, that is only at the global minima.

The construction is inspired by Does $f$ have a critical point if $f(x, y) \to +\infty$ on all horizontal lines and $f(x, y) \to -\infty$ on all vertical lines?. We have $f(x, y) = g(\phi(x, y))$ where:

• $g(u, v) = (u^2-1)^2 + v^2$ has two global minima, but also an additional critical point at $(0, 0)$, and
• $ \phi(x, y) = ( x , e^y-x^2)$ is a diffeomorphism from the plane onto the set $\{ (u, v) \mid v > -u^2 \}$. The image is chosen such that it contains the minima of the function $g$, but not its critical point.

With respect to the “connected lakes” approach: The level sets $$ L(z) = \{ (x, y) \mid f(x, y) \le z \} $$ connect the minima $(-1, 0)$ and $(1, 0)$ exactly if $z > 1$. The infimum of such levels is therefore $m=1$, but $L(1)$ does not connect the minima (it does not contain the y-axis). Therefore this approach does not lead to a candidate for a critical point.

The above approach can also be used to construct a counterexample with bounded derivatives. Set $f(x, y) = g(\phi(x, y))$ with

• $g(u, v) = \frac{(u^2-1)^2}{1+u^4} + \frac{v^2}{1+v^2}$, which has two global minima at $(\pm 1, 0)$, one critical point at $(0, 0)$, and bounded derivatives.
• $\phi(x, y) = (x, \log(1+e^y) +1 -\sqrt{1+x^2} )$, which is a diffeomorphism from $\Bbb R^2$ with bounded derivatives onto the set $\{ (u, v) \mid v > 1- \sqrt{1+v^2} \}$, which contains the points $(\pm 1, 0)$ but not the point $(0, 0)$.

With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is $$ f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, . $$ $f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$.

If the gradient $$ \nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(e^y-x^2) \bigr) $$ is zero then $e^y =x^2$ and $x(x^2-1) = 0$. $x= 0 $ is not possible, so that the gradient is zero only if $x=\pm1$ and $y=0$, that is only at the global minima.

The construction is inspired by Does $f$ have a critical point if $f(x, y) \to +\infty$ on all horizontal lines and $f(x, y) \to -\infty$ on all vertical lines?. We have $f(x, y) = g(\phi(x, y))$ where:

• $g(u, v) = (u^2-1)^2 + v^2$ has two global minima, but also an additional critical point at $(0, 0)$, and
• $ \phi(x, y) = ( x , e^y-x^2)$ is a diffeomorphism from the plane onto the set $\{ (u, v) \mid v > -u^2 \}$. The image is chosen such that it contains the minima of the function $g$, but not its critical point.

With respect to the “connected lakes” approach: The level sets $$ L(z) = \{ (x, y) \mid f(x, y) \le z \} $$ connect the minima $(-1, 0)$ and $(1, 0)$ exactly if $z > 1$. The infimum of such levels is therefore $m=1$, but $L(1)$ does not connect the minima (it does not contain the y-axis). Therefore this approach does not lead to a candidate for a critical point.

The above approach can also be used to construct a counterexample with bounded derivatives. Set $f(x, y) = g(\phi(x, y))$ with

• $g(u, v) = \frac{(u^2-1)^2}{1+u^4} + \frac{v^2}{1+v^2}$, which has two global minima at $(\pm 1, 0)$, one critical point at $(0, 0)$, and bounded derivatives.
• $\phi(x, y) = (x, \log(1+e^y) +1 -\sqrt{1+x^2} )$, which is a diffeomorphism from $\Bbb R^2$ with bounded derivatives onto the set $\{ (u, v) \mid v > 1- \sqrt{1+v^2} \}$, which contains the points $(\pm 1, 0)$ but not the point $(0, 0)$.

With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is $$ f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, . $$ $f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$.

If the gradient $$ \nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(e^y-x^2) \bigr) $$ is zero then $e^y =x^2$ and $x(x^2-1) = 0$. $x= 0 $ is not possible, so that the gradient is zero only if $x=\pm1$ and $y=0$, that is only at the global minima.

The construction is inspired by Does $f$ have a critical point if $f(x, y) \to +\infty$ on all horizontal lines and $f(x, y) \to -\infty$ on all vertical lines?. We have $f(x, y) = g(\phi(x, y))$ where:

• $g(u, v) = (u^2-1)^2 + v^2$ has two global minima, but also an additional critical point at $(0, 0)$, and
• $ \phi(x, y) = ( x , e^y-x^2)$ is a diffeomorphism from the plane onto the set $\{ (u, v) \mid v > -u^2 \}$. The image is chosen such that it contains the minima of the function $g$, but not its critical point.

With respect to the “connected lakes” approach: The level sets $$ L(z) = \{ (x, y) \mid f(x, y) \le z \} $$ connect the minima $(-1, 0)$ and $(1, 0)$ exactly if $z > 1$. The infimum of such levels is therefore $m=1$, but $L(1)$ does not connect the minima (it does not contain the y-axis). Therefore this approach does not lead to a candidate for a critical point.

