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Prove or disprove. Let $f: \mathbb{R}^{2} \to \mathbb{R}$, $g(x) = f(x,0)$ e $h(y) = f(0,y)$. If $x = 0$ is local minimum point of $g$ and $y = 0$ is local minimum point of $h$, then $(0,0)$ is local minimum point of $f$.

I couldn't prove it to be true. I don't know what I can use, since there are not many assumptions about $f$. So I'm trying to get a counterexample, but I was not successful.

First, I tried to get a differentiable function that had no local minimum and after, to get $g$ and $h$. But in all cases, at least one of the two don't satisfy the hypotheses.

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Let us consider $f:\mathbb{R}^2 \to \mathbb{R},~ f(x,y):=-xy$. Then $h=g\equiv 0$, but $f$ has no local minimum in $(0,0)$.

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