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I want to show that the multivariablefunction $$f(x,y)=2x^4+5y^4-|x|-\sqrt{|x|+|y|}$$ has no global minimum.

For that do we calculate the critical points to get the desired result?

Or do we suppose that there is a global minimum at a point $(a,b)$ and then we have to show that there is a point $(x,y)$ with $f(x,y)\leq f(a,b)$ ?

Mary Star
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    You mean global maximum instead? – Michael Hoppe Jun 17 '21 at 16:10
  • Ah does the function has a global minimum? I want to check the existence of global minimum. I checked in wolfram and there I got no global minimum. @MichaelHoppe – Mary Star Jun 17 '21 at 16:11
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    In the last line, you probably mean $f(a,b)$ and not $f(x,b)$? – bigbang Jun 17 '21 at 16:14
  • Yes, I edited it! @bigbang – Mary Star Jun 17 '21 at 16:14
  • Which is the correct way to check the existence of global minimum? According to Wolfram there is no global minimum, so which is the correct way? To we have to find a counterexample? @bigbang – Mary Star Jun 17 '21 at 16:16
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    I expect that the absolute value is making WA unhappy. Best to do it in regions where you can remove the absolute value. – lulu Jun 17 '21 at 16:21
  • So there is a global minimum? @lulu – Mary Star Jun 17 '21 at 16:23
  • Just informally: The quartic terms clearly dominate if $|x|,|y|$ are large, and inside a big circle there must be a minimum, so the function should have an absolute minimum. Quick search finds $F(.72, .29)=-1.42088495$ which looks pretty good. – lulu Jun 17 '21 at 16:23
  • A little more effort improves this to $F(0.72063121,0.291799834)=-1.420895163$. You can do it with WA if you use the substitution $|x|=X^2, |y|=Y^2$. As a general rule, try to avoid absolute values and such...the fact that the derivative of the absolute is singular at $0$ causes a great deal of grief. – lulu Jun 17 '21 at 16:33
  • @lulu did you enter the formula correctly? See my answer for what should be the real correct minimum. – Parcly Taxel Jun 17 '21 at 16:55
  • @ParclyTaxel Ah, you are right. I dropped the initial factor of $2$. My general comments still hold, but the numerical results are wrong. – lulu Jun 17 '21 at 16:56
  • here is the correct equation done by WA. I note that this does tax the standard computing time so you might need to run it repeatedly. Note that "no global solution found" is its default mode...if it also says that "standard computing time exceeded" then you should disregard it. In any case, note that it matches the result obtained by @ParclyTaxel – lulu Jun 17 '21 at 17:47

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Note that $f(-x,y)=f(x,-y)=f(x,y)$, so we can focus on the positive quadrant $x,y\ge0$ and remove the absolute value signs: $$f(x,y)=2x^4+5y^4-x-\sqrt{x+y}\qquad x,y\ge0$$ In any direction the quartic terms dominate, so there is no global maximum. On the other hand, $f$ in the positive quadrant is a sum of continuous convex functions ($2x^4,5y^4,-x,-\sqrt{x+y}$) and hence is continuous strictly convex, so there is a unique (up to sign) global minimum. An exact minimisation performed by Mathematica gives the global minimum in the positive quadrant as $$f(0.576690\dots,0.298945\dots)=-1.25130\dots$$

Parcly Taxel
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You can argue using the properties of continuous functions - as $x,y \to \infty$, $f \to \infty$, hence there is a closed ball $b(r) = \{ |(x,y)| \leq r\}$ such that if $x \in \mathbb R^2 \setminus b(r)$ then $$ f(x,y) \geq f(0,0) $$ Now consider the restriction of $f$ to $b(r)$. Since $f$ is continuous and $b(r)$ is compact, the image $f(b(r))$ is also compact and hence the restriction of $f$ is bounded and attains its bounds. In particular there is a point $(x,y)$ such that $f(x,y) \leq f(x',y')$ for all $(x',y') \in b(r)$. If $(x',y') \in \mathbb R^2 \setminus b(r)$ then $$ f(x,y) \leq f(0,0) \leq f(x',y') $$ Hence $(x,y)$ is a global minimum.

Holmes
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