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Proof global minimum of $f(x,y)=x^2+y^4-y^2$ is $\frac{1}{2}$ where $f:$ $\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$. Is my proof attempt correct?

Define $g(y):=f(0,y)$. Differentiating $g$ after $y$ we find $4y^{3}-2y$. By setting $4y^{3}-2y=0$ we find that $y=0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$. Differentiating again we find $\frac{d}{dy}\frac{dg}{dy}=12y^{2}-2$. $\frac{d}{dy}\frac{dg}{dy}(0)<0$ and $\frac{d}{dy}\frac{dg}{dy}(\frac{1}{\sqrt{2}})>0$ we find that only $\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}$ are a local minimum. Notice that $g$ is continuous and differentiable thus we find that the minimum of the local minimums is the global minimum. Since $f(\frac{1}{\sqrt{2}})=f(-\frac{1}{\sqrt{2}})=-\frac{1}{4}$ this is the global minimum of $g$. We also have $f(x,y) \geq g(y) \geq -\frac{1}{4}$ because $x^2$ is always positive. Which yields to the result that $-\frac{1}{4}$ is the global minimum of the function $f$

Iwan5050
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    Almost perfect. "... because $x^2$ is always positive." NO. Because $x^2$ is never negative. – DanielWainfleet Jun 29 '22 at 15:57
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    I'd like to write $f(x, y) = x^2+(y^2-\frac{1}{2})^2-\frac{1}{4}$ to prove the minimum. There's no need of calculus, though. – Riemann Jun 29 '22 at 15:58
  • this is a smart idea I tried this but I could not succeedty – Iwan5050 Jun 29 '22 at 16:00
  • It is much easier to see that $$f(x,y)=x^2 + \left(y^2-\frac 12\right)^2-\frac14\ .$$Usually, one has also to "check the boundary" for a minimum, this is not done above. Should be done (and can be easily done) to have a complete argumentation. – dan_fulea Jun 29 '22 at 16:00
  • @dan_fulea, what do you mean by "boundary". It looks like there are no restrictions on the domain: $(x,y) \in \mathbb{R} \times \mathbb{R}$. – Doug Jun 29 '22 at 16:38
  • Sometimes, but not in this case, an infimum is taken when values $(x,y)$ have one or both components going to $\pm \infty$ (in a more or less coordinated manner). In such a case, we would have no global minimum at a critical point. At any rate, a full proof going as in the posted solution should mention this aspect. To understand why, just do the same for a function like $x\to exp(-x^2)p(x)$ where $p$ is a polynomial with only positive values on $\Bbb R$. For instance $p(x)=x^4+3x^2+1$. Just plot and do what you did... – dan_fulea Jun 29 '22 at 19:21

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