If i have function $f(x,y)=(1+y)^3x^2+y^2$ and i want to find out is there global minima for this function when $(x,y)\in \mathbb{R}$ I can find out where $f(x,y)$ has critical points by computing gradient $\nabla f(x,y)=0$ I did find that one of those critical points is point $(0,0)$. I compute hessian matrix and it's eigenvalues when $(x,y)\rightarrow (0,0)$. This tells me point $(0,0)$ is local minima since the hessian matrix is Positive-definite matrix. Now how do i find out if the point is global minima ?

Computations are:

$$ \nabla f(x,y)=\begin{bmatrix} 2x(1+y)^2 \\ 2(x^2(x+1)+y) \end{bmatrix} $$ $$ H(x,y)=\begin{bmatrix} 2(y+1)^2 & 4x(y+1) \\ 4x(y+1) & 2x^2+2\end{bmatrix} $$ $$ H(0,0)=\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} $$ $$\lambda_1=2 $$ $$ \lambda_2=2 $$

The only critical point should be $(x,y)\rightarrow (0,0)$ ?

$$ \lim_{(x,y)\rightarrow (0,0)}(1+y)^3x^2+y^2=0 $$ also limit can be evaluated at (0,0) this would make this function continous with all $\mathbb{R}$ and this function should be also defined with all $\mathbb{R}$ ? another thing is i don't know how something like limit does exists for multivariable case can be proven ?

Anyway if i was able to prove that this function has it's only critical point at $(0,0)$ and it is defined with all $\mathbb{R}$ this would mean that point (0,0) is global minima since no such other point exists that could be maxima or minima ?

  • $\begingroup$ compare the values of all local minima, and the behavior of the function as $|(x,y)|\to\infty$ in each direction $\endgroup$ – Masacroso Jan 28 '18 at 13:52
  • $\begingroup$ this is only a local Minimum! $\endgroup$ – Dr. Sonnhard Graubner Jan 28 '18 at 13:53

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: enter image description here

and for a further look:

enter image description here

  • $\begingroup$ your using $\epsilon,\delta$ definition to prove that $(0,0)$ is local minima ? When you hold $x$ constant it is only necessary to show that @Mostafa Ayaz $\epsilon$ can get arbitrary small when $y\rightarrow -\infty$ which would mean that such values that are smaller than values in point $(0,0)$ exist $\rightarrow$ meaning point $(0,0)$ is only local minima ? Is this correct interpretation of your answer ? I am trying to understand this. $\endgroup$ – Tuki Jan 28 '18 at 14:14
  • $\begingroup$ That's true in fact for showing a point to be local minima we must find a neighborhood where the value of the function in that minima is not greater than the value in whole points of the neghborhood. $\endgroup$ – Mostafa Ayaz Jan 28 '18 at 14:19

It does not exist.

Try $x=1$ and $y\rightarrow-\infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.