Given a multivariable function $f(x_1,…,x_n)$ and assume that $\frac{\partial f}{\partial x_i}$ are monotone increasing with respect to $x_i$. This means the diagonal of Hessian matrix are positive. I wonder if there are more than one isolate local minimum.
Global minimum of multivariable function with monotone increasing partial derivative.
hessian-matrixmultivariable-calculusoptimizationpartial derivative
Related Solutions
There is no global minima. To check this take $y\to-\infty$ and $x=1$ and you get $f(x,y)\to-\infty$ but $(0,0)$ is a local minimum since for every $0<\epsilon<1$:$$\forall |x|<\epsilon ,|y|<\epsilon\qquad\qquad \epsilon^2(1+(1+\epsilon)^3)<2\epsilon^2<f(x,y)<\epsilon^2(1+(1+\epsilon)^3)<9\epsilon^2$$for a close look:
and for a further look:
To find the global extrema of continuous $f$ on a compact subset $M$ of $\mathbb{R}^2$, you do two things:
1. Interior. Find all critical points of $f$ in the interior of $M$.
2. Boundary. Find potential global extrema of $f$ on the boundary of $M$. If this boundary is given by a constraint (e.g., $x^2+y^2-4x=5$, as yours is), then you may use Lagrange multipliers.
Evaluate $f$ at all the points found in the two steps above. The largest (respectively, smallest) of these values is the global maximum (resp., global minimum) value of $f$ on $M$.
You have completed part 1.
(Note: Since you're asking about global extrema on a compact subset $M$ of $\mathbb{R}^2$, there is absolutely no reason to do anything for local extrema. If you wanted to know what happens at the critical point $(3,2)$, then fine---test it using the second-derivative test. But you are asking about global extrema; the second-derivative test is useless for this task.)
Best Answer
The answer is yes. Here there are two examples.
Consider the function $$f(x,y)=(x+y)^2$$ then $f_x(x,y)=f_y(x,y)=2(x+y)$ are both increasing with respect $x$ and $y$, but $f$ has infinite minimum points (not isolated) along the line $y=-x$.
For two isolated local minumum points, consider the function $$f(x,y)=(x^2+y^2)((x-1)^2+(y-1)^2)$$ which attains its minimum value $0$ at $(0,0)$ and $(1,1)$. It is easy to verify that $$f_{xx}(x,y)=2((x-1)^2+(y-1)^2)+8x(x-1)+2(x^2+y^2)=3(2x-1)^2+(2y-1)^2\geq 0$$ which implies that $f_x$ is increasing. By symmetry we find that $f_y$ is increasing too.