The above approach can also be used to construct a counterexample with bounded derivatives. Set $f(x, y) = g(\phi(x, y))$ with

• $g(u, v) = \frac{(u^2-1)^2}{1+u^4} + \frac{v^2}{1+v^2}$, which has two global minima at $(\pm 1, 0)$, one critical point at $(0, 0)$, and bounded derivatives.
• $\phi(x, y) = (x, \log(1+e^y) +1 -\sqrt{1+x^2} )$, which is a diffeomorphism from $\Bbb R^2$ with bounded derivatives onto the set $\{ (u, v) \mid v > 1- \sqrt{1+v^2} \}$, which contains the points $(\pm 1, 0)$ but not the point $(0, 0)$.

With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is $$ f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, . $$ $f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$.

If the gradient $$ \nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(e^y-x^2) \bigr) $$ is zero then $e^y =x^2$ and $x(x^2-1) = 0$. $x= 0 $ is not possible, so that the gradient is zero only if $x=\pm1$ and $y=0$, that is only at the global minima.

The construction is inspired by Does $f$ have a critical point if $f(x, y) \to +\infty$ on all horizontal lines and $f(x, y) \to -\infty$ on all vertical lines?. We have $f(x, y) = g(\phi(x, y))$ where:

• $g(u, v) = (u^2-1)^2 + v^2$ has two global minima, but also an additional critical point at $(0, 0)$, and
• $ \phi(x, y) = ( x , e^y-x^2)$ is a diffeomorphism from the plane onto the set $\{ (u, v) \mid v > -u^2 \}$. The image is chosen such that it contains the minima of the function $g$, but not its critical point.

With respect to the “connected lakes” approach: The level sets $$ L(z) = \{ (x, y) \mid f(x, y) \le z \} $$ connect the minima $(-1, 0)$ and $(1, 0)$ exactly if $z > 1$. The infimum of such levels is therefore $m=1$, but $L(1)$ does not connect the minima (it does not contain the y-axis). Therefore this approach does not lead to a candidate for a critical point.

The above approach can also be used to construct a counterexample with bounded derivatives. Set $f(x, y) = g(\phi(x, y))$ with

• $g(u, v) = \frac{(u^2-1)^2}{1+u^4} + \frac{v^2}{1+v^2}$, which has two global minima at $(\pm 1, 0)$, one critical point at $(0, 0)$, and bounded derivatives.
• $\phi(x, y) = (x, \log(1+e^y) +1 -\sqrt{1+x^2} )$, which is a diffeomorphism from $\Bbb R^2$ with bounded derivatives onto the set $\{ (u, v) \mid v > 1- \sqrt{1+v^2} \}$, which contains the points $(\pm 1, 0)$ but not the point $(0, 0)$.

With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is $$ f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, . $$ $f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$.

If the gradient $$ \nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(e^y-x^2) \bigr) $$ is zero then $e^y =x^2$ and $x(x^2-1) = 0$. $x= 0 $ is not possible, so that the gradient is zero only if $x=\pm1$ and $y=0$, that is only at the global minima.

The construction is inspired by Does $f$ have a critical point if $f(x, y) \to +\infty$ on all horizontal lines and $f(x, y) \to -\infty$ on all vertical lines?. We have $f(x, y) = g(\phi(x, y))$ where:

• $g(u, v) = (u^2-1)^2 + v^2$ has two global minima, but also an additional critical point at $(0, 0)$, and
• $ \phi(x, y) = ( x , e^y-x^2)$ is a diffeomorphism from the plane onto the set $\{ (u, v) \mid v > -u^2 \}$. The image is chosen such that it contains the minima of the function $g$, but not its critical point.

With respect to the “connected lakes” approach: The level sets $$ L(z) = \{ (x, y) \mid f(x, y) \le z \} $$ connect the minima $(-1, 0)$ and $(1, 0)$ exactly if $z > 1$. The infimum of such levels is therefore $m=1$, but $L(1)$ does not connect the minima (it does not contain the y-axis). Therefore this approach does not lead to a candidate for a critical point.

The above approach can also be used to construct a counterexample with bounded derivatives. Set $f(x, y) = g(\phi(x, y))$ with

• $g(u, v) = \frac{(u^2-1)^2}{1+u^4} + \frac{v^2}{1+v^2}$, which has two global minima at $(\pm 1, 0)$, one critical point at $(0, 0)$, and bounded derivatives.
• $\phi(x, y) = (x, \log(1+e^y) +1 -\sqrt{1+x^2} )$, which is a diffeomorphism from $\Bbb R^2$ with bounded derivatives onto the set $\{ (u, v) \mid v > 1- \sqrt{1+v^2} \}$, which contains the points $(\pm 1, 0)$ but not the point $(0, 0)$.