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Problem

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 ?

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  • $\begingroup$ 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
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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

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  • $\begingroup$ 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
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It does not exist.

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

